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            "content": "高等代数笔记"
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            "content": "作者：Reichtum",
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            "content": "组织：ZheJiang University",
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            "content": "时间：September 20, 2022",
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            "content": "版本：0.1",
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            "content": "There was no royal road to geometry. ——Euclid",
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            "content": "第1章 多项式 1\n\n1.1 一元多项式 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1\n1.1.1 整除理论 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1\n1.1.2 最大公因式 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2\n1.1.3 互素 . . . . . . . . . . . . . . . . . . 3\n1.1.4 不可约多项式与因式分解 4\n1.1.5 余数定理及其应用 5\n1.1.6 重因式与重根 6\n1.1.7 整数与多项式 7\n1.2 特殊数域上的一元多项式理论 8\n1.2.1 复数域上的一元多项式 8\n1.2.2 实数域上的一元多项式 8\n1.2.3 有理数域上一元多项式因式分解 8\n1.2.4 一元有理多项式的根 10\n1.3 多元多项式 10\n1.3.1 基本概念 10\n1.3.2 对称多项式 11\n1.3.3 判别式 12\n1.3.4 结式 12\n第2章 矩阵 14\n2.1 矩阵基础 14\n2.1.1 矩阵概念与运算 14\n2.1.2 矩阵的逆 14\n2.1.3 伴随矩阵的性质 15\n2.1.4 广义逆矩阵 16\n2.1.5 Moore-Penrose广义逆 17\n2.2 分块矩阵 18\n2.2.1 常见矩阵拆分技巧 18\n2.2.2 分块初等变换 19\n2.2.3 分块打洞技巧总结 19\n2.2.4 分块矩阵行列式 20\n2.2.5 分块矩阵求逆 20\n2.3 矩阵的秩 21\n2.3.1 秩的概念 21\n2.3.2 初等变换标准型与矩阵分解 21\n2.3.3 利用初等变换标准型进行矩阵分解 22\n2.3.4 秩不等式 22\n2.3.5 空间分解法研究矩阵秩 24\n2.4 矩阵的迹 25\n2.5 几种技巧矩阵 25"
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            "content": "目录"
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            "content": "2.5.1 初等矩阵 25  \n2.5.2 基本矩阵：仅一个元素为1的矩阵 25  \n2.5.3 置换矩阵 26  \n2.6 特殊矩阵性质总结 27  \n2.6.1 对角矩阵 27  \n2.6.2 上、下三角矩阵 27  \n2.6.3 实对称矩阵 28  \n2.6.4 反对称矩阵 28  \n2.6.5 Hermite矩阵 30  \n2.6.6 正交矩阵 30  \n2.6.7 Unitary矩阵 31  \n2.6.8 幂零矩阵 31  \n2.6.9 幂等矩阵 31  \n2.6.10 对角占优矩阵 33  \n\n第3章 行列式 34  \n3.1 行列式基本概念与性质 34  \n3.2 行列式展开 34  \n3.3 特殊的行列式 34  \n3.3.1 爪型行列式 34  \n3.3.2 大对角型(三斜线) 35  \n3.3.3 行和相同型 35  \n3.3.4 循环型 35  \n3.3.5 行列式加边法 36  \n3.4 Cramer法则 36  \n\n第4章 线性方程组 37  \n4.1 线性方程组解的判定定理 37  \n4.1.1 解存在判定定理 37  \n4.1.2 线性方程组同解 37  \n4.1.3 解个数判定定理 37  \n4.2 线性方程组的求解 37  \n4.2.1 Ax=0求解 37  \n4.2.2 Ax=b求解 38  \n4.2.3 线性方程组公共解的计算 39  \n4.3 最小二乘解 39  \n4.3.1 \\(A^T A\\) 的研究 39  \n4.3.2 正规方程组 39  \n\n第5章 线性空间与线性变换 41  \n5.1 线性空间 41  \n5.1.1 线性空间的基本概念 41  \n5.1.2 线性相关、线性无关与向量组的秩 41  \n5.1.3 线性空间同构 42  \n5.1.4 基、坐标 43  \n5.1.5 坐标变换 44"
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            "content": "5.1.6 和与交、维数定理 45  \n5.1.7 直和理论 47  \n5.1.8 商空间 48  \n5.2 线性映射 48  \n5.2.1 线性映射的概念 48  \n5.2.2 核与像 48  \n5.2.3 幂等变换的核与像 50  \n5.2.4 一般线性变换的核像分解 50  \n5.3 线性变换 51  \n5.3.1 线性变换基本概念 51  \n5.3.2 线性变换在基下的矩阵、变换后的坐标 51  \n5.3.3 线性变换在不同基下的矩阵：相似关系 52  \n5.3.4 相似矩阵的性质 53  \n5.4 特征值、相似对角化 53  \n5.4.1 特征值与特征向量 53  \n5.4.2 特征值与特征向量进一步结论 54  \n5.4.3 相似的进一步研究 55  \n5.4.4 相似对角化 56  \n5.4.5 相似上三角型 57  \n5.5 最小多项式 58  \n5.5.1 最小多项式与基本性质 58  \n5.5.2 Hamilton-Cayley 定理 58  \n5.5.3 最小多项式与相似对角化 59  \n\n6章 Jordan 标准型 60  \n6.1 Jordan 标准形：空间分解法 60  \n6.1.1 空间第二分解定理：循环不变子空间分解 60  \n6.1.2 核像升链与根子空间 60  \n6.1.3 空间第一分解定理：根子空间分解 61  \n6.1.4 Jordan 标准形：空间分解法 62  \n6.2 空间分解法：零化多项式角度 62  \n6.2.1 根子空间分解：矩阵角度 62  \n6.2.2 根子空间分解：线性变换角度 63  \n6.2.3 Jordan 阵与循环不变子空间 64  \n6.3 Jordan 标准形：\\(\\lambda\\) 矩阵法 65  \n6.3.1 \\(\\lambda\\) 矩阵与相抵 65  \n6.3.2 行列式因子、不变因子 66  \n6.3.3 初等因子与 Jordan 标准型 67  \n6.4 \\(\\lambda\\) 矩阵的进一步研究 68  \n6.4.1 Jordan 标准型的直接计算 68  \n6.4.2 \\(k\\) 阶 Jordan 块的个数 70  \n6.4.3 分块矩阵的 Jordan 阵 71  \n6.4.4 过渡矩阵的计算 71"
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            "content": "第7章 双线性函数与二次型"
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            "content": "7.1 双线性函数 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73\n7.1.1 双线性函数概念 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73\n7.1.2 双线性函数度量矩阵 . . . . . . . . . . . . . . . . . . . . . . . . . 73\n7.1.3 双线性函数不同基下的矩阵：合同 ..... 73\n7.2 二次型 ..... 74\n7.2.1 二次型概念及其矩阵 ..... 74\n7.2.2 二次型的标准型存在与配方法 ..... 75\n7.2.3 二次型的标准型：初等变换法 ..... 77\n7.2.4 二次型的标准型：正交变换法 ..... 77\n7.2.5 R, C 上的规范型 ..... 79\n7.3 (半) 正定二次型的研究 ..... 79\n7.3.1 正定二次型的概念与判别 ..... 79\n7.3.2 半正定二次型的概念与判别 ..... 81\n7.3.3 \\(C^T C\\) 的 (半) 正定性 ..... 82\n7.3.4 正定矩阵同时对角化 ..... 83\n7.3.5 正定矩阵的四则运算性质 ..... 84\n7.3.6 正定矩阵分解进阶 ..... 84\n7.3.7 负定二次型 ..... 85\n8 章 内积空间 ..... 86\n8.1 Euclid 空间 ..... 86\n8.1.1 Euclid 空间与正交 ..... 86\n8.1.2 标准正交基 ..... 87\n8.1.3 正交和与正交补 ..... 88\n8.1.4 正交变换 ..... 89\n8.1.5 镜面反射 ..... 90\n8.2 Unitary 空间 ..... 90\n8.2.1 Unitary 空间概念 ..... 90\n9 章 对偶空间 ..... 91\n9.1 对偶空间与对偶基 ..... 91"
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            "content": "第1章 多项式"
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            "content": "1.1 一元多项式"
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            "content": "1.1.1 整除理论"
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            "content": "定义 1.1 (整除)"
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            "content": "\\(g(x), f(x)\\) 是 \\(\\mathbb{P}\\) 上的多项式，若 \\(\\exists h(x)\\) 使得 \\(f(x) = g(x)h(x)\\)，则称 \\(g(x)\\) 整除 \\(f(x)\\)，记为 \\(g(x)|f(x)\\)。称 \\(g(x)\\) 是 \\(f(x)\\) 的因式，\\(f(x)\\) 是 \\(g(x)\\) 的倍式。",
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            "content": "定理 1.1 (整除的数域不变性)"
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            "content": "\\(\\mathbb{P}, \\mathbb{F}\\) 为两个数域，\\(\\mathrm{P} \\subseteq \\mathbb{F}\\)，\\(f(x), g(x) \\in \\mathbb{P}[x]\\) (在小数域中)，则 \\(\\mathbb{P}\\) 中 \\(g(x)|f(x)\\) 当且仅当 \\(\\mathbb{F}\\) 中 \\(g(x)|f(x)\\)",
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            "content": "笔记该定理的意义在于可以在最大的数域(方便因式分解)中证明整除，即可证明小数域中整除。",
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            "content": "定理 1.2 (带余除法)"
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            "content": "对于 \\(\\forall f(x), g(x) \\in \\mathbb{P}[x]\\)，存在唯一 \\(q(x), r(x) \\in \\mathbb{P}[x]\\) 使得 \\(f(x) = q(x)g(x) + r(x)\\)。这里 \\(r(x) = 0\\) 或者 \\(\\partial(r(x)) < \\partial(g(x))\\)",
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            "content": "练习1.1带余除法的计算 \\((1)f(x) = x^{4} + 2x^{3} - 5x + 7, g(x) = x^{2} - 3x + 1\\)，求 \\(g(x)\\) 除 \\(f(x)\\) 的商式与余式解(1)商为 \\(x^{2} + 5x + 14\\)，余式为 \\(32x - 7\\)",
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            "content": "练习 1.2 带余除法的应用 (1) 设  \\( g(x)=ax+b \\) ， \\( f(x)\\in\\mathbb{P}[x] \\) ，证明： \\( g(x)|f^{2}(x) \\)  当且仅当  \\( g(x)|f(x) \\)",
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            "content": "(2) \\( f(x), h(x), g(x) \\) 为三个非零多项式，证明： \\( h(x)|(f(x)-g(x)) \\) 当且仅当 \\( f(x), g(x) \\) 除以 \\( h(x) \\) 的余式相同",
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            "content": "\\[\nf (x), g (x)\n\\]"
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            "content": "\\[\ng ^ {m} (x) | f ^ {m} (x)\n\\]"
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            "content": "\\[\ng (x) | f (x)\n\\]"
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            "content": "证明 (1) 右推左显然。左推右: \\( f^2(x) = g(x)h(x) \\)，若 \\( f(x) = q(x)g(x) + r(x) \\)，则 \\( f^2(x) = q^2(x)g^2(x) + 2q(x)g(x)r(x) + r^2(x) \\)，根据 \\( g(x)|f(x) \\) 得到 \\( g(x)|r^2(x) \\)，显然 \\( r(x) = 0 \\)",
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            "content": "(2) 设  \\( f(x)=q_{1}(x)h(x)+r_{1}(x), g(x)=q_{1}(x)h(x)+r_{2}(x) \\) ，相减得到",
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            "content": "\\[\nf (x) - g (x) = h (x) \\left[ q _ {1} (x) - q _ {2} (x) \\right] + r _ {1} (x) - r _ {2} (x)\n\\]"
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            "content": "\\[\n\\left\\{ \\begin{array}{l} f (x) = q _ {1} (x) g (x) + r _ {1} (x) \\\\ g (x) = q _ {2} (x) r _ {1} (x) + r _ {2} (x) \\\\ r _ {1} (x) = q _ {3} (x) r _ {2} (x) + r _ {3} (x) \\\\ \\vdots \\\\ r _ {s - 3} (x) = q _ {s - 1} (x) r _ {s - 2} (x) + r _ {s - 1} (x) \\\\ r _ {s - 2} (x) = q _ {s} (x) r _ {s - 1} (x) + r _ {s} (x) \\\\ r _ {s - 1} (x) = q _ {s + 1} (x) r _ {s} (x) + 0 \\end{array} \\right.\n\\]"
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            "content": "则此时可以得到 \\(r_s(x)\\) 是 \\(f(x), g(x)\\) 的最大公因式，并且从倒数第二个等式开始往回推，得到",
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            "content": "\\[\n\\begin{array}{r l} r _ {s} (x) & = r _ {s - 2} (x) - q _ {s} (x) r _ {s - 1} (x) \\\\ & = r _ {s - 2} (x) - q _ {s} (x) (r _ {s - 3} (x) - q _ {s - 1} (x) r _ {s - 2} (x)) \\\\ & = \\dots \\dots \\\\ & = u (x) f (x) + v (x) g (x) \\end{array}\n\\]"
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            "content": "即逐个将前面的等式代入方程，逐个消去  \\( r_{s-2}(x), \\cdots, r_{2}(x), r_{1}(x) \\) ，最终只剩下  \\( g(x), f(x) \\)",
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            "content": "笔记 最大公因式 \\(d(x) = u(x)f(x) + v(x)g(x)\\) 的 \\(u(x)\\) 和 \\(v(x)\\) 不是唯一的，如果要唯一则需要加上额外的限制条件。",
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            "content": "(1) \\( d(x)|d_{1}(x) \\) ： \\( d(x) \\) 是左侧最大公因式，其是右侧每个部分的最大公因式，因此 \\( d(x)|d_{1}(x) \\)",
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            "content": "\\[\n\\begin{array}{c} \\text {   ①   } \\\\ q _ {1} (x) = x - \\frac {1}{3} \\left| \\begin{array}{l l} f (x) & \\frac {1}{3} g (x) \\\\ x ^ {4} + 3 x ^ {3} - x ^ {2} - 4 x - 3 & x ^ {3} + \\frac {1 0}{3} x ^ {2} + \\frac {2}{3} x - 1 \\\\ x ^ {4} + \\frac {1 0}{3} x ^ {3} + \\frac {2}{3} x ^ {2} - x & x ^ {3} + 5 x ^ {2} + 6 x \\\\ \\hline - \\frac {1}{3} x ^ {3} - \\frac {5}{3} x ^ {2} - 3 x - 3 & - \\frac {5}{3} x ^ {2} - \\frac {1 6}{3} x - 1 \\\\ - \\frac {1}{3} x ^ {3} - \\frac {1 0}{9} x ^ {2} - \\frac {9}{9} x + \\frac {1}{3} & - \\frac {5}{3} x ^ {2} - \\frac {2 5}{3} x - 1 0 \\\\ \\hline r _ {1} (x) = - \\frac {5}{9} x ^ {2} - \\frac {2 5}{9} x - \\frac {1 0}{3} & r _ {2} (x) = 3 x + 9 \\\\ - \\frac {9}{5} r _ {1} (x) = x ^ {2} + 5 x + 6 & \\frac {1}{3} r _ {2} (x) = x + 3. \\end{array} \\right| \\quad \\text {   ②   } \\\\ q _ {2} (x) = x - \\frac {5}{3} \\end{array}\n\\]"
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            "content": "\\[\n\\begin{array}{r l} & {\\mathrm{因此}: ① f (x) = \\frac {1}{3} g _ {1} (x) g (x) + r _ {1} (x) \\quad ② g (x) = g _ {2} (x) (- \\frac {9}{5} r _ {1} (x)) + r _ {2} (x)} \\\\ & {③ - \\frac {9}{5} r _ {1} (x) = (x + 2) \\frac {1}{3} r _ {2} (x) \\qquad r _ {2} (x) \\mathrm{即大公因式}} \\\\ & {\\Rightarrow ① ^ {\\prime}: r _ {1} (x) = f (x) - \\frac {1}{3} g _ {1} (x) g (x), ② ^ {\\prime} r _ {2} (x) = g (x) + \\frac {9}{5} g _ {2} (x) r _ {1} (x)} \\\\ & {\\mathrm{将} r _ {1} (x) \\mathrm{代入} ② ^ {\\prime} \\mathrm{中即可}} \\end{array}\n\\]"
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            "content": "具体用 cramer 法则求出即可，如  \\( f(x)=\\frac{df_{1}(x)-bg_{1}(x)}{ad-bc}, g(x)=\\frac{ag_{1}(x)-cf_{1}(x)}{ad-bc} \\) ，因此与 (1) 同理",
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            "content": "推论 1.1 (加减保最大公因式)"
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            "content": "\\[\n(f (x), f (x) - g (x)) = (f (x), f (x) + g (x)) = (f (x), g (x))\n\\]"
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            "content": "定理 1.5 (最小公倍式的表示)"
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            "content": "\\(f(x), g(x)\\) 的首 1 最小公倍式 \\([f(x), g(x)] = \\frac{f(x)g(x)}{(f(x), g(x))}\\)",
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            "content": "证明 设 \\((f(x), g(x)) = d(x), f(x) = d(x)f_1(x), g(x) = d(x)g_1(x)\\)，且 \\((f_1(x), g_1(x)) = 1\\)，记 \\(s(x) = \\frac{f(x)g(x)}{d(x)} = d(x)f_1(x)g_1(x) = f(x)g_1(x) = f_1(x)g(x)\\)，下面说明 \\(s(x)\\) 是最小公倍式：(1)是公倍式：显然(2)最小：设 \\(h(x)\\) 也是公倍式，设 \\(h(x) = f(x)m(x)\\)，根据 \\(g(x)|h(x)\\) 得到 \\(d(x)g_1(x)|d(x)f_1(x)m(x)\\)，根据 \\(g_1(x)\\) 与 \\(f_1(x)\\) 互素，得到 \\(g_1(x)|m(x)\\)，因此 \\(g_1(x)f(x)|f(x)m(x)\\)，得到 \\(s(x)|h(x)\\)，因此 \\(s(x)\\) 是最小的。",
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            "content": "1.1.3 互素"
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            "content": "定义 1.3 (互素)"
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            "content": "若 \\(f(x), g(x) \\in \\mathbb{P}[x]\\) 的最大公因式为1，则称它们互素。",
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            "content": "定理 1.6 (互素充要条件)"
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            "content": "\\(f(x), g(x) \\in \\mathbb{P}[x]\\) 互素当且仅当 \\(\\exists u(x), v(x) \\in \\mathbb{P}[x]\\) 使得 \\(u(x)f(x) + v(x)g(x) = 1\\)",
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            "content": "证明 左推右显然 (根据最大公因式定理)。右推左：设 \\((f(x), g(x)) = d(x)\\)，则 \\(d(x)|f(x), d(x)|g(x)\\)，设 \\(f(x) = d(x)f_1(x), g(x) = d(x)g_1(x)\\)，从而",
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            "content": "因此 \\(d(x), u(x)f_1(x) + v(x)g_1(x)\\) 均为常数，故 \\(f(x), g(x)\\) 互素笔记上述 \\(u(x), v(x)\\) 在无条件限制下也是不一定唯一的。",
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            "content": "练习1.4互素相关应用(1)若 \\(f(x),g(x)\\) 互素，证明： \\(f(x^{m}),g(x^{m})\\) 互素",
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            "content": "(2) 重点：若 \\((f(x), g(x)) = d(x)\\)，证明：\\((f(x^m), g(x^m)) = d(x^m)\\)",
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            "content": "(3) 设 \\((f(x), g(x)) = 1, (f(x), h(x)) = 1\\)，证明：\\((f(x), g(x)h(x)) = 1\\)",
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            "content": "证明 (1) 由于 \\( f(x), g(x) \\) 互素，得到 \\( u(x)f(x) + v(x)g(x) = 1 \\)，从而 \\( u(x^{m})f(x^{m}) + v(x^{m})g(x^{m}) = 1 \\)，由于是互素充要条件，因此 \\( f(x^{m}), g(x^{m}) \\) 互素",
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            "content": "\\[\nu (x ^ {k}) m (x ^ {k}) + v (x ^ {k}) n (x ^ {k}) = 1\n\\]"
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            "content": "得到 \\(m(x^{k}), n(x^{k})\\) 互素，而 \\(f(x^{k}) = d(x^{k})m(x^{k}), g(x^{k}) = d(x^{k})n(x^{k})\\) ，因此 \\((f(x^{k}), g(x^{k})) = d(x^{k})\\)",
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            "content": "(3) 根据互素，有 \\( u_{1}(x)f(x) + v_{1}(x)g(x) = 1, u_{2}(x)f(x) + v_{2}(x)h(x) = 1 \\)，两式相乘得到",
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            "content": "\\[\n\\left[ u _ {1} (x) u _ {2} (x) f (x) + v _ {1} (x) g (x) + v _ {2} (x) h (x) \\right] f (x) + v _ {1} (x) v _ {2} (x) g (x) h (x) = 1\n\\]"
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            "content": "因此根据充要条件得到 \\( (f(x), g(x)h(x)) = 1 \\)",
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            "content": "练习 1.5 互素充要条件推广  \\( f(x) \\) ,  \\( g(x) \\)  是次数大于 0 的互素多项式，则存在唯一的  \\( u(x) \\) ,  \\( v(x) \\)  使得  \\( u(x)f(x) + v(x)g(x) = 1 \\) ，这里  \\( \\partial(u(x)) < \\partial(g(x)) \\) ,  \\( \\partial(v(x)) < \\partial(f(x)) \\)",
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            "content": "证明 (1) 存在性：由于 \\( f(x), g(x) \\) 互素，故存在 \\( a(x), b(x) \\) 使得 \\( a(x)f(x) + b(x)g(x) = 1 \\)，显然 \\( a(x), g(x) \\) 也互素，作带余除法 \\( a(x) = q(x)g(x) + u(x) \\)，代入公式得到",
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            "content": "\\[\n(q (x) g (x) + u (x)) f (x) + b (x) g (x) = 1 \\Rightarrow u (x) f (x) + (b (x) + q (x) f (x)) g (x) = 1\n\\]"
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            "content": "记 \\(v(x) = b(x) + q(x)f(x)\\)，下面证明 \\(\\partial(v(x)) < \\partial(f(x))\\)。反设 \\(\\partial(v(x)) \\geq \\partial(f(x))\\)，根据 \\(\\partial(u(x)) < \\partial(g(x))\\) 可知 \\(\\partial(v(x)g(x)) > \\partial(u(x)f(x))\\)，显然它们的和不为 1，矛盾。",
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            "content": "设 \\(p(x) \\in \\mathbb{P}[x]\\) 且 \\(\\partial(p(x)) \\geq 1\\)，若 \\(p(x)\\) 不能表示成 \\(\\mathbb{P}[x]\\) 中两个次数小于 \\(p(x)\\) 的多项式的乘积，则称 \\(p(x)\\) 为 \\(\\mathbb{P}[x]\\) 中的不可约多项式。",
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            "content": "- \\(p(x)\\) 与 \\(\\forall f(x) \\in \\mathbb{P}[x]\\) 要么互素，要么 \\(p(x) | f(x)\\)",
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            "content": "1.1 一元多项式"
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            "content": "(2) 已知  \\( f_{1}(x) \\) ,  \\( f_{2}(x) \\) ,  \\( \\cdots \\) ,  \\( f_{s}(x) \\)  两两互素，记  \\( F_{i}(x) = f_{1}(x) \\cdots f_{i-1}(x) f_{i+1}(x) \\cdots f_{s}(x) \\) ，证明  \\( F_{1}(x) \\) ,  \\( F_{2}(x) \\) ,  \\( \\cdots \\) ,  \\( F_{s}(x) \\)  互素",
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            "content": "(3) 稍加变化: 复数域上的多项式 \\(f(x)\\) 没有重因式, 证明 \\((f(x) + 2f'(x), 3f^2(x)) = 1\\)",
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            "content": "证明 (1) 反设有不可约多项式 \\(q(x)\\) 作为公因式，则 \\(q(x)\\) 同时是 \\(f(x)g(x), f(x) + g(x)\\) 因式。根据前者得到 \\(q(x)\\) 至少整除 \\(f(x), g(x)\\) 之一，不妨设 \\(q(x)|f(x)\\)，根据第二个式子得到 \\(q(x)\\) 也一定整除 \\(g(x)\\)，否则无法得到多项式。因此矛盾。",
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            "content": "(2) 同种方法，随便挑 \\(F_{i}(x), F_{j}(x)\\) 说一说就行",
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            "content": "(3) 同样设不可约  \\( q(x) \\) ，根据  \\( 3f^{2}(x) \\)  可知  \\( q(x)|f(x) \\) ，再根据  \\( q(x)|f(x)+2f'(x) \\)  可知  \\( q(x)|f'(x) \\) （否则不是多项式），根据重因式充要条件可知。",
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            "content": "练习 1.7 证明不可约多项式 (1) \\( p(x) \\) 是次数大于 0 的多项式，若对任意多项式 \\( f(x), g(x) \\)，根据 \\( p(x)|f(x)g(x) \\) 可以得到 \\( p(x)|f(x) \\) 或者 \\( p(x)|g(x) \\)，则 \\( p(x) \\) 是不可约多项式。",
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            "content": "证明 (1) 反设 \\( p(x) \\) 不是不可约多项式，则存在 \\( f(x), g(x) \\) 使得 \\( p(x) = f(x)g(x) \\)，此时 \\( p(x)|f(x)g(x) \\)，但是 \\( p(x) \\nmid f(x), p(x) \\nmid g(x) \\)",
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            "content": "定理 1.8 (因式分解唯一性)"
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            "content": "\\(f(x)\\) 是 \\(\\mathbb{P}\\) 上次数大于等于1的多项式，则 \\(f(x)\\) 可唯一分解为 \\(\\mathbb{P}\\) 上有限个不可约多项式乘积：",
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            "content": "\\[\nf (x) = c p _ {1} ^ {r _ {1}} (x) p _ {2} ^ {r _ {2}} (x) \\dots p _ {s} ^ {r _ {s}} (x)\n\\]"
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            "content": "1.1.5 余数定理及其应用"
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            "content": "定理 1.9 (余数定理)"
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            "content": "\\(f(x)\\in \\mathbb{P}[x]\\) ，对 \\(\\forall \\alpha \\in \\mathbb{P}\\) ，总有",
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            "content": "\\[\nf (x) = (x - \\alpha) q (x) + f (\\alpha)\n\\]"
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            "content": "证明 用带余除法得到  \\( f(x)=(x-\\alpha)q(x)+c \\) ，而代入  \\( x=\\alpha \\)  得到  \\( f(\\alpha)=0+c \\) ，即  \\( c=f(\\alpha) \\) ，因此",
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            "content": "\\[\nf (x) = (x - \\alpha) q (x) + f (\\alpha)\n\\]"
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            "content": "推论 1.2"
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            "content": "\\(\\alpha \\in \\mathbb{P}\\) 是多项式 \\(f(x)\\) 的根当且仅当 \\((x - \\alpha) \\mid f(x)\\)",
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            "content": "笔记 余数定理的主要价值在于 \\(\\mathbb{P}\\) 上的因式等价于 \\(\\mathbb{P}\\) 上的根，两者可以灵活互换。判断重根可以转换为判断重因式，判断整除可以转换为根。",
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            "content": "练习 1.8 余数定理的应用 (1) 若  \\( (x-1)\\mid f(x^{m}) \\) ，证明  \\( (x^{m}-1)\\mid f(x^{m}) \\)",
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            "content": "(2) 证明 \\(x^{2} + x + 1 \\mid x^{3m} + x^{3n + 1} + x^{3p + 2}\\)",
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            "content": "(4) 已知 \\(x^{2} + 1 \\mid f_{1}(x^{4}) + xf_{2}(x^{4})\\)，证明 1 是 \\(f_{1}(x), f_{2}(x)\\) 的公共根。",
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            "content": "1.1 一元多项式"
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            "content": "1.1.6 重因式与重根"
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            "content": "定理 1.10 (重因式性质汇总)"
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            "angle": 0,
            "content": "\\(f(x)\\) 是次数大于0的多项式， \\(p(x)\\) 是不可约多项式，则",
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            "type": "text",
            "bbox": [
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            "content": "- 核心：若 \\(p(x)\\) 是 \\(f(x)\\) 的 \\(k\\) 重因式，则其是 \\(f'(x)\\) 的 \\(k - 1\\) 重因式。",
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        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- 无重因式充要：\\(f(x)\\) 没有重因式当且仅当 \\(f(x), f'(x)\\) 互素。",
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        },
        {
            "type": "text",
            "bbox": [
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            "content": "- 不可约因式： \\(\\frac{f(x)}{(f(x), f'(x))}\\) 是 \\(f(x)\\) 所有不可约因式 (重数均为 1) 乘积的常数倍",
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            "content": "笔记 重因式与重根的关系：重因式不随数域而改变，而重根会随着数域而改变。例如 \\((x^{2} + 1)^{2}\\)，其在 \\(\\mathbb{C},\\mathbb{R},\\mathbb{Q}\\) 上都有重因式，但是只在 \\(\\mathbb{C}\\) 上有重根。因此有重根一定有重因式，有重因式不一定有重根。",
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            "content": "练习1.9用根判断因式(1)重点：设 \\(h(x), f(x), g(x)\\) 为实系数多项式满足以下等式，证明 \\(x^{2} + 1|f(x), x^{2} + 1|g(x)\\)",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} (x ^ {2} + 1) h (x) + (x + 1) f (x) + (x - 2) g (x) = 0 \\\\ (x ^ {2} + 1) k (x) + (x - 1) f (x) + (x + 2) g (x) = 0 \\end{array} \\right.\n\\]"
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            "content": "证明 (1) 要证明 \\(\\mathbb{R}\\) 上 \\((x^2 + 1)|f(x)\\)，只需要证明 \\(\\mathbb{C}\\) 上 \\((x - i)|f(x), (x + i)|f(x)\\)，等价于 \\(x = \\pm i\\) 在 \\(\\mathbb{C}\\) 上为 \\(f(x)\\) 的根。代入 \\(x = i\\) 得到：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} (i + 1) f (i) + (i - 2) g (i) = 0 \\\\ (i - 1) f (i) + (i + 2) g (i) = 0 \\end{array} \\right.\n\\]"
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            "content": "解得 \\( f(i) = g(i) = 0 \\) 。同理可证 \\( f(-i) = g(-i) = 0 \\)",
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            "content": "定理 1.11 (重根性质汇总)"
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            "angle": 0,
            "content": "\\(f(x)\\) 是 \\(\\mathbb{P}[x]\\) 上次数大于0的多项式，则",
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            "content": "- 重根：\\(x_0\\) 是 \\(f(x)\\) 的 \\(k\\) 重根当且仅当 \\(f(x_0) = f'(x_0) = \\cdots = f^{(k-1)}(x_0) = 0\\)，而 \\(f^{(k)}(x_0) \\neq 0\\)",
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            "content": "- 重根判定定理：\\(p(x)\\) 是 \\(f(x)\\) 重因式当且仅当其是 \\(f(x), f'(x)\\) 公因式。",
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            "bbox": [
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            "content": "笔记 \\(\\mathbb{C}\\) 上的重根问题等价于 \\(\\mathbb{C}\\) 上的重因式问题，因此判断多项式是否有重根时，可判断其在 \\(\\mathbb{C}\\) 上是否有重因式。重因式的好处是可以用多项式理论，重根的好处是可以带进去算，要相互补充。",
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            "content": "练习 1.10 判定重根存在性 (1) 证明  \\( f(x)=1+x+\\frac{x^{2}}{2!}+\\cdots+\\frac{x^{n}}{n!} \\)  没有重根",
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            "angle": 0,
            "content": "6"
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            "content": "(2) 证明  \\( f(x)=x^{n}+nx^{n-1}+n(n-1)x^{n-2}+\\cdots+n! \\)  无重根",
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            "content": "证明 (1) 直接证重根不好证，转换为重因式，若在 \\(\\mathbb{C}\\) 上无重因式则肯定无重根。根据 \\(f'(x) = 1 + x + \\frac{x^2}{2!} + \\cdots + \\frac{x^{n-1}}{(n-1)!}\\) 得到",
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            "content": "\\[\n(f (x), f ^ {\\prime} (x)) = (f (x), f (x) - f ^ {\\prime} (x)) = (f (x), \\frac {x ^ {n}}{n !}) = (f (x), x ^ {n})\n\\]"
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            "content": "由于 \\(x^n\\) 项的原因，\\((f(x), f'(x)) = x^k, k \\geq 0\\)。而 \\(f(0) = 1\\)，这说明 \\(x\\) 不是 \\(f(x)\\) 的因式，从而 \\(k = 0\\)，因此 \\(f(x), f'(x)\\) 互素，无重根。",
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            "content": "(2) 也用无重因式证明",
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            "content": "练习 1.11 重根、重因式相互转化 (1) 求  \\( x^{5} + x^{4} - 6x^{3} - 14x^{2} - 11x - 3 \\)  的重根和重数",
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            "content": "(2) 重点: 求 \\(A, B\\) 使得 \\((x - 1)^{2} | Ax^{4} + Bx^{2} + 1\\)",
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            "content": "(3) 重点：求 \\(t\\) 使得 \\(f(x) = x^{3} - 3x^{2} + tx - 1\\) 有重根",
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            "content": "(4) 重点 ZJU2018.2: 证明 \\(x^{3} + px + q\\) 有重根的充要条件 \\(4p^{3} + 27q^{2} = 0\\)",
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            "content": "1.2 特殊数域上的一元多项式理论"
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            "content": "(5)ZJU2021: 求 \\(t\\) 为何值时, \\(f(x) = x^3 + 6x^2 + tx + 8\\) 有重根, 并求重根解 (1) 可以尝试猜。一般用辗转相除法算 \\(f(x), f'(x)\\) 的最大公因式, 最终答案为 \\((x + 1)^4(x - 3)\\)",
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            "content": "(2) 根据重根与重因式关系，在 \\(\\mathbb{C}\\) 上必有 \\(f(1) = 0, f'(1) = 0\\)，因此得出 \\(A + B + 1 = 0, 4A + 2B = 0\\)，得到 \\(A = -2, B = 1\\)",
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            "content": "(3) 同理考虑 \\(\\mathbb{C}\\) 上的因式分解，\\(f'(x) = 3x^2 - 6x + t\\)，因此考虑 \\(f(x)\\) 与 \\(\\frac{1}{3} f'(x) = x^2 - 2x + \\frac{t}{3}\\) 辗转相除法，第一步得出 \\(f(x) = \\frac{1}{3} f'(x)(x - 1) + (\\frac{2}{3} t - 2)x + (\\frac{t}{3} - 1)\\)，如果 \\(t = 3\\) 则直接整除有重根，如果 \\(t \\neq 3\\)，则接着算，\\(\\frac{1}{3} f' = (\\frac{1}{\\frac{2}{3} t - 2})r_1(x)(x - \\frac{5}{2}) + \\frac{t}{3} + \\frac{5}{4}\\)，则需要 \\(\\frac{t}{3} + \\frac{5}{4} = 0\\)，得到 \\(t = -\\frac{15}{4}\\)",
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            "content": "(4) 在 \\(\\mathbb{C}\\) 上考虑，有重根即有重因式，\\(f'(x) = 3x^2 + p\\)，先有 \\(f(x) = \\frac{1}{3} f'(x)x + \\frac{2}{3} px + q\\)，若 \\(p = q = 0\\)，则成立，下面考虑 \\(p \\neq 0\\)。得到 \\(r_1(x) = \\frac{3}{2p} r_1(x)(x - \\frac{3q}{2p}) + \\frac{p}{3} + \\frac{9q^2}{4p^2}\\)，因此得到 \\(4p^3 + 27q^2 = 0\\)",
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            "content": "(5)t=12 时有 3 重 x=-2 根，t=-15 时有 2 重 x=1 根。",
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            "content": "练习 1.12 导数重根反推函数重根？ \\( x_{0} \\) 是 \\( f'(x) \\) 的 k 重根，举例说明其不能推出 \\( x_{0} \\) 是 \\( f(x) \\) 的 \\( k+1 \\) 重根，该命题需要 \\( x_{0} \\) 是 \\( f(x) \\) 的根。",
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            "content": "解例如 \\(f(x) = x^{k + 1} + 1\\) ， \\(x_0 = 0\\)",
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            "content": "笔记上面的结论非常重要，在利用导数验证根的重数时，一定要验证其是 \\( f(x) \\) 本身的根。",
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            "content": "1.1.7 整数与多项式"
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            "content": "定理 1.12 (整数与多项式)"
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            "content": "定理1.12（整数与多项式）\\(m,n\\) 为正整数，\\((m,n) = d\\) 的充要条件是 \\((x^{m} - 1, x^{n} - 1) = x^{d} - 1\\)",
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            "content": "证明 (1) 左推右是公因式：设 m = kd，则  \\( x^{m} - 1 = (x^{d})^{k} - 1 = (x^{d} - 1)(1 + x^{d} + \\cdots + x^{(k-1)d}) \\) ，这说明  \\( x^{d} - 1|x^{m} - 1 \\) ，另一侧同理，因此是公因式",
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            "content": "(2) 最大: 设 \\(p(x)\\) 是 \\(x^n - 1, x^m - 1\\) 的公因式, 下面证明 \\(p(x)|x^d - 1\\) 。由于 \\(x^n - 1\\) 在 \\(\\mathbb{C}\\) 上无重根 (因为和导数互素), 因此 \\(p(x)\\) 也没有重根, 从而 \\(p(x) = a(x - x_1) \\cdots (x - x_s)\\), 这里 \\(x_1 \\sim x_s\\) 是互异复数, 由于 \\(x_i^n = x_i^m = 1\\), 得到 \\(x_i^d = x_i^{um + vn} = (x_i^m)^u(x_i^n)^v = 1\\), 因此 \\(x - x_i|x^d - 1\\), \\((x - x_1) \\cdots (x - x_s)|x^d - 1\\), 因此 \\(p(x)|x^d - 1\\)",
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            "content": "(3) 右推左：若 \\((m, n) = d_{1}\\)，则根据前面的结论 \\((x^{m} - 1, x^{n} - 1) = x^{d_{1}} - 1 = x^{d} - 1\\)，因此 \\(d_{1} = d\\)",
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            "content": "练习1.13整除与多项式 \\((1)d, n \\in \\mathbb{N}^{+}\\)，证明 \\(x^{d} - 1|x^{n} - 1\\) 的充要条件为 \\(d|n\\)"
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            "content": "(2)m,n 是大于 1 的自然数，证明：m,n 互素的充要条件是  \\( 1 + x + x^{2} + \\cdots + x^{m-1} \\)  与  \\( 1 + x + x^{2} + \\cdots + x^{n-1} \\)  互素",
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            "content": "(3) \\( g(x)=1+x+x^{2}+\\cdots+x^{n},f(x)=g(x^{k}),(k,n+1)=1 \\) ，证明： \\( g(x)|f(x) \\)",
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            "content": "证明 (1) 本题即前面题目的特例。",
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            "content": "右推左：设  \\( x^{d}-1=(x-x_{1})\\cdots(x-x_{d})(\\mathbb{C} \\)  上)，因此  \\( x_{i}^{d}=1 \\) ，由于 n=dk，因此  \\( x_{i}^{n}=(x_{i}^{d})^{k}=1 \\) ，因此  \\( x^{d}-1|x^{n}-1 \\) 。",
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            "content": "左推右：设  \\( x^{d}-1=(x-x_{1})\\cdots(x-x_{d}) \\) ，由于  \\( x^{d}-1|x^{n}-1,\\quad x_{i}^{d}=x_{i}^{n}=1 \\) ，假设  \\( n=kd+r, r\\neq0 \\) ，则  \\( x_{i}^{n}=x_{i}^{kd+r}=x_{i}^{r} \\) ，由于  \\( x_{i}=1 \\)  只有一重，其他  \\( x_{i}\\neq1 \\) ，对于这些  \\( x_{i} \\)  有  \\( x_{i}^{r}\\neq1 \\) ，因此矛盾。",
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            "content": "1.2 特殊数域上的一元多项式理论"
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            "content": "1.2 特殊数域上的一元多项式理论"
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            "content": "1.2.1 复数域上的一元多项式"
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            "content": "任意次数大于等于1的多项式在 \\(\\mathbb{C}\\) 上至少有一个根。",
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            "content": "定理 1.14 (复数域因式分解)"
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            "content": "\\(\\mathbb{C}\\) 上任意次数大于等于1的多项式可唯一地分解为一次因式的乘积，因此 \\(n\\) 次多项式在 \\(\\mathbb{C}\\) 上恰有 \\(n\\) 个复根。",
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            "content": "1.2.2 实数域上的一元多项式"
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            "angle": 0,
            "content": "\\(\\mathbb{R}\\) 上的多项式总能唯一地分解为一次因式和不可约二次因式的乘积。",
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            "content": "1.2.3 有理数域上一元多项式因式分解"
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            "content": "复不可约多项式是一次的，实不可约多项式是一次或二次的，下面将说明有理不可约多项式的次数可以是任意的。",
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            "content": "定义 1.5 (本原多项式)"
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            "content": "若整系数多项式  \\( g(x)=b_{n}x^{n}+\\cdots+b_{1}x+b_{0} \\)  的系数  \\( b_{0}\\sim b_{n} \\)  是互素的 (整体互素而非两两互素)，即",
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            "content": "\\[\n(b _ {1}, \\dots , b _ {n}) = 1\n\\]"
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            "content": "则称 \\(g(x)\\) 是一个本原多项式",
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            "content": "两个本原多项式的乘积仍为本原多项式"
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            "content": "证明 设 \\( f(x) = \\sum_{k=0}^{n} a_k x^k, g(x) = \\sum_{k=0}^{m} b_k x^k \\)，设 \\( h(x) = f(x)g(x) = \\sum_{k=0}^{n+m} d_k x^k \\)。假设 \\( h(x) \\) 不是本原多项式，则存在素数 \\( p \\) 整除所有 \\( d_k \\)，由于 \\( f(x), g(x) \\) 本原，因此从常数项开始搜索，存在 \\( i, j \\)，使得",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} p \\mid a _ {0}, \\dots , p \\mid a _ {i - 1}, p \\nmid a _ {i} \\\\ p \\mid b _ {0}, \\dots , p \\mid b _ {j - 1}, p \\nmid b _ {j} \\end{array} \\right.\n\\]"
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            "content": "而 \\(d_{i + j} = a_i b_j + \\sum_{k=0, k \\neq i}^{i+j} a_k b_{i+j-k}\\)，除了第一项外均至少有一项被 \\(p\\) 整除，第一项不被 \\(p\\) 整除，因此 \\(p \\nmid d_{i+j}\\)，这与假设矛盾。",
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            "content": "定理 1.17 (有理多项式变本原多项式)"
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            "content": "设 \\(f(x)\\) 是有理多项式，\\(g(x)\\) 为本原多项式，则 \\(f(x)\\) 可表示为有理数 \\(r\\) 与本原多项式 \\(g(x)\\) 的乘积",
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            "content": "1.2 特殊数域上的一元多项式理论"
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            "content": "练习1.14 本原多项式 (1) \\( f(x), g(x) \\) 是本原多项式，\\( g(x) \\mid f(x) \\)，证明：\\( \\frac{f(x)}{g(x)} \\) 也是本原多项式 (2) \\( f(x) \\) 是本原多项式。证明：\\( f(x+1) \\) 也是本原多项式",
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            "content": "(2) \\( f(x) \\) 是本原多项式，证明： \\( f(x+1) \\) 也是本原多项式。",
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            "content": "证明 (1) 设 \\( f(x) = g(x)q(x) \\)，由于 \\( f(x), g(x) \\) 为本原多项式，\\( q(x) \\) 必然是有理系数的，设 \\( q(x) = rq_1(x) \\)，这里 \\( q_1(x) \\) 为本原多项式，因此",
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            "content": "\\[\nf (x) = r g (x) q _ {1} (x)\n\\]"
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            "content": "根据 Gauss 引理可知 \\( g(x)q_{1}(x) \\) 仍为本原多项式，因此 \\( r = \\pm 1 \\)，因此 \\( q(x) = \\pm q_{1}(x) \\) 是本原多项式。",
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            "content": "(2) 直接展开得到  \\( f(x+1)=b_{n}x^{n}+\\cdots+b_{1}x+b_{0} \\) ，其中  \\( b_{i}=\\sum_{k=i}^{n}\\binom{k}{i}a_{k} \\) 。下面证明互素，假设  \\( (b_{0},b_{1},\\cdots,b_{n})=d \\) ，则",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} d \\mid b _ {n} \\Rightarrow d \\mid a _ {n} \\\\ d \\mid b _ {n - 1} \\Rightarrow d \\mid \\binom {n} {1} a _ {n} + a _ {n - 1} \\Rightarrow d \\mid a _ {n - 1} \\end{array} \\right.\n\\]"
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            "content": "同理可以得到  \\( d \\mid a_{n-2}, \\cdots, d \\mid a_{1}, d \\mid a_{0} \\) ，这与  \\( f(x) \\)  是本原多项式矛盾。",
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            "content": "定理 1.18 (整系数多项式因式分解)"
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            "content": "整系数多项式若能分解为两个较低次有理多项式乘积，则一定也能分解为两个较低次整系数多项式乘积",
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            "content": "证明 设 \\( f(x) = g(x)h(x) \\)，这里 \\( g(x), h(x) \\in \\mathbb{Q}[x] \\)，则存在本原多项式 \\( f_{1}(x), h_{1}(x), g_{1}(x) \\) 使得",
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            "content": "\\[\nf (x) = a f _ {1} (x), g (x) = r g _ {1} (x), h (x) = s h _ {1} (x), a \\in \\mathbb {Z}, r, s \\in \\mathbb {Q}\n\\]"
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            "content": "从而 \\(af_{1}(x) = rsg_{1}(x)h_{1}(x)\\)，根据Gauss引理可知 \\(g_{1}(x)h_{1}(x)\\) 是本原多项式，不妨设最高项系数均正，则 \\(f_{1}(x) = g_{1}(x)h_{1}(x), a = rs\\)，得到 \\(rs \\in \\mathbb{Z}\\)，于是",
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            "content": "\\[\nf (x) = (r s g _ {1} (x)) h _ {1} (x)\n\\]"
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            "content": "其中 \\(rsg_{1}(x), h_{1}(x)\\) 都是整系数多项式。",
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            "content": "练习 1.15 整系数多项式整数根 (1) 若  \\( \\alpha \\neq \\pm1 \\)  为整系数多项式  \\( f(x) \\)  的整数根，证明： \\( \\frac{f(1)}{\\alpha-1}, \\frac{f(-1)}{\\alpha+1} \\)  都是整数 (2) ZJU2017.1:  \\( f(x) \\)  为整系数多项式， \\( f(0), f(1) \\)  为奇数，证明： \\( f(x) \\)  无整数根",
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            "content": "(3)ZJU2019.4:  \\( a_{1}, \\cdots, a_{n} \\)  为互不相同的整数,  \\( a_{1} \\cdots a_{n} + 1 \\)  不是某个整数的平方, 证明  \\( f(x) = (x + a_{1}) \\cdots (x + a_{n}) + 1 \\)  不能表示为 Q 上两个次数大于等于 1 的多项式的乘积。",
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            "content": "证明 (1) 由于 \\(\\alpha\\) 是 \\(f(x)\\) 的根，因此 \\(f(x) = (x - \\alpha)(b_{n-1}x^{n-1} + \\cdots + b_1x + b_0)\\)，这里 \\(b_0, \\cdots, b_{n-1}\\) 也都是整数，因此",
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            "content": "\\[\nf (1) = (1 - \\alpha) (b _ {n - 1} + \\dots + b _ {1} + b _ {0}) \\Rightarrow \\frac {f (1)}{\\alpha - 1} = - b _ {0} - \\dots - b _ {n - 1} \\in \\mathbb {Z}\n\\]"
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            "content": "(2) 若 \\( f \\) 有整数根，则 \\( f(x) = (x - a)f_0(x) \\)，这里 \\( f_0(x) \\) 为整系数多项式，而 \\( f(0) = (-a)f_0(0), f(1) = (1 - a)f_0(1) \\)，而根据 \\( f(0), f(1) \\) 为奇数得到 \\( -a, 1 - a \\) 同时为奇数，矛盾。",
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            "content": "(3) 若 \\( f(x) \\) 在 \\( Q[x] \\) 上可约，则 \\( f(x) \\) 在 \\( \\mathbb{Z}[x] \\) 也可约，不妨设 \\( f(x) = g(x)h(x), g(x), h(x) \\in \\mathbb{Z}[x] \\)，则 \\( f(-a_i) = g(-a_i)h(-a_i) = 1 \\)，这说明 \\( g(-a_i) = h(-a_i) = \\pm 1 \\)，因此 \\( g(x) - h(x) \\) 有 \\( n \\) 个不同的根，而 \\( \\partial(g(x) - h(x)) < n \\)，因此 \\( g(x) \\equiv h(x) \\)，根据条件 \\( f(x) = g^2(x) \\)，常数项为整数的平方，这与条件不是平方矛盾。",
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            "content": "1.3 多元多项式"
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            "content": "1.2.4 一元有理多项式的根"
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            "content": "定理 1.19 (有理多项式的根)"
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            "content": "\\( f(x)=a_{n}x^{n}+\\cdots+a_{1}x+a_{0} \\) 是整系数多项式，若既约分数 \\( \\alpha=\\frac{q}{p} \\) 是 \\( f(x) \\) 的有理根，则必有 \\( q\\mid a_{0},p\\mid a_{n} \\)",
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            "content": "证明 假设 \\(\\frac{q}{p}\\) 是解，则代入多项式得到：",
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            "content": "\\[\na _ {n} \\frac {q ^ {n}}{p ^ {n}} + \\dots + a _ {1} \\frac {q}{p} + a _ {0} = 0\n\\]"
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            "content": "两边乘上 \\(p^n\\) 并移项得到：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} a _ {n} q ^ {n} = p (- a _ {n - 1} q ^ {n - 1} - a _ {n - 2} p q ^ {n - 2} - \\dots - a _ {0} p ^ {n - 1}) \\\\ a _ {0} p ^ {n} = q (- a _ {n} q ^ {n - 1} - a _ {n - 1} p q ^ {n - 2} - \\dots - a _ {n - 1} p ^ {n - 1}) \\end{array} \\right.\n\\]"
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            "content": "因为 \\((p,q) = 1\\) ，因此 \\((p,q^n) = 1,(q,p^n) = 1\\) ，因此根据上面两个式子得到",
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            "content": "\\[\np \\mid a _ {0}, q \\mid a _ {n}\n\\]"
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            "content": "练习 1.16 计算有理根 (1) 证明:  \\( f(x) = x^{3} + 2x^{2} + x + 1 \\)  在有理数域上不可约",
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            "content": "(2) 求 \\( f(x) = 3x^{4} + 5x^{3} + x^{2} + 5x - 2 \\) 的有理根",
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            "content": "证明 (1) 有理数域不可约即无有理根。而有理根只可能是 \\(\\pm 1, f(\\pm 1) \\neq 0\\)，因此不可约。",
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            "content": "(2) 全部列出来验证，结果为 \\(\\frac{1}{3}, -2\\)",
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            "content": "定理 1.20 (Eisenstein 判别法)"
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            "content": "\\( f(x)=a_{n}x^{n}+\\cdots+a_{1}x+a_{0} \\) 是整系数多项式，若存在素数 p 使得：(1) \\( p\\nmid a_{n}(2)p\\mid a_{n-1},\\cdots,a_{0}(3)p^{2}\\nmid a_{0} \\) ，则 \\( f(x) \\) 在有理数域上是不可约多项式。",
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            "content": "练习 1.17 使用 Eisenstein 判别法 证明在 Q 不可约",
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            "content": "(1) \\( f(x)=2x^{4}+3x^{3}-9x^{2}-3x+6 \\)",
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            "content": "(2) \\( f(x)=x^{n}+2x+2 \\)",
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            "content": "(2) 取 \\(p = 2\\)",
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            "content": "对于多元多项式  \\( p(x_{1}, x_{2}, \\cdots, x_{n}) \\)  的乘积项，先按  \\( x_{1} \\)  次数由高到低排序，再根据  \\( x_{2}, x_{3}, \\cdots \\)  次数排序，将字典排序得到的第一项称为首项。",
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            "content": "\\( f(x_{1},\\cdots,x_{n}),g(x_{1},\\cdots,x_{n}) \\)  为多元多项式，则它们乘积的首项等于首项的乘积。",
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            "content": "若多元多项式  \\( f(x_{1},\\cdots,x_{n}) \\)  互换任意  \\( x_{i},x_{j} \\)  后多项式保持不变，则称其为对称多项式",
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            "content": "定理 1.22 (对称多项式的性质)"
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            "content": "对称多项式的和、差、乘积、多项式复合仍为对称多项式。",
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            "content": "定义 1.8 (初等对称多项式)"
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            "content": "将形如  \\( \\sigma_{i}=\\sum_{k_{1}<k_{2}<\\cdots<k_{i}}x_{k_{1}}x_{k_{2}}\\cdots x_{k_{i}} \\)  的对称多项式称为初等对称多项式：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\sigma_ {1} = x _ {1} + x _ {2} + \\dots + x _ {n} \\\\ \\sigma_ {2} = x _ {1} x _ {2} + x _ {1} x _ {3} + \\dots + x _ {n - 1} x _ {n} \\\\ \\quad \\vdots \\\\ \\sigma_ {n} = x _ {1} x _ {2} \\dots x _ {n} \\end{array} \\right.\n\\]"
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            "content": "定理 1.23 (Vieta 定理)"
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            "content": "对于  \\( f(x)=x^{n}+a_{n-1}x^{n-1}+\\cdots+a_{0} \\) ，假设其解为  \\( x_{1},\\cdots,x_{n} \\) ，则",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} a _ {n - 1} = - \\sigma_ {1} \\\\ a _ {n - 2} = \\sigma_ {2} \\\\ \\vdots \\\\ a _ {n - i} = (- 1) ^ {i} \\sigma_ {i} \\\\ \\vdots \\\\ a _ {0} = (- 1) ^ {n} \\sigma_ {n} \\end{array} \\right.\n\\]"
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            "content": "定理 1.24 (复合定理)"
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            "content": "任意对称多项式  \\( f(x_{1},\\cdots,x_{n}) \\)  可唯一地写成初等多项式的多项式  \\( \\varphi(\\sigma_{1},\\cdots,\\sigma_{n}) \\)",
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            "content": "证明 设  \\( f(x_{1},\\cdots,x_{n}) \\)  的首项为  \\( ax_{1}^{l_{1}}x_{2}^{l_{2}}\\cdots x_{n}^{l_{n}} \\) ，且  \\( l_{1}\\geq l_{2}\\geq\\cdots l_{n} \\) （根据对称多项式可交换性可知该性质），做多项式  \\( \\varphi_{1}=a\\sigma_{1}^{l_{1}-l_{2}}\\sigma_{2}^{l_{2}-l_{3}}\\cdots\\sigma_{n}^{l_{n}} \\) ，得到  \\( \\varphi_{1} \\)  的首项为  \\( ax_{1}^{l_{1}-l_{2}}(x_{1}x_{2})^{l_{2}-l_{3}}\\cdots(x_{1}\\cdots x_{n})^{l_{n}}=ax_{1}^{l_{1}}\\cdots x_{n}^{l_{n}} \\) ，取  \\( f_{1}(x)=f(x)-\\varphi_{1} \\) ，以此类推可以得到结果。",
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            "content": "练习 1.18 (1) 将  \\( f(x_{1}, x_{2}, x_{3}) = x_{1}^{3} + x_{2}^{3} + x_{3}^{3} \\)  表达为  \\( \\sigma_{1}, \\sigma_{2}, \\sigma_{3} \\)  的多项式",
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            "content": "解(1)根据首项可知 \\(l_{1} = 3, l_{2} = l_{3} = 0\\)，\\(\\varphi_{1} = \\sigma_{1}^{l_{1}}\\sigma_{2}^{l_{2}}\\sigma_{3}^{l_{3}} = (x_{1} + x_{2} + x_{3})^{3}\\)，得到",
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            "content": "\\[\nf _ {1} (x) = f (x) - \\varphi_ {1} = - 3 (x _ {1} ^ {2} x _ {2} + x _ {1} ^ {2} x _ {3} + \\dots) - 6 x _ {1} x _ {2} x _ {3}\n\\]"
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            "content": "1.3 多元多项式"
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            "content": "对称多项式理论可应用于判断一元高次多项式重根的存在性，这就导出了判别式的概念。",
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            "content": "给定多项式  \\( f(x)=x^{n}+a_{n-1}x^{n-1}+\\cdots+a_{0}\\in\\mathbb{C}[x] \\) ，设  \\( x_{1},\\cdots,x_{n} \\)  是  \\( f(x) \\)  的根，则定义判别式为",
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            "content": "\\[\nD (f) := \\prod_ {1 \\leq i <   j \\leq n} (x _ {i} - x _ {j}) ^ {n}\n\\]"
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            "content": "显然 \\(f(x)\\) 有重根当且仅当 \\(D(f) = 0\\)",
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            "content": "由于  \\( D(f) \\)  是  \\( x_{1},\\cdots,x_{n} \\)  的对称多项式，因此可以表示为初等对称多项式的多项式，再根据 Vieta 定理即可表达为  \\( a_{0},\\cdots,a_{n-1} \\)  的多项式",
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            "content": "练习 1.19 判别式的计算 (1) 计算  \\( f(x)=x^{2}+px+q \\)  的判别式",
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            "content": "解(1)设 \\(g(x_{1},x_{2}) = (x_{1} - x_{2})^{2}\\) ，可以将 \\(g(x_{1},x_{2})\\) 表达为初等对称多项式的多项式：",
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            "content": "\\[\ng (x _ {1}, x _ {2}) = \\sigma_ {1} ^ {2} - 4 \\sigma_ {2}\n\\]"
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            "content": "而根据 Vieta 定理， \\( p = -\\sigma_{1}, q = \\sigma_{2} \\) ，因此  \\( D(f) = p^{2} - 4q \\)",
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            "content": "1.3.4 结式"
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            "content": "给定多项式  \\( f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\\cdots+a_{0}, g(x)=b_{m}x^{m}+\\cdots+b_{0} \\) ，虽然可以用辗转相除法计算它们的最大公因式，但判断它们是否互素有更好的解决办法，这就是结式",
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            "content": "定义 1.10 (结式)"
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            "content": "考虑两个多项式  \\( f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\\cdots+a_{0}, g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\\cdots+b_{0} \\) ，则定义两个矩阵：",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c c c c c} a _ {n} & a _ {n - 1} & a _ {n - 2} & \\dots & a _ {0} & 0 & \\dots & 0 \\\\ 0 & a _ {n} & a _ {n - 1} & a _ {n - 2} & \\dots & a _ {0} & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\ddots & \\ddots & & \\ddots & \\vdots \\\\ 0 & 0 & \\dots & a _ {n} & a _ {n - 1} & a _ {n - 2} & \\dots & a _ {0} \\end{array} \\right] _ {m \\times (n + m)}\n\\]"
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            "content": "\\[\nB = \\left[ \\begin{array}{c c c c c c c c} b _ {m} & b _ {m - 1} & b _ {m - 2} & \\dots & b _ {0} & 0 & \\dots & 0 \\\\ 0 & b _ {m} & b _ {m - 1} & b _ {m - 2} & \\dots & b _ {0} & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\ddots & \\ddots & & \\ddots & \\vdots \\\\ 0 & 0 & \\dots & b _ {m} & b _ {m - 1} & b _ {m - 2} & \\dots & b _ {0} \\end{array} \\right] _ {n \\times (n + m)}\n\\]"
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            "content": "则定义它们的结式 \\(R(f,g) = \\left| \\begin{array}{l}A\\\\ B \\end{array} \\right|\\)",
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            "content": "给定  \\( f(x)=a_{n}x^{n}+\\cdots+a_{0}, g(x)=b_{m}x^{m}+\\cdots+b_{0}, a_{n}, b_{m}\\neq0 \\) ，则 f, g 不互素的充要条件为  \\( R(f,g)=0 \\)",
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            "content": "证明 设 \\(\\partial (f) = n, \\partial (g) = m\\)，首先 \\(f(x), g(x)\\) 不互素当且仅当 \\(\\exists u(x), v(x)\\) 满足 \\(\\partial (u) < \\partial (g), \\partial (v) < \\partial (f)\\)，使得",
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            "content": "1.3 多元多项式"
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            "content": "\\(u(x)f(x) = v(x)g(x)\\) ，不妨设",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} u (x) = u _ {m - 1} x ^ {m - 1} + \\dots + u _ {0} \\\\ v (x) = v _ {n - 1} x ^ {n - 1} + \\dots + v _ {0} \\end{array} \\right.\n\\]"
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            "content": "将 \\(u(x)f(x) = v(x)g(x)\\) 展开对比系数得到",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} a _ {n} u _ {m - 1} = b _ {m} v _ {n - 1} \\\\ a _ {n - 1} u _ {m - 1} + a _ {n} u _ {m - 2} = b _ {m - 1} v _ {n - 1} + b _ {m} v _ {n - 2} \\\\ \\vdots \\\\ a _ {0} u _ {0} = b _ {0} v _ {0} \\end{array} \\right.\n\\]"
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            "content": "将上述线性方程组看成  \\( u_{0}, \\cdots, u_{m-1}, v_{0}, \\cdots, v_{n-1} \\)  的线性方程组，则系数矩阵的转置  \\( C^{T} \\)  为结式定义中的矩阵：",
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            "content": "\\[\nC ^ {T} = \\binom{A}{- B}\n\\]"
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            "content": "因此有解当且仅当 \\(|C^T| = 0\\) ，也等价于 \\(R(f,g) = 0\\)",
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            "content": "定理 1.27 (结式计算公式)"
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            "content": "给定前面的  \\( f(x), g(x) \\) ， \\( \\alpha_{1}, \\cdots, \\alpha_{n} \\)  和  \\( \\beta_{1}, \\cdots, \\beta_{m} \\)  分别是  \\( f(x), g(x) \\)  的复根，则",
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            "content": "\\[\nR (f, g) = a _ {n} ^ {m} b _ {m} ^ {n} \\prod_ {i = 1} ^ {n} \\prod_ {j = 1} ^ {m} (\\alpha_ {i} - \\beta_ {i})\n\\]"
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            "content": "定理 1.28 (判别式计算公式)"
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            "content": "\\(\\mathbb{C}[x]\\) 中多项式 \\(f(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + \\dots +a_{0}, a_{n}\\neq 0\\) ，则 \\(f(x)\\) 的判别式",
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            "content": "\\[\nD (f) = (- 1) ^ {\\frac {n (n - 1)}{2}} a _ {0} ^ {- (2 n - 1)} R (f, f ^ {\\prime})\n\\]"
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            "content": "第2章 矩阵"
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            "content": "2.1 矩阵基础"
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            "content": "2.1.2 矩阵的逆"
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            "content": "定义 2.1 (矩阵的逆)"
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            "content": "若 \\(AB = BA = E\\) ，则称 \\(B\\) 为 \\(A\\) 的逆，记 \\(A^{-1}\\)",
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            "content": "定理 2.1 (矩阵逆的基本性质)"
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            "content": "\\[\n(1) (k A) ^ {- 1} = k ^ {- 1} A ^ {- 1} (2) (A B) ^ {- 1} = B ^ {- 1} A ^ {- 1} (3) \\left(A ^ {T}\\right) ^ {- 1} = \\left(A ^ {- 1}\\right) ^ {T} (4) | A ^ {- 1} | = | A | ^ {- 1}\n\\]"
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            "content": "证明 (2) \\((AB)^{-1}AB = E\\) ，因此 \\((AB)^{-1} = B^{-1}A^{-1}\\)",
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            "content": "(3) \\((A^{-1})^T A^T = (AA^{-1})^T = E\\)",
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            "content": "(4) \\( |A^{-1}A|=|A^{-1}|\\cdot|A|=1 \\) ，因此 \\( |A^{-1}|=|A|^{-1} \\)",
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            "content": "矩阵 \\(A\\) 可以的充要条件如下：",
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            "content": "- \\(A\\) 可表示为初等矩阵的积",
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            "content": "定理 2.3 (利用初等行变换求逆)"
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            "content": "若 \\(A\\) 可逆，则利用初等行变换可以得到 \\((A,E)\\rightarrow (E,A^{-1})\\) 。特别地， \\((A,B)\\rightarrow (E,A^{-1}B)\\)",
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            "content": "笔记 同样可以通过初等列变换得到： \\( \\begin{pmatrix} A \\\\ B \\end{pmatrix} \\rightarrow \\begin{pmatrix} E \\\\ BA^{-1} \\end{pmatrix} \\)",
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            "content": "练习 2.1 利用初等变换求逆 (1) 计算下面矩阵的逆:",
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            "content": "解结果为 \\(\\frac{1}{3}\\left[ \\begin{array}{ccc}0 & 1 & 1\\\\ 0 & 1 & -2\\\\ -3 & 2 & -1 \\end{array} \\right]\\)",
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            "content": "(3) 根据 \\(B(A^{-1} + B^{-1})A = (A + B)\\) 得到",
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            "content": "\\[\n(A ^ {- 1} + B ^ {- 1}) B = A ^ {- 1} (A + B) \\Rightarrow (A ^ {- 1} + B ^ {- 1}) B (A + B) ^ {- 1} = A ^ {- 1}\n\\]"
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            "content": "因此 \\((A^{-1} + B^{-1})^{-1} = B(A + B)^{-1}A\\)",
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            "content": "\\(A = (a_{ij})\\) 为 \\(n\\times n\\) 矩阵，划去 \\(i\\) 行 \\(j\\) 列的行列式称为余子式 \\(M_{ij}\\) ，代数余子式为 \\((-1)^{i + j}M_{ij}\\)",
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            "content": "\\( A = (a_{ij})_n \\) ， \\( A_{ij} \\)  是  \\( a_{ij} \\)  的代数余子式，则其伴随为 (注意行列颠倒!!):",
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            "content": "\\[\nA ^ {*} = \\left[ \\begin{array}{c c c c} A _ {1 1} & A _ {2 1} & \\dots & A _ {n 1} \\\\ A _ {1 2} & A _ {2 2} & \\dots & A _ {n 2} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A _ {1 n} & A _ {2 n} & \\dots & A _ {n n} \\end{array} \\right]\n\\]"
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            "content": "\\(A\\) 及其伴随矩阵 \\(A^{*}\\) 满足 \\(AA^{*} = |A|E\\) ，若 \\(A\\) 可逆，则 \\(A^{-1} = \\frac{1}{|A|} A^{*}\\)",
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            "content": "证明 考虑 \\(B = AA^{*}\\)，则 \\(B_{ij} = \\sum_{k=1}^{n} a_{ik} A_{kj}\\)，根据行列式的展开式可知",
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            "content": "\\[\nB _ {i j} = \\left\\{ \\begin{array}{l l} | A |, & i = j \\\\ 0, & i \\neq j \\end{array} \\right.\n\\]"
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            "content": "2. \\((A^{*})^{-1} = (A^{-1})^{*}\\)",
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            "content": "2.1 矩阵基础"
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            "content": "\\[\n(3) (A B) ^ {*} = | A B | (A B) ^ {- 1} = | B | (B) ^ {- 1} | A | A ^ {- 1} = B ^ {*} A ^ {*}\n\\]"
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            "content": "定理 2.6 (伴随矩阵的行列式)",
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            "content": "\\[\n| A ^ {*} | = | A | ^ {n - 1}\n\\]"
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            "angle": 0,
            "content": "证明 \\(|A^{*}| = \\left| |A| A^{-1} \\right| = |A|^{n} \\cdot |A^{-1}| = |A|^{n - 1}\\)",
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            "content": "定理 2.7 (伴随矩阵的秩)"
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            "content": "\\(A^{*}\\) 为 \\(A\\) 的伴随矩阵，则",
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            "content": "\\[\nr (A ^ {*}) = \\left\\{ \\begin{array}{l l} n, & r (A) = n \\\\ 1, & r (A) = n - 1 \\\\ 0, & r (A) <   n - 1 \\end{array} \\right.\n\\]"
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            "content": "证明 \\(r(A) = n\\) 时，\\(|A^{*}| = |A|^{n - 1} \\neq 0\\)，因此满秩。\\(r(A) = n - 1\\) 时，由于 \\(AA^{*} = |A|E = O\\)，根据秩不等式：",
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            "content": "\\[\nr (A A ^ {*}) \\geq r (A) + r (A ^ {*}) - n = r (A ^ {*}) - 1 \\Rightarrow r (A ^ {*}) \\leq 1\n\\]"
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            "content": "而由于 \\(r(A) = n - 1\\) ，其至少有一个 \\(n - 1\\) 阶代数余子式不为0，因此 \\(r(A^{*})\\geq 1\\) ，综上 \\(r(A^{*}) = 1\\) 。其余显然",
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            "content": "定理 2.8 (伴随矩阵的特征值)"
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            "content": "矩阵 A 的特征值为  \\( \\lambda_{1},\\cdots,\\lambda_{n} \\) ，则  \\( A^{*} \\)  的特征值  \\( \\lambda_{j}^{*}=\\prod_{i\\neq j}\\lambda_{i} \\)",
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            "content": "证明 考虑 \\(A = P^{-1}JP\\)，由于 \\(J\\) 是上三角阵，根据初等变换法，\\(J^{-1}\\) 是上三角阵，\\(J^{*} = |J|J^{-1}\\) 也是上三角阵，而 \\(J\\) 对角元素的代数余子式为 \\(\\prod_{i\\neq j}\\lambda_i\\)，因此 \\(J^{*}\\) 的对角元为 \\(\\prod_{i\\neq j}\\lambda_i\\)，代数余子式在初等变换下不变 (位置可能变化，",
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            "content": "但值整体相同)，因此结论成立。",
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            "content": "2.1.4 广义逆矩阵"
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            "content": "定义 2.4 (广义逆)"
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            "angle": 0,
            "content": "\\(A_{m\\times n}\\) 是数域 \\(\\mathbb{P}\\) 上的矩阵，若存在 \\(G_{n\\times m}\\) 使得",
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            "content": "\\[\nA G A = A\n\\]"
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            "content": "则称 \\(G\\) 是 \\(A\\) 的广义逆矩阵。",
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            "content": "笔记 广义逆矩阵一般不唯一。"
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            "content": "定理 2.9 (广义逆的表达式)"
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            "content": "设 \\(A_{m\\times n}\\) 的秩为 \\(r\\) ，设 \\(A = P\\left( \\begin{array}{cc}E_r & O\\\\ O & O \\end{array} \\right)Q\\) ，其中 \\(P_{m\\times m},Q_{n\\times n}\\) 为可逆矩阵，则 \\(A\\) 全部的广义逆矩阵为",
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            "content": "16"
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            "content": "\\[\nG = Q ^ {- 1} \\left( \\begin{array}{c c} E _ {r} & C \\\\ D & F \\end{array} \\right) P ^ {- 1}\n\\]"
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            "content": "这里 C, D, F 为任意  \\( r \\times (m - r) \\) ,  \\( (n - r) \\times r \\) ,  \\( (m - r) \\times (n - r) \\)  阶矩阵。",
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            "content": "证明 (1) 验证 \\(G\\) 是广义逆: 直接乘起来就行, 得到 \\(AGA = A\\)",
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            "content": "2.1 矩阵基础"
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            "content": "(2) 验证广义逆都是 \\(G\\)：设 \\(G = Q^{-1}\\left( \\begin{array}{cc}B & C\\\\ D & F \\end{array} \\right)P^{-1}\\)，乘法得到",
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            "content": "\\[\nA G A = P \\left( \\begin{array}{c c} B & O \\\\ O & O \\end{array} \\right) Q\n\\]"
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            "content": "因此得到 \\(B = E_{r}\\)",
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            "content": "定理 2.10 (非方阵线性方程组解的表达式)"
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            "content": "\\(A_{m\\times n}\\) ， \\(G\\) 是 \\(A\\) 的广义逆，则 \\(AX = b\\) 的通解可表示为",
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            "content": "\\[\nX = G b + (E _ {n} - G A) Y\n\\]"
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            "content": "这里 \\(Y\\) 是任一 \\(n\\) 维向量",
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            "content": "证明 (1) 证明是解：设 \\(Z\\) 是 \\(AZ = b\\) 的解，则此时有 \\(b = AZ = AGAZ = AGb\\)。将 \\(X\\) 表达式代入 \\(AX\\) 验证：",
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            "content": "\\[\nA X = A G b + (A - A G A) Y = b + O = b\n\\]"
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            "content": "(2) 证明解都有该形式: 设 \\(X_{0}\\) 是 \\(AX = b\\) 的解, 则",
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            "content": "\\[\nX _ {0} = G b + X _ {0} - G b = G b + X _ {0} - G A X _ {0} = G b + (E _ {n} - G A) X _ {0}\n\\]"
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            "content": "2.1.5 Moore-Penrose 广义逆"
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            "content": "定义 2.5 (共轭转置)"
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            "content": "\\(A\\) 是 \\(\\mathbb{C}\\) 上 \\(n\\times m\\) 矩阵，则定义 \\(A^{H} = (\\overline{A})^{T}\\)",
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            "content": "定义 2.6 (Moore-Penrose 广义逆)"
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            "content": "\\(A\\) 是 \\(\\mathbb{C}\\) 上 \\(m \\times n\\) 阶矩阵，若存在 \\(\\mathbb{C}\\) 上 \\(n \\times m\\) 阶矩阵 \\(G\\) 满足：",
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            "content": "1. AGA = A",
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            "content": "2. \\(GAG = G\\)",
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            "content": "3. \\((AG)^{H} = AG\\)",
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            "content": "4. \\((GA)^{H} = GA\\)",
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            "content": "则称 \\(G\\) 为 \\(A\\) 的M-P广义逆",
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            "content": "定理 2.11 (M-P 广义逆的存在唯一性)"
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            "content": "\\[\nA ^ {+} = C ^ {H} (C C ^ {H}) ^ {- 1} (B ^ {H} B) ^ {- 1} B ^ {H}\n\\]"
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            "content": "\\[\nG = C ^ {H} (C C ^ {H}) ^ {- 1} (B ^ {H} B) ^ {- 1} B ^ {H}\n\\]"
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            "content": "2.2 分块矩阵"
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            "content": "唯一性：设 \\(X\\) 是另一个广义逆，则进行下面步骤，本质是用 \\(AXA = A\\) 消去 \\(X\\)",
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            "content": "\\[\n\\begin{array}{l} \\boldsymbol {X} = \\boldsymbol {X A X} = \\boldsymbol {X} (\\boldsymbol {A X}) ^ {\\mathrm{H}} = \\boldsymbol {X X} ^ {\\mathrm{H}} \\boldsymbol {A} ^ {\\mathrm{H}} = \\boldsymbol {X X} ^ {\\mathrm{H}} (\\boldsymbol {A G A}) ^ {\\mathrm{H}} \\\\ \\quad = \\boldsymbol {X X} ^ {\\mathrm{H}} \\boldsymbol {A} ^ {\\mathrm{H}} (\\boldsymbol {A G}) ^ {\\mathrm{H}} = \\boldsymbol {X} (\\boldsymbol {A X}) ^ {\\mathrm{H}} (\\boldsymbol {A G}) ^ {\\mathrm{H}} = \\boldsymbol {X A X A G} \\\\ \\quad = \\boldsymbol {X A G} = \\boldsymbol {X} (\\boldsymbol {A G A}) \\boldsymbol {G} = (\\boldsymbol {X A}) ^ {\\mathrm{H}} (\\boldsymbol {G A}) ^ {\\mathrm{H}} \\boldsymbol {G} \\\\ \\quad = (\\boldsymbol {G A X A}) ^ {\\mathrm{H}} \\boldsymbol {G} = (\\boldsymbol {G A}) ^ {\\mathrm{H}} \\boldsymbol {G} = (\\boldsymbol {G A}) \\boldsymbol {G} \\\\ \\quad = \\boldsymbol {G}. \\end{array}\n\\]"
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            "content": "练习2.3计算M-P逆 \\((1)A = \\left( \\begin{array}{ccc}1 & 0 & -1\\\\ 1 & 2 & 0\\\\ 0 & 2 & 1 \\end{array} \\right)\\) 的M-P逆",
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            "content": "解(1) 先进行分解，得到",
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            "content": "\\[\nB = \\left( \\begin{array}{c c} 1 & 0 \\\\ 1 & 1 \\\\ 0 & 1 \\end{array} \\right), C = \\left( \\begin{array}{c c c} 1 & 0 & - 1 \\\\ 0 & 2 & 1 \\end{array} \\right)\n\\]"
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            "content": "最终结果得到",
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            "content": "\\[\nA ^ {+} = \\frac {1}{9} \\left( \\begin{array}{c c c} 3 & 2 & - 1 \\\\ 0 & 2 & 2 \\\\ - 3 & - 1 & 2 \\end{array} \\right)\n\\]"
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            "content": "定理 2.12 (最小二乘解的表达式)",
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            "content": "\\(AX = b\\) 最小二乘解为 \\(X = A^{+}b + (E_{n} - A^{+}A)Y\\) ，其中 \\(A^{+}b\\) 是唯一的极小最小二乘解。",
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            "content": "2.2 分块矩阵"
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            "content": "18"
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            "content": "2.2.1 常见矩阵拆分技巧"
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            "content": "定理 2.13 (点态分解)"
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            "content": "\\(A\\) 是 \\(n \\times n\\) 矩阵，则",
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            "content": "A = \\left[ \\begin{array}{c c c c} a _ {1} b _ {1} & a _ {1} b _ {2} & \\dots & a _ {1} b _ {n} \\\\ a _ {2} b _ {1} & a _ {2} b _ {2} & \\dots & a _ {2} b _ {n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a _ {n} b _ {1} & a _ {n} b _ {2} & \\dots & a _ {n} b _ {n} \\end{array} \\right] \\Rightarrow A = \\left( \\begin{array}{c} a _ {1} \\\\ a _ {2} \\\\ \\vdots \\\\ a _ {n} \\end{array} \\right) \\left( \\begin{array}{c c c c} b _ {1} & b _ {2} & \\dots & b _ {n} \\end{array} \\right)\\tag{❤️}"
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            "content": "定理 2.14 ( \\( A^{T}A \\) 的和式表达)"
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            "content": "\\(A\\) 为 \\(n\\times m\\) 矩阵，将 \\(A\\) 做下面的行划分，则 \\(A^T A = \\sum_{i = 1}^{n}\\alpha_i^T\\alpha_i\\)",
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            "content": "\\[\nA = \\left( \\begin{array}{c} \\alpha_ {1} \\\\ \\alpha_ {2} \\\\ \\vdots \\\\ \\alpha_ {s} \\end{array} \\right)\n\\]"
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            "content": "#"
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            "content": "2.2 分块矩阵"
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            "content": "2.2.2 分块初等变换"
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            "content": "定义 2.7 (分块矩阵初等变换)"
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            "content": "分块矩阵初等变换分为行变换和列变换：",
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            "content": "- 行变换: (1) 左乘矩阵 \\(P\\), 加到另一行 (2) 互换两行位置 (3) 将可逆矩阵左乘至一行",
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            "content": "- 列变换：(1) 右乘矩阵 \\(P\\)，加到另一列 (2) 互换两列位置 (3) 将可逆矩阵右乘至一列",
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            "content": "定义 2.8 (分块初等矩阵)"
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            "content": "对应三种分块初等变换有三种分块初等矩阵，常用的即加至另一行/列的初等分块矩阵：",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c} E & 0 \\\\ P & E \\end{array} \\right]\n\\]"
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            "content": "定理 2.15 (初等变换的分块初等矩阵表达)"
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            "content": "分块初等矩阵左乘一个矩阵即做行变换，右乘即做列变换：",
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            "content": "\\[\n\\left[ \\begin{array}{c c} E & 0 \\\\ P & E \\end{array} \\right] \\left[ \\begin{array}{c c} A _ {1} & A _ {2} \\\\ A _ {3} & A _ {4} \\end{array} \\right] = \\left[ \\begin{array}{c c} A _ {1} & A _ {2} \\\\ P A _ {1} + A _ {3} & P A _ {2} + A _ {4} \\end{array} \\right]\n\\]"
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            "content": "2.2.3 分块打洞技巧总结"
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            "content": "定理 2.16 (分块打洞)"
        },
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            "content": "对于分块矩阵  \\( \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix} \\)  可通过分块初等变换转换为：",
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        },
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            "bbox": [
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & B \\\\ O & D - C A ^ {- 1} B \\end{array} \\right] \\quad \\left[ \\begin{array}{c c} A - B D ^ {- 1} C & O \\\\ O & D \\end{array} \\right]\n\\]"
        },
        {
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            "angle": 0,
            "content": "记忆方式：除去被减的位置，减去的内容顺序都是顺时针转的。",
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        },
        {
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            "angle": 0,
            "content": "推论 2.1 (两种特殊形式)"
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            "bbox": [
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            "angle": 0,
            "content": "对称形式和归纳形式：",
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            "bbox": [
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            "angle": 0,
            "content": "- 对称形式:",
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            "type": "equation",
            "bbox": [
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & B \\\\ B ^ {T} & D \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A & O \\\\ O & D - B ^ {T} A ^ {- 1} B \\end{array} \\right]\n\\]"
        },
        {
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            "content": "归纳形式：",
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & \\alpha \\\\ \\beta^ {T} & a _ {n n} \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A & \\alpha \\\\ 0 & a _ {n n} - \\beta^ {T} A ^ {- 1} \\alpha \\end{array} \\right]\n\\]"
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            "content": "练习 2.4 分块矩阵与数学归纳法 (1)n 阶方阵 A 的顺序主子式非零，证明：存在 n 阶下三角矩阵 B 使得 BA 是上三角阵",
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            "content": "(2) \\(A\\) 是 \\(n\\) 阶实方阵, \\(A\\) 所有顺序主子式大于 0 , 且非主对角元素为负, 证明: \\(A^{-1}\\) 的元素均为正数",
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            "content": "证明 \\((1)n = 1\\) 显然成立。设 \\(n - 1\\) 时成立，考虑如下分块，由于 \\(A\\) 顺序主子式非零，因此 \\(A_{1}\\) 顺序主子式非零，存在 \\(n - 1\\) 阶下三角矩阵 \\(B_{1}\\) 满足 \\(B_{1}A_{1}\\) 为上三角阵，因此做对应分块初等变换",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c} A _ {1} & \\alpha \\\\ \\beta^ {T} & a _ {n n} \\end{array} \\right] \\Rightarrow \\left[ \\begin{array}{c c} B _ {1} & \\\\ & 1 \\end{array} \\right] \\cdot \\left[ \\begin{array}{c c} E _ {n - 1} & O \\\\ - \\beta^ {T} A _ {1} ^ {- 1} & 1 \\end{array} \\right] \\cdot \\left[ \\begin{array}{c c} A _ {1} & \\alpha \\\\ \\beta^ {T} & a _ {n n} \\end{array} \\right] = \\left[ \\begin{array}{c c} B _ {1} A _ {1} & B _ {1} \\alpha \\\\ O & a _ {n n} - \\beta^ {T} A _ {1} ^ {- 1} \\alpha \\end{array} \\right]\n\\]"
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            "content": "2.2 分块矩阵"
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            "content": "(2) \\(n = 1\\) 显然。设 \\(n - 1\\) 时成立，对 \\(A\\) 做下述分块，则 \\(A_{1}^{-1}\\) 元素都是正的",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c} A _ {1} & \\alpha \\\\ \\beta^ {T} & a _ {n n} \\end{array} \\right] \\Rightarrow \\left[ \\begin{array}{c c} E _ {n - 1} & O \\\\ - \\beta^ {T} A _ {1} ^ {- 1} & 1 \\end{array} \\right] \\left[ \\begin{array}{c c} A _ {1} & \\alpha \\\\ \\beta^ {T} & a _ {n n} \\end{array} \\right] \\left[ \\begin{array}{c c} E _ {n - 1} & - A _ {1} ^ {- 1} \\alpha \\\\ O & 1 \\end{array} \\right] = \\left[ \\begin{array}{c c} A _ {1} & O \\\\ O & a _ {n n} - \\beta^ {T} A _ {1} ^ {- 1} \\alpha \\end{array} \\right]\n\\]"
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            "content": "根据 \\(|A| > 0, |A_1| > 0\\) 得到 \\(a_{nn} - \\beta^T A_1^{-1}\\alpha > 0\\) 。记上述等式为 \\(BAC = D\\) ，则 \\(A^{-1} = CD^{-1}B\\) ，根据矩阵乘法乘开即可获得结论。",
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            "content": "2.2.4 分块矩阵行列式"
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            "content": "定理 2.17 (分块对角矩阵的行列式)"
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            "content": "根据 Laplace 多行展开定理可以得到:",
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            "content": "\\[\n\\left| \\begin{array}{c c} A & 0 \\\\ B & C \\end{array} \\right| = | A | | B |,\n\\]"
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            "content": "\\left| \\begin{array}{c c c c} A _ {1 1} & 0 & \\dots & 0 \\\\ A _ {2 1} & A _ {2 2} & \\dots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A _ {n 1} & A _ {n 2} & \\dots & A _ {n n} \\end{array} \\right| = | A _ {1 1} | | A _ {2 2} | \\dots | A _ {n n} |\\tag{❤️}"
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            "content": "定理 2.18 (一般分块矩阵的行列式)"
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            "content": "设 \\(A, D\\) 分别为 \\(r, s\\) 阶矩阵，\\(A, D\\) 可逆，则",
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            "content": "\\[\n\\left| \\begin{array}{c c} A & B \\\\ C & D \\end{array} \\right| = | D | | A - B D ^ {- 1} C | = | A | | D - C A ^ {- 1} B |\n\\]"
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            "content": "证明 (1) 如果 \\(D\\) 可逆，则把 \\(B\\) 消掉 (左乘 \\(BD^{-1}\\))，取行列式即可",
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & B \\\\ C & D \\end{array} \\right] \\Rightarrow \\left[ \\begin{array}{c c} A - B D ^ {- 1} C & 0 \\\\ C & D \\end{array} \\right]\n\\]"
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            "content": "定理 2.19 (行列式计算技巧 1: 配凑分块矩阵)"
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            "content": "矩阵 A 可分解为  \\( A = J - u^{T}u \\) ，要计算  \\( |A| \\) ，则可以构造分块矩阵：",
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            "content": "\\[\n\\left| \\begin{array}{c c} J & u \\\\ u ^ {T} & 1 \\end{array} \\right| = | 1 | \\cdot | J - u ^ {T} u | = | J | \\cdot | 1 - u J ^ {- 1} u ^ {T} | \\Rightarrow | A | = | J | \\cdot | 1 - u J ^ {- 1} u ^ {T} |\n\\]"
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            "content": "注意这里 \\(u\\) 是列向量",
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            "content": "2.2.5 分块矩阵求逆"
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            "content": "一般分块矩阵求逆都先转换为分块对角矩阵，具体如下。",
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            "content": "定理 2.20 (上三角分块矩阵求逆)"
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            "content": "当下面的 \\(A_{1}, A_{2}\\) 都是方阵，则 \\(A\\) 可逆当且仅当 \\(A_{1}, A_{2}\\) 都可逆，且逆矩阵如下：",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c} A _ {1} & A _ {3} \\\\ O & A _ {2} \\end{array} \\right] \\Rightarrow A ^ {- 1} = \\left[ \\begin{array}{c c} A _ {1} ^ {- 1} & - A _ {1} ^ {- 1} A _ {3} A _ {2} ^ {- 1} \\\\ 0 & A _ {2} ^ {- 1} \\end{array} \\right]\n\\]"
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            "content": "证明 根据分块矩阵初等变换可知",
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            "content": "\\[\n\\left[ \\begin{array}{c c} E & - A _ {2} ^ {- 1} A _ {3} \\\\ O & E \\end{array} \\right] \\left[ \\begin{array}{c c} A _ {1} & A _ {3} \\\\ O & A _ {2} \\end{array} \\right] = \\left[ \\begin{array}{c c} A _ {1} & O \\\\ O & A _ {2} \\end{array} \\right] \\Rightarrow \\left[ \\begin{array}{c c} E & - A _ {2} ^ {- 1} A _ {3} \\\\ O & E \\end{array} \\right] ^ {- 1} \\left[ \\begin{array}{c c} A _ {1} & A _ {3} \\\\ O & A _ {2} \\end{array} \\right] ^ {- 1} = \\left[ \\begin{array}{c c} A _ {1} ^ {- 1} & O \\\\ O & A _ {2} ^ {- 1} \\end{array} \\right]\n\\]"
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            "content": "因此可以得到结论。",
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            "content": "2.3 矩阵的秩"
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            "content": "定理 2.21 (一般分块矩阵可逆条件)"
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            "content": "分块矩阵 \\(H = \\left( \\begin{array}{ll}A & B\\\\ C & D \\end{array} \\right)\\) 可逆当且仅当 \\(A - BD^{-1}C\\) 和 \\(D - CA^{-1}B\\) 都可逆",
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            "content": "解 先做交换行的变换: \\(\\left[ \\begin{array}{ll} O & A \\\\ B & O \\end{array} \\right]\\left[ \\begin{array}{ll} O & E \\\\ E & O \\end{array} \\right] = \\left[ \\begin{array}{l l} A & \\\\ & B \\end{array} \\right]\\), 两侧求逆即可得到:",
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            "content": "\\[\n\\left[ \\begin{array}{c c} O & A \\\\ B & O \\end{array} \\right] ^ {- 1} = \\left[ \\begin{array}{c c} A ^ {- 1} & O \\\\ O & B ^ {- 1} \\end{array} \\right] \\left[ \\begin{array}{c c} O & E \\\\ E & O \\end{array} \\right] = \\left[ \\begin{array}{c c} O & B ^ {- 1} \\\\ A ^ {- 1} & O \\end{array} \\right]\n\\]"
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            "content": "2.3 矩阵的秩"
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            "content": "2.3.1 秩的概念"
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            "content": "定义 2.9 (矩阵的秩)"
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            "content": "矩阵的秩等于:",
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            "content": "- 矩阵行列向量的秩",
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            "content": "- 最大非零子式的阶数",
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            "content": "设 \\(A\\) 是秩为 \\(r\\) 的 \\(s \\times n\\) 矩阵，则存在 \\(s\\) 阶可逆矩阵 \\(P\\) 和 \\(n\\) 阶可逆矩阵 \\(Q\\) 使得",
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            "content": "\\[\nA = P \\left[ \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right] Q\n\\]"
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            "content": "笔记 如果一个题目中没给出条件，或者只给出了矩阵的秩，则往往需要用到初等变换标准型",
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            "content": "定理 2.23 (行列满秩的标准型)"
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            "content": "\\(A\\) 是 \\(m\\times n\\) 矩阵， \\(A\\) 列满秩当且仅当存在 \\(m\\) 阶可逆矩阵 \\(P\\) 使得 \\(A = P\\left[ \\begin{array}{c}E_{n}\\\\ O \\end{array} \\right]\\) 。 \\(A\\) 行满秩当且仅当存在\\(n\\) 阶可逆矩阵 \\(Q\\) ，使得 \\(A = \\left[ \\begin{array}{ll}E_m & O \\end{array} \\right]Q\\)",
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            "content": "2.3 矩阵的秩"
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            "content": "2.3.3 利用初等变换标准型进行矩阵分解"
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            "content": "定理 2.24 (矩阵分解)"
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            "content": "\\(s\\times n\\) 矩阵 \\(\\left[ \\begin{array}{cc}E_r & O\\\\ O & O \\end{array} \\right]\\) 可分解为",
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            "content": "\\[\n\\left[ \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right] = \\left[ \\begin{array}{c} E _ {r} \\\\ O \\end{array} \\right] \\left[ \\begin{array}{c c} E _ {r} & O \\end{array} \\right]\n\\]"
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            "content": "若为方阵时，可以分解为：",
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            "content": "\\[\n{\\left[ \\begin{array}{l l} E _ {r} & O \\\\ O & O \\end{array} \\right]} = {\\left[ \\begin{array}{l l} E _ {r} & O \\\\ O & O \\end{array} \\right]} {\\left[ \\begin{array}{l l} E _ {r} & O \\\\ O & O \\end{array} \\right]}\n\\]"
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            "content": "练习 2.6 行列满秩分解 (1) 证明：任意非零矩阵都可以分解为列满秩和行满秩矩阵的乘积",
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            "content": "(2) 设 \\(B_{1}, B_{2}\\) 为数域 \\(P\\) 上 \\(s \\times n\\) 列满秩矩阵，证明：存在 \\(P\\) 上 \\(s\\) 阶可逆矩阵 \\(C\\)，使得 \\(B_{2} = CB_{1}\\) (即 \\(B_{1}, B_{2}\\) 只需要行变换就可以相互转换)",
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            "content": "(3) \\( C_{1}, C_{2} \\) 为P上 \\( s \\times n \\) 行满秩矩阵，证明： \\( C_{1}, C_{2} \\) 可以仅通过列变换相互转换。",
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            "content": "证明 (1) 使用等价标准型:",
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            "content": "\\[\nA = P \\left[ \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right] Q = P \\left[ \\begin{array}{c} E _ {r} \\\\ O \\end{array} \\right] \\left[ \\begin{array}{c c} E _ {r} & O \\end{array} \\right] Q\n\\]"
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            "content": "(2) 根据列满秩初等标准型，\\(\\exists P\\) 使得 \\(PB_{1} = \\left[ \\begin{array}{c} E_{n} \\\\ O \\end{array} \\right]\\)，同理存在 \\(Q\\) 使得 \\(QB_{2} = \\left[ \\begin{array}{c} E_{n} \\\\ O \\end{array} \\right]\\)，结论成立。",
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            "content": "练习 2.7 幂等矩阵分解 (1) 证明：任意方阵可以分解为可逆矩阵与幂等矩阵乘积",
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            "content": "证明 (1) 利用等价标准型得到:",
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            "content": "\\[\nA = P \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) Q = P Q Q ^ {- 1} \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) Q\n\\]"
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            "content": "PQ 为可逆矩阵，后面为幂等",
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            "content": "引理 2.1 (乘可逆矩阵不改变秩)"
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            "content": "\\(P, Q\\) 为可逆矩阵，则 \\(r(A) = r(PAQ)\\)",
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            "content": "证明 相当于做了一系列初等变换，因此不改变秩",
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            "content": "练习 2.8 非方阵情况 A, B 分别为  \\( m \\times n, n \\times s \\)  矩阵，证明：",
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            "content": "2.3 矩阵的秩"
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            "content": "引理 2.2 (分块矩阵的秩)"
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            "content": "\\[\n\\begin{array}{l} \\bullet r \\left( \\begin{array}{c c} A & O \\\\ O & B \\end{array} \\right) = r (A) + r (B) \\\\ \\bullet r \\left( \\begin{array}{c c} A & O \\\\ C & B \\end{array} \\right) \\geq r (A) + r (B) \\end{array}\n\\]"
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            "content": "证明 (1) 根据初等变换等价标准型有 \\(P_{1}AQ_{1} = \\left( \\begin{array}{cc}E_{r_1} & O\\\\ O & O \\end{array} \\right),P_{2}BQ_{2} = \\left( \\begin{array}{cc}E_{r_2} & O\\\\ O & O \\end{array} \\right)\\)，因此",
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            "content": "\\[\n\\left( \\begin{array}{c c} P _ {1} & O \\\\ O & P _ {2} \\end{array} \\right) \\left( \\begin{array}{c c} A & O \\\\ O & B \\end{array} \\right) \\left( \\begin{array}{c c} Q _ {1} & O \\\\ O & Q _ {2} \\end{array} \\right) = \\left( \\begin{array}{c c c c} E _ {r _ {1}} & O & O & O \\\\ O & O & O & O \\\\ O & O & E _ {r _ {2}} & O \\\\ O & O & O & O \\end{array} \\right)\n\\]"
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            "content": "(2) 根据初等变换等价标准型有 \\(P_{1}AQ_{1} = \\left( \\begin{array}{cc}E_{r_1} & O\\\\ O & O \\end{array} \\right), P_{2}BQ_{2} = \\left( \\begin{array}{cc}E_{r_2} & O\\\\ O & O \\end{array} \\right)\\)，因此",
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            "content": "\\[\n\\left( \\begin{array}{c c} P _ {1} & O \\\\ O & P _ {2} \\end{array} \\right) \\left( \\begin{array}{c c} A & O \\\\ C & B \\end{array} \\right) \\left( \\begin{array}{c c} Q _ {1} & O \\\\ O & Q _ {2} \\end{array} \\right) = \\left( \\begin{array}{c c c c} E _ {r _ {1}} & O & O & O \\\\ O & O & O & O \\\\ C _ {1 1} & C _ {1 2} & E _ {r _ {2}} & O \\\\ C _ {2 1} & C _ {2 2} & O & O \\end{array} \\right) \\to \\left( \\begin{array}{c c c c} E _ {r _ {1}} & O & O & O \\\\ O & O & O & O \\\\ O & O & E _ {r _ {2}} & O \\\\ O & C _ {2 2} & O & O \\end{array} \\right)\n\\]"
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            "content": "定理 2.25 (矩阵秩基本不等式)"
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            "content": "•  \\( r(AB) \\leq \\min\\{r(A), r(B)\\} \\)",
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            "content": "•  \\( r(A+B) \\leq r(A) + r(B) \\)",
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            "content": "- Sylvester 不等式: \\(A, B\\) 是 \\(n\\) 阶方阵, 则 \\(r(A) + r(B) \\leq r(AB) + n\\)",
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            "content": "- Frobenius 不等式: \\(A, B, C\\) 是 \\(n\\) 阶方阵, 则 \\(r(ABC) \\geq r(AB) + r(BC) - r(B)\\)",
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            "content": "- \\(A, B\\) 为 \\(s \\times n, n \\times m\\) 阶矩阵，\\(AB = O\\)，则 \\(r(A) + r(B) \\neq n\\)",
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            "content": "- \\(A\\) 是 \\(m \\times n\\) 的实矩阵，则 \\(r(A^T A) = r(A)\\)",
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            "content": "证明 (1) 将 \\(B\\) 写为行向量形式: \\(B = \\left( \\begin{array}{c} \\beta_{1} \\\\ \\beta_{2} \\\\ \\vdots \\\\ \\beta_{n} \\end{array} \\right)\\), 则",
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            "content": "\\[\nA B = \\left( \\begin{array}{c c c} a _ {1 1} & \\dots & a _ {1 n} \\\\ \\vdots & & \\vdots \\\\ a _ {n 1} & \\dots & a _ {n n} \\end{array} \\right) \\left( \\begin{array}{c} \\beta_ {1} \\\\ \\vdots \\\\ \\beta_ {n} \\end{array} \\right) = \\left( \\begin{array}{c} a _ {1 1} \\beta_ {1} + \\dots + a _ {1 n} \\beta_ {n} \\\\ \\vdots \\\\ a _ {n 1} \\beta_ {1} + \\dots + a _ {n n} \\beta_ {n} \\end{array} \\right)\n\\]"
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            "content": "因此 AB 每个行向量都可被 B 行向量线性表出，从而  \\( r(AB) \\leq r(B) \\) ，同理可知  \\( r(AB) \\leq r(A) \\)",
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            "content": "(2) 用列向量角度， \\( A + B \\) 的列向量显然可以被 \\( A, B \\) 的列向量线性表出，从而结论成立。",
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            "content": "(3) 是 (4) 的推论",
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            "content": "(4)Frobenius: 即构造  \\( r(ABC) + r(B) \\geq r(AB) + r(BC) \\)",
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            "content": "\\[\n\\left[ \\begin{array}{c c} A B C & \\\\ & B \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A B C & O \\\\ B C & B \\end{array} \\right] \\to \\left[ \\begin{array}{c c} O & - A B \\\\ B C & B \\end{array} \\right]\n\\]"
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            "content": "根据分块矩阵秩不等式可知",
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            "content": "练习 2.9 秩不等式的推广 (1) A, B 为 n 阶方阵，则  \\( r(A - ABA) = r(A) + r(E - BA) - n \\)",
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            "content": "2.3 矩阵的秩"
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            "content": "(2) \\(A, B\\) 是 \\(n\\) 阶方阵，\\(AB = BA = O, r(A^2) = r(A)\\)，证明：\\(r(A + B) = r(A) + r(B)\\) 证明 (1) 等号需要构造单位矩阵，即等式中的 \\(n\\) 用 \\(E_n\\) 代替：",
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & \\\\ & E _ {n} - B A \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A & O \\\\ B A & E _ {n} - B A \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A & A \\\\ B A & E _ {n} \\end{array} \\right] \\to \\left[ \\begin{array}{c c} A - A B A & O \\\\ O & E _ {n} \\end{array} \\right]\n\\]"
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            "content": "从右往左推非常难，因此一般从无 \\(E_{n}\\) 的一边推另一边(后面的练习也会体现这一点)",
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            "content": "(2) 注意从复杂的一侧 \\(A + B\\) 开始，否则 \\(A, B\\) 不知道怎么搞，",
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            "content": "\\[\n\\left( \\begin{array}{c c} A + B & O \\\\ O & O \\end{array} \\right) \\to \\left( \\begin{array}{c c} A + B & O \\\\ A ^ {2} + A B & O \\end{array} \\right) = \\left( \\begin{array}{c c} A + B & O \\\\ A ^ {2} & O \\end{array} \\right) \\to \\left( \\begin{array}{c c} A + B & A ^ {2} \\\\ A ^ {2} & A ^ {3} \\end{array} \\right)\n\\]"
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            "content": "由于 \\(r(A) = r(A^2)\\)，因此矩阵方程 \\(A^2 X = A\\) 有解，故 \\(\\exists P\\) 使得 \\(A^2 P = A\\)，故第二列可右乘 \\(P\\) 消去第一列：",
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            "content": "\\[\n\\left( \\begin{array}{c c} A + B & A ^ {2} \\\\ A ^ {2} & A ^ {3} \\end{array} \\right) \\to \\left( \\begin{array}{c c} B & A ^ {2} \\\\ O & A ^ {3} \\end{array} \\right)\n\\]"
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            "content": "因此根据分块矩阵的秩得到 \\(r(A + B) \\geq r(A^3) + r(B) = r(A) + r(B)\\)，而不等式另一侧显然.",
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            "content": "练习 2.10 用秩不等式研究矩阵性质 (1) 证明零矩阵:  \\( A_{m \\times n}, B_{n \\times s}, r(B) = n \\) ，证明：若 AB = O，则 A = O",
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            "content": "(2) 判定可逆: \\(n\\) 为奇数, \\(A, B\\) 为 \\(n\\) 阶矩阵, \\(A^2 = O\\), 证明: \\(AB - BA\\) 不可逆证明 (1) 由于 \\(r(A) + r(B) \\leq r(AB) + n\\), 而 \\(r(B) = n\\), 得到 \\(r(A) = 0\\), 即 \\(A = O\\)",
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            "content": "(2) \\( n=2k+1 \\) ，根据 \\( A^{2}=O \\) 得到 \\( 2r(A)\\leq2k+1+0 \\) ，从而 \\( r(A)\\leq k \\) ，此时",
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            "content": "\\[\nr (A B - B A) \\leq r (A B) + r (B A) \\leq r (A) + r (A) \\leq 2 k\n\\]"
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            "content": "因此不可逆",
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            "content": "2.3.5 空间分解法研究矩阵秩"
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            "content": "定理 2.26 (多项式矩阵的秩关系)"
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            "content": "\\(f(x) = f_{1}(x)f_{2}(x)\\) 是多项式， \\((f_{1}(x),f_{2}(x)) = 1\\) ，则 \\(f(A) = O\\) 当且仅当 \\(r(f_{1}(A)) + r(f_{2}(A)) = n\\) 。常见的推论如下：",
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            "content": "- \\(A^2 = A\\) 当且仅当 \\(r(A - E) + r(A) = n\\)",
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            "content": "- \\(A^2 = E\\) 当且仅当 \\(r(A - E) + r(A + E) = n\\)",
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            "content": "\\[\n\\left[ \\begin{array}{c c} f _ {1} (A) & \\\\ & f _ {2} (A) \\end{array} \\right] \\to \\left[ \\begin{array}{c c} f _ {1} (A) & f _ {1} (A) u (A) + v (A) f _ {2} (A) \\\\ O & f _ {2} (A) \\end{array} \\right] = \\left[ \\begin{array}{c c} f _ {1} (A) & E _ {n} \\\\ O & f _ {2} (A) \\end{array} \\right]\n\\]"
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            "content": "因此得到  \\( r(f_{1}(A))+r(f_{2}(A))=n+r(f(A)) \\)",
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            "content": "2.4 矩阵的迹"
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            "content": "因此推出：",
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            "content": "\\[\n(n - r (A)) + (n - r (B)) \\leq (n - r (A B)) + (n - r (A + B))\n\\]"
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            "content": "化简得到结论。",
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            "content": "笔记 这种 \\(r(A), r(B), r(A + B), r(AB)\\) 相关问题绝大多数从线性空间角度都能迎刃而解",
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            "content": "\\[\n\\begin{array}{l} \\bullet \\operatorname{tr} (A) = \\operatorname{tr} (k A) \\\\ \\bullet \\operatorname{tr} (A + B) = \\operatorname{tr} (A) + \\operatorname{tr} (B) \\\\ \\bullet \\operatorname{tr} (A B) = \\operatorname{tr} (B A) \\end{array}\n\\]"
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            "content": "若 \\(A\\) 为实方阵，则 \\(A = O\\) 当且仅当 \\(\\operatorname{tr}(A^T A) = 0\\)",
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            "content": "单位矩阵通过一次初等变换得到的矩阵称为初等矩阵。记作 \\(P(i,j), P(i(c)), P(i,j(k))\\) 表示 \\(i,j\\) 行互换，\\(i\\) 行乘 \\(c\\)，\\(P_{ij} = k\\) 的矩阵",
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            "content": "2.5 几种技巧矩阵"
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            "content": "练习2.12用基础矩阵研究幂次(1)设 \\(A = \\left( \\begin{array}{cccc}0 & 1 & 0 & \\dots & 0\\\\ 0 & 0 & 1 & \\dots & 0\\\\ \\vdots & \\vdots & \\vdots & & \\vdots \\\\ 0 & 0 & 0 & \\dots & 1\\\\ 1 & 0 & 0 & \\dots & 0 \\end{array} \\right)\\) ，证明： \\(A^k = \\left( \\begin{array}{cc}O & E_{n - k}\\\\ E_k & O \\end{array} \\right)\\) 。",
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            "content": "(2) \\(A\\) 是 \\(n\\) 阶上三角阵且主对角上元素均为0，证明 \\(A^n = O\\)",
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            "content": "证明 (1) 写为  \\( A = (e_{n}, e_{1}, \\cdots, e_{n-1}) \\) ，则  \\( A^{2} = (A e_{n}, A e_{1}, \\cdots, A e_{n-1}) \\) 。显然  \\( Ae_{i} \\)  为 A 的第 i 列，因此",
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            "content": "\\[\nA ^ {2} = (e _ {n - 1}, e _ {n}, e _ {1} \\dots , e _ {n - 2})\n\\]"
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            "content": "以此类推容易得到结论",
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            "content": "(2) \\( A=\\sum_{i<j}a_{ij}E_{ij} \\) ，而 \\( j\\neq k \\) 时 \\( E_{ij}E_{kl}=O \\) ，因此 \\( A^{n} \\) 展开式中非零项只能是 \\( E_{ij_{1}}E_{j_{1}j_{2}}\\cdots E_{j_{n-1}j_{n}} \\) 这样的项，但满足 \\( 1\\leq i<j_{1}<\\cdots<j_{n}\\leq n \\) 的项不存在，因此 \\( A^{n}=O \\)",
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            "content": "练习2.13用二次型研究矩阵性质(1)证明： \\(n\\) 阶对称矩阵 \\(A\\) 为零矩阵当且仅当 \\(\\forall \\alpha \\in \\mathbb{R}^n\\) 有 \\(\\alpha^T A\\alpha = 0\\)",
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            "content": "(2) 证明： \\(n\\) 阶方阵 \\(A\\) 是反对称矩阵当且仅当 \\(\\forall \\alpha \\in \\mathbb{R}^n\\) 有 \\(\\alpha^T A\\alpha = 0\\)",
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            "content": "证明 (1) 左推右显然。先设 \\(\\alpha = e_i\\)，则 \\(e_i^T Ae_i = a_{ii} = 0\\)，再取 \\(\\alpha = e_i + e_j\\)，则",
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            "content": "\\[\n0 = (e _ {i} + e _ {j}) ^ {T} A (e _ {i} + e _ {j}) = a _ {i i} + a _ {j j} + a _ {i j} + a _ {j i}\n\\]"
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            "content": "而 \\(a_{ij} = a_{ji}\\)，因此得到 \\(a_{ij} = 0, \\forall i, j\\) (2) 左推右用 \\(\\alpha^T A\\alpha = (\\alpha^T A\\alpha)^T = -\\alpha^T A\\alpha\\) 即可。右推左：根据条件得到 \\(\\alpha^T A^T\\alpha = 0\\)，因此 \\(\\alpha^T (A + A^T)\\alpha = 0\\)，由于 \\(A + A^T\\) 为对称矩阵，根据上题结论得到 \\(A + A^T = O\\)，因此 \\(A^T = -A\\)",
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            "content": "2.5.3 置换矩阵"
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            "content": "每行只有一个元素为 1，每列也只有一个元素为 1，则称该矩阵为置换矩阵。一般写为  \\( P = [\\epsilon_{i_{1}}, \\cdots, \\epsilon_{i_{n}}] = [\\epsilon_{j_{1}}^{T}, \\cdots, \\epsilon_{j_{n}}^{T}]^{T} \\) ，其中  \\( \\epsilon_{i} \\)  表示仅第 i 元素为 1 的列向量， \\( i_{k}, j_{k} \\)  分别表示列、行的重排顺序。",
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            "content": "定理 2.30 (置换矩阵是正交矩阵)"
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            "content": "置换矩阵是正交矩阵。",
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            "content": "证明 证明 \\(P^T P = I\\)"
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            "content": "定理 2.31 (置换矩阵的作用)"
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            "content": "左乘置换矩阵相当于把行原本的  \\( j_{1}, j_{2}, \\cdots, j_{n} \\)  行换到  \\( 1, 2, \\cdots, n \\)  行，右乘相当于把列原本的  \\( i_{1}, \\cdots, i_{n} \\)  列换到  \\( 1, 2, \\cdots, n \\)  列。",
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            "content": "定理 2.32 (置换矩阵正交相似)"
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            "content": "\\( i_{1}\\cdots i_{n} \\) 是 \\( 1\\sim n \\) 的排列， \\( A=(a_{ij}) \\) ，取 \\( P=[\\epsilon_{i_{1}},\\cdots,\\epsilon_{i_{n}}] \\) ，则：",
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            "content": "\\[\nP ^ {T} A P = P ^ {- 1} A P = \\left[ \\begin{array}{c c c c} a _ {i _ {1} i _ {1}} & a _ {i _ {1} i _ {2}} & \\dots & a _ {i _ {1} i _ {n}} \\\\ a _ {i _ {2} i _ {1}} & a _ {i _ {2} i _ {2}} & \\dots & a _ {i _ {2} i _ {n}} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a _ {i _ {n} i _ {1}} & a _ {i _ {n} i _ {2}} & \\dots & a _ {i _ {n} i _ {n}} \\end{array} \\right]\n\\]"
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "练习 2.14 置换矩阵正交相似例子  \\( i_{1}\\cdots i_{n} \\)  是  \\( 1,2,\\cdots,n \\)  的排列， \\( A=\\mathrm{diag}\\{\\lambda_{1},\\cdots,\\lambda_{n}\\},B=\\mathrm{diag}\\{\\lambda_{i_{1}},\\cdots,\\lambda_{i_{n}}\\} \\) ，证明 A,B 是正交相似的",
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        },
        {
            "type": "text",
            "bbox": [
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                0.33,
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            "angle": 0,
            "content": "证明 取  \\( P = [\\epsilon_{i_{1}}, \\cdots, \\epsilon_{i_{n}}] \\)  即可。",
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        {
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            "content": "定义 2.13 (特殊的置换矩阵)"
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            "content": "特殊置换矩阵 \\(K = \\left[ \\begin{array}{cccc}  &  &  & 1 \\\\  &  & 1 &  \\\\  & \\ddots &  &  \\\\ 1 &  &  &  \\\\ \\end{array} \\right]\\)，\\(K^T AK\\) 的功能是绕中心转180度(行列全部倒排)",
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        {
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            "angle": 0,
            "content": "2.6 特殊矩阵性质总结"
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            "content": "2.6.1 对角矩阵"
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            "content": "定理 2.33 (对角矩阵乘法)"
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            "content": "矩阵 \\(A\\) 右乘一个对角阵，则每列乘上对角相应元素。\\(A\\) 左乘对角阵，则每行乘上对角相应元素。",
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            "content": "定理 2.34 (对角矩阵求逆)",
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            "content": "\\[\n\\operatorname{diag} \\left\\{d _ {1}, \\dots , d _ {n} \\right\\} ^ {- 1} = \\operatorname{diag} \\left\\{d _ {1} ^ {- 1}, \\dots , d _ {n} ^ {- 1} \\right\\}\n\\]"
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            "content": "2.6.2 上、下三角矩阵"
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            "content": "定理 2.35 (上下三角乘法、逆、伴随)"
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            "angle": 0,
            "content": "\\(A, B\\) 是上三角矩阵，则",
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            "angle": 0,
            "content": "- \\(AB\\) 仍然是上三角矩阵，且 \\(AB\\) 对角元为 \\(a_{11}b_{11},\\dots ,a_{nn}b_{nn}\\)",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(A^{-1}\\) 也是上三角阵，且 \\(A^{-1}\\) 对角元为 \\(a_{11}^{-1}, \\cdots, a_{nn}^{-1}\\)",
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            "content": "- \\(A^{*}\\) 也是上三角阵，且 \\(A^{*}\\) 的对角元为 \\(a_{22}a_{33}\\cdots a_{nn}, a_{11}a_{33}\\cdots a_{nn}, \\cdots, a_{11}a_{22}\\cdots a_{n-1,n-1}\\) (也是特征值)",
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        },
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            "content": "证明 (1) 令 \\(C = AB\\)，若 \\(c_{ij} = \\sum_{k=1}^{n} a_{ik} b_{kj} = \\sum_{k=1}^{i-1} a_{ik} b_{kj} + \\sum_{k=i}^{n} a_{ik} b_{kj}\\)，当 \\(i > j\\) 时，显然 \\(c_{ij} = 0\\)。而 \\(c_{ii} = \\sum_{k=1}^{i-1} a_{ik} b_{ki} + a_{ii} b_{ii} + \\sum_{k=i+1}^{n} a_{ik} b_{ki} = a_{ii} b_{ii}\\)",
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            "content": "(2) 根据行变换可以将 \\((A, I) \\to (I, A^{-1})\\)，当 \\(A\\) 为上三角时，先用最后一行消去 \\(A\\) 最后一列其余行的元素，以此类推可以发现 \\(A^{-1}\\) 为上三角，根据矩阵乘法显然有 \\(A^{-1}\\) 对角为 \\(A\\) 对角的逆。",
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            "content": "(3) 根据 \\(A^{*} = |A|A^{-1}\\) 即可",
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            "content": "27"
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            "content": "定理 2.36 (n 个上三角矩阵的乘积)"
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            "content": "\\( A_{1},\\cdots,A_{n} \\)  时 n 个对角为 0 的 n 阶上三角矩阵，证明  \\( A_{1}A_{2}\\cdots A_{n}=O \\)",
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            "content": "证明 用数学归纳法，设 \\(n - 1\\) 时成立，取 \\(A_{n} = \\left[ \\begin{array}{cc}B_{n - 1} & *\\\\ 0 & 0 \\end{array} \\right]\\) ，此时",
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            "content": "\\[\nA _ {1} \\dots A _ {n} = \\left[ \\begin{array}{c c} B _ {1} \\dots B _ {n - 1} & * \\\\ 0 & 0 \\end{array} \\right] \\cdot \\left[ \\begin{array}{c c} B _ {n} & * \\\\ 0 & 0 \\end{array} \\right] = \\left[ \\begin{array}{c c} 0 & * \\\\ 0 & 0 \\end{array} \\right] \\left[ \\begin{array}{c c} B _ {n} & * \\\\ 0 & 0 \\end{array} \\right] = O\n\\]"
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            "content": "28"
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "2.6.3 实对称矩阵"
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            "content": "定义 2.14 (实对称矩阵)"
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            "angle": 0,
            "content": "每个元素都是实数，且满足 \\(A^T = A\\) 的矩阵 \\(A\\) 称为实对称矩阵",
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        },
        {
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            "content": "定理 2.37 (实对称矩阵的矩阵性质)"
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            "angle": 0,
            "content": "实对称矩阵 A 具有以下性质：",
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            "type": "text",
            "bbox": [
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            "content": "- \\(A^{-1}\\) 也是实对称矩阵",
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        },
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            "bbox": [
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            "content": "证明 (1) 由于 \\((AA^{-1})^T = E_n\\)，得到 \\((A^{-1})^T A^T = E_n\\)，而 \\(A\\) 实对称，因此 \\((A^{-1})^T A = E_n\\)，这说明 \\((A^{-1})^T = A^{-1}\\)",
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        {
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            "content": "定理 2.38 (实对称矩阵的特征值性质)"
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            "angle": 0,
            "content": "实对称矩阵有如下性质：",
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        },
        {
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            "content": "- 每个特征值都是实数",
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        },
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            "type": "text",
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            "angle": 0,
            "content": "- 不同特征值的特征向量正交",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- 一定正交相似 (相似且合同) 于对角矩阵",
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        {
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            "content": "证明 (1) 共轭转置法：设 \\(\\lambda \\in \\mathbb{C}\\) 是特征值，\\(\\xi \\in \\mathbb{C}^n\\) 是特征向量，一方面根据 \\(A\\xi = \\lambda\\xi\\) 两侧同时左乘 \\(\\overline{\\xi}^T\\) 得到",
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            "angle": 0,
            "content": "\\[\n\\overline {{\\xi}} ^ {T} A \\xi = \\lambda \\overline {{\\xi}} ^ {T} \\xi\n\\]"
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "另一方面直接对 \\(A\\xi = \\lambda \\xi\\) 取共轭转置，再同时右乘 \\(\\xi\\) 得到",
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        },
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            "bbox": [
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            "content": "\\[\n\\overline {{\\xi}} ^ {T} A = \\overline {{\\xi}} ^ {T} \\overline {{\\lambda}} \\Rightarrow \\overline {{\\xi}} ^ {T} A \\xi = \\overline {{\\lambda}} \\overline {{\\xi}} ^ {T} \\xi\n\\]"
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            "angle": 0,
            "content": "因此比较得到 \\(\\lambda = \\overline{\\lambda}\\)",
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            "type": "text",
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            "content": "(2) 不同特征值的特征向量正交：设 \\(\\lambda_{1}, \\lambda_{2}\\) 是两个互异特征值，\\(\\xi_{1}, \\xi_{2}\\) 分别属于 \\(\\lambda_{1}, \\lambda_{2}\\) 的实特征向量，则",
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            "angle": 0,
            "content": "\\[\n\\lambda_ {1} (\\xi_ {1}, \\xi_ {2}) = (\\lambda_ {1} \\xi_ {1}) ^ {T} \\xi_ {2} = (A \\xi_ {1}) ^ {T} \\xi_ {2} = \\xi_ {1} ^ {T} (A \\xi_ {2}) = \\lambda_ {2} (\\xi_ {1}, \\xi_ {2})\n\\]"
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            "content": "从而得到 \\((\\lambda_{2} - \\lambda_{1})(\\xi_{1},\\xi_{2}) = 0\\) ，而 \\(\\lambda_1\\neq \\lambda_2\\) ，从而 \\((\\xi_1,\\xi_2) = 0\\)",
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            "content": "(3) 做数学归纳法，n=1 显然，假设 n-1 成立，任取特征值  \\( \\lambda \\) ，对应单位特征向量  \\( \\alpha_{1} \\) ，将  \\( \\alpha_{1} \\)  扩充为  \\( R^{n} \\)  的一组标准正交基  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) ，令  \\( T=(\\alpha_{1},\\cdots,\\alpha_{n}) \\) ，则",
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            "content": "\\[\nA T = T \\left( \\begin{array}{c c} \\lambda & \\alpha^ {T} \\\\ 0 & A _ {n - 1} \\end{array} \\right) \\Rightarrow T ^ {T} A T = \\left( \\begin{array}{c c} \\lambda & \\alpha^ {T} \\\\ 0 & A _ {n - 1} \\end{array} \\right)\n\\]"
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            "content": "由于 \\(T^T A T\\) 对称，因此 \\(\\alpha^T = 0\\) ，再根据归纳假设，存在正交矩阵 \\(P_{1}\\) 使得 \\(P_{1}^{T}A_{n - 1}P_{1}\\) 为对角阵，因此令 \\(P = \\mathrm{diag}(1,P_1)\\) ，此时",
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            "content": "\\[\nP ^ {T} T ^ {T} A T P = \\left( \\begin{array}{c c} \\lambda & 0 \\\\ 0 & P ^ {T} A _ {n - 1} P \\end{array} \\right)\n\\]"
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            "angle": 0,
            "content": "显然 \\(TP = Q\\) ，则 \\(Q\\) 也是正交阵。",
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            "content": "2.6.4 反对称矩阵"
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            "content": "定理 2.39 (反对称矩阵的行列式)"
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            "content": "奇数阶反对称矩阵的行列式必为 0。",
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            "content": "证明 根据 \\(A^T = -A\\) 得到 \\(|A| = (-1)^n |A|\\)，当 \\(n\\) 为奇数时 \\(|A| = -|A|\\)，因此 \\(|A| = 0\\)",
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "定理 2.40 (反对称矩阵的矩阵性质)"
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            "content": "\\(A\\) 为反对称矩阵，则",
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            "content": "- \\(A^{-1}\\) 为反对称矩阵",
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            "content": "证明 (1) 类似实对称矩阵的"
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            "content": "定理 2.41 (反对称矩阵的二次型)"
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            "content": "\\(A\\) 为反对称矩阵当且仅当任意 \\(X \\in \\mathbb{R}^n\\)，有 \\(X^T A X = 0\\)",
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            "content": "证明 在基础矩阵技巧一节中给出过证明"
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            "content": "定理 2.42 (反对称矩阵的特征值、特征向量)"
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            "content": "反对称矩阵 A 的特征值是 0 或者纯虚数。且纯虚数对应特征向量的实部和虚部的实向量等长、正交。",
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            "content": "证明 (1) 设 \\(A\\) 的特征值 \\(\\lambda = a + bi\\)，对应特征向量 \\(x = u + vi\\)，\\(A(u + vi) = (a + bi)(u + vi)\\)，因此 \\(Au + iAv = (au - bv) + (bu + av)i\\)，实虚部分离得到：",
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            "content": "\\[\nA u = a u - b v, A v = b u + a v \\Rightarrow u ^ {T} A u = a u ^ {T} u - b u ^ {T} v, v ^ {T} A v = b v ^ {T} u + a v ^ {T} v\n\\]"
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            "content": "因此得到 \\(u^T Au + v^T Av = a(|u|^2 + |v|^2)\\)，由于 \\(u^T Au \\in \\mathbb{C}\\) 是一个数，因此",
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            "content": "\\[\nu ^ {T} A u = (u ^ {T} A u) ^ {T} = u ^ {T} A ^ {T} u = - u ^ {T} A u \\Rightarrow u ^ {T} A u = 0\n\\]"
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            "content": "同理有 \\(v^{T}Av = 0\\) ，因此 \\(a(|u|^2 +|v|^2) = 0\\) ，由于 \\(u + vi\\neq 0\\) ，得到 \\(a = 0\\) ，即 \\(\\lambda = a + bi = bi\\)",
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            "content": "(2) 设 \\(\\alpha + \\beta i\\) 是特征值 \\(ki\\) 的特征向量，即 \\(A(\\alpha + i\\beta) = ki(\\alpha + i\\beta)\\)，对比实部虚部，得到",
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            "content": "\\[\nA \\alpha = - k \\beta , A \\beta = k \\alpha\n\\]"
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            "content": "由于 \\(\\alpha^T A\\alpha = 0\\) ，得到 \\(\\alpha^T A\\alpha = \\alpha^T (-k\\beta) = -k\\alpha^T\\beta = 0\\) ，这说明 \\(\\alpha, \\beta\\) 正交，再由于",
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            "content": "\\[\n\\alpha^ {T} A \\beta + \\beta^ {T} A \\alpha = \\left\\{ \\begin{array}{l} \\alpha^ {T} A \\beta - (\\alpha^ {T} A \\beta) ^ {T} = 0 \\\\ k (\\alpha^ {T} \\alpha - \\beta^ {T} \\beta) \\end{array} \\right.\n\\]"
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            "content": "得到 \\(\\alpha^T\\alpha -\\beta^T\\beta = 0\\) ，即模长相等",
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            "content": "- \\(A\\) 正交相似于如下准对角阵",
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            "type": "header",
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            "angle": 0,
            "content": "2.6 特殊矩阵性质总结"
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            "content": "2.6.5 Hermite 矩阵"
        },
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            "angle": 0,
            "content": "定义 2.15 (Hermite 矩阵)",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "\\(\\overline{A}^T = A\\) 的矩阵称为Hermite矩阵",
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        },
        {
            "type": "text",
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            "content": "定理 2.45 (Hermite 矩阵的性质)",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "Hermite 矩阵的特征值都是实数",
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        },
        {
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            "bbox": [
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            "content": "2.6.6 正交矩阵"
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            "angle": 0,
            "content": "定义 2.16 (正交矩阵)",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "若矩阵 \\(T\\) 满足 \\(TT^{T} = I\\) ，则称 \\(T\\) 为正交矩阵。",
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            "content": "定理 2.46 (正交矩阵的性质)"
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            "content": "\\(A, B\\) 为 \\(n\\) 阶正交矩阵，则",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- 求逆: \\(A^{-1} = A^{T}\\)",
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            "type": "text",
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            "angle": 0,
            "content": "- \\(A^{-1}, A^T, AB\\) 都是正交矩阵",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(S = \\left[ \\begin{array}{ll}A & 0\\\\ 0 & B \\end{array} \\right]\\) 也是正交矩阵",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(A\\) 正交当且仅当 \\(A\\) 的列向量是 \\(\\mathbb{R}^n\\) 的一组标准正交基。",
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            "content": "证明 (1) 显然",
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            "angle": 0,
            "content": "(2) \\((AB)^T AB = I\\)",
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            "angle": 0,
            "content": "(3) 直接用 \\(S^T S\\) 验证即可",
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            "type": "page_number",
            "bbox": [
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            "content": "30"
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            "type": "text",
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            "angle": 0,
            "content": "(4) \\( A=\\left[\\alpha_{1},\\cdots,\\alpha_{n}\\right] \\) ，计算 \\( A^{T}A=I \\) 可得出结论。",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "练习 2.15 正交矩阵练习 b 是单位列向量，证明存在 n 阶对称正交矩阵 A，使得 b 为 A 的第一列证明",
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            "content": "定理 2.47 (分块上三角正交矩阵)"
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "若 \\(A = \\left[ \\begin{array}{ll}A_{1} & A_{2}\\\\ 0 & A_{4} \\end{array} \\right]\\) 是正交矩阵，则 \\(A_{2} = 0\\) ， \\(A_{1},A_{4}\\) 是正交矩阵",
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            "bbox": [
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            "angle": 0,
            "content": "证明 根据分块矩阵转置以及乘法可以得到结论",
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            "content": "\\[\nA ^ {T} A = \\left[ \\begin{array}{c c} A _ {1} ^ {T} & 0 \\\\ A _ {2} ^ {T} & A _ {4} ^ {T} \\end{array} \\right] \\left[ \\begin{array}{c c} A _ {1} & A _ {2} \\\\ 0 & A _ {4} \\end{array} \\right]\n\\]"
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            "content": "定理 2.48 (正交矩阵的行列式与特征值)"
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            "content": "\\(A\\) 是 \\(n\\) 阶正交矩阵，则 \\(A\\) 的特征值模长为1，复特征值成对出现，\\(|A| = \\pm 1\\) ，进一步地",
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "证明 (1) 主命题: \\(A\\alpha = \\lambda \\alpha\\)，两侧取共轭得到 \\(A\\overline{\\alpha} = \\overline{\\lambda}\\overline{\\alpha}\\)，两侧取转置得到 \\(\\overline{\\alpha}^T A^T = \\overline{\\lambda}\\overline{\\alpha}^T\\)，两侧右乘 \\(\\alpha\\) 得到 \\(\\overline{\\alpha}^T A^{-1}\\alpha = \\overline{\\lambda}\\overline{\\alpha}^T\\alpha\\)，因此 \\(\\frac{1}{\\lambda}\\overline{\\alpha}^T\\alpha = \\overline{\\lambda}\\overline{\\alpha}^T\\alpha\\)，因此 \\(\\lambda \\overline{\\lambda} = 1\\)，即模长为 1。根据 \\(|A^T A| = 1\\) 可知行列式为 \\(\\pm 1\\)",
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            "content": "(2) \\( |A|=-1 \\)  情况：根据  \\( |A| \\)  是特征值乘积可知至少有 1 个 -1",
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            "content": "(3) \\( |A|=1 \\) 情况：同理",
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            "content": "笔记在计算 \\(A\\) 的特征值时，如果 \\(A\\) 的特殊性需要 \\(A^T\\) 来体现，则经常两侧取共轭转置(例如实对称、反对称、正交矩阵)。",
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            "content": "定理 2.49 (n 阶正交矩阵的标准型)"
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            "content": "\\(A\\) 是 \\(n\\) 阶正交矩阵，则 \\(A\\) 总是正交相似于",
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            "content": "\\[\n\\operatorname{diag} \\left\\{\\lambda_ {1}, \\dots , \\lambda_ {r}, \\left( \\begin{array}{c c} \\cos \\theta_ {1} & - \\sin \\theta_ {1} \\\\ \\sin \\theta_ {1} & \\cos \\theta_ {1} \\end{array} , \\dots , \\left( \\begin{array}{c c} \\cos \\theta_ {m} & - \\sin \\theta_ {m} \\\\ \\sin \\theta_ {m} & \\cos \\theta_ {m} \\end{array} \\right)\\right) \\right\\}\n\\]"
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            "content": "2.6.7 Unitary 矩阵"
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            "content": "定理 2.50 (Unitary 矩阵的性质)",
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            "content": "Unitary 矩阵的特征值模长均为 1",
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            "content": "2.6.8 幂零矩阵"
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            "content": "若存在整数 l，使得  \\( A^{l}=O \\) ，则称 A 为幂零矩阵。",
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            "content": "定理 2.51 (幂零矩阵充要条件)"
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            "content": "\\(A\\) 是幂零矩阵当且仅当"
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            "content": "1. 特征值: \\(A\\) 的特征值只有 0",
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            "content": "2. 特征多项式: \\(A\\) 的特征多项式为 \\(x^{n}\\)",
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            "content": "3. 最小多项式： \\(A\\) 的最小多项式为 \\(x^{k}, k \\leq n\\)",
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            "content": "4. 迹 (重点): 对  \\( \\forall k > 0 \\) ,  \\( \\operatorname{tr}(A^{k}) = 0 \\)",
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            "content": "2.6.9 幂等矩阵"
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            "content": "定义 2.18 (幂等矩阵)"
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            "content": "若 \\(n\\) 阶矩阵 \\(A\\) 满足 \\(A^2 = A\\) ，则称 \\(A\\) 为幂等矩阵",
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            "content": "定理 2.52 (幂等矩阵充要条件)"
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            "angle": 0,
            "content": "\\(n\\) 阶方阵 \\(A\\) 是幂等矩阵当且仅当 \\(r(A) + r(A - E) = n\\)",
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            "content": "证明 由于 \\(A^2 = A\\)，即 \\(A(A - E) = O\\)，根据空间分解研究矩阵秩定理（具体见线性空间-直和与空间分解）",
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            "content": "31"
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            "content": "\\[\nn = n - r (A) + n - r (A - E)\n\\]"
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            "content": "因此 \\(r(A) + r(A - E) = n\\)",
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "定理 2.53 (幂等矩阵的性质)"
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            "content": "\\(A\\) 为 \\(n\\) 阶幂等矩阵，则",
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "- \\(A\\) 的特征值只有1和0",
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            "angle": 0,
            "content": "- \\(A\\) 一定可以对角化，且标准型为 \\(\\mathrm{diag}\\{E_r, O\\}\\)",
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            "content": "\\[\nr (A) = \\operatorname{tr} (A)\n\\]"
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            "content": "证明 (1) 设 \\(\\lambda\\) 是 \\(A\\) 的特征值，故 \\(A\\alpha = \\lambda \\alpha\\) ，\\(A^2\\alpha = \\lambda^2\\alpha\\) ，由于 \\(A = A^2\\) ，得到 \\(\\lambda^2 = \\lambda\\) ，因此 \\(\\lambda = 0,1\\)",
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            "content": "(2) 根据最小多项式理论， \\( f(A)=A(A-E)=O \\) ，因此最小多项式只有一次项乘积，故可对角化。",
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            "content": "(3) 根据秩、迹在相似下的不变性可知",
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            "content": "定理 2.54 (加减乘法的幂等性)"
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            "content": "\\(A, B\\) 为两个 \\(n\\) 阶幂等矩阵，则",
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            "content": "1. \\(A + B\\) 为幂等矩阵当且仅当 \\(AB = BA = O\\)",
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            "angle": 0,
            "content": "2. \\(A - B\\) 为幂等矩阵当且仅当 \\(AB = BA = B\\)",
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            "content": "3. 若 \\(AB = BA\\)，则 \\(AB\\) 为幂等矩阵。反之不一定",
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            "content": "4. 如 E-A-B 可逆，则  \\( r(A)=r(B) \\)",
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            "content": "证明 (1) 右推左：由于 \\(AB = BA = O\\)，因此 \\((A + B)^2 = A^2 + AB + BA + B^2 = A + B\\)。左推右：\\((A + B)^2 = A + AB + BA + B = A + B\\)，因此 \\(AB + BA = O\\)，得到",
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            "content": "\\[\nA B = A B ^ {2} = (- B A) B = - B A B = B (B A) = B A\n\\]"
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            "content": "练习 2.16 幂等矩阵分解 (1) 证明：任意方阵都可以分解为可逆矩阵与幂等矩阵的乘积",
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            "content": "证明 (1) 利用等价标准型得到:",
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            "content": "\\[\nA = P \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) Q = P Q Q ^ {- 1} \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) Q\n\\]"
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            "content": "PQ 为可逆矩阵，后面为幂等",
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            "content": "定理 2.55 (幂等矩阵列的性质)"
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            "content": "\\( A_{1},\\cdots,A_{m} \\)  为 n 阶方阵，且  \\( A_{1}+A_{2}+\\cdots+A_{m}=E_{n} \\) ，则下面三个命题等价：",
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            "content": "- \\(A_{1}, \\cdots, A_{m}\\) 为幂等矩阵",
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            "content": "\\[\n\\bullet \\mathrm{r} (A _ {1}) + \\dots + \\mathrm{r} (A _ {m}) = n\n\\]"
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            "content": "- \\(\\forall i \\neq j\\) 有 \\(A_{i}A_{j} = 0\\)",
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            "content": "证明 \\((1) \\Rightarrow (2)\\): 由于 \\(r(A_{i}) = \\operatorname{tr}(A_{i})\\), 因此",
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            "content": "\\[\nr \\left(A _ {1}\\right) + \\dots + r \\left(A _ {m}\\right) = \\operatorname{tr} \\left(A _ {1}\\right) + \\dots + \\operatorname{tr} \\left(A _ {m}\\right) = \\operatorname{tr} \\left(A _ {1} + \\dots + A _ {m}\\right) = \\operatorname{tr} \\left(E _ {n}\\right) = n\n\\]"
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            "content": "(2) \\( \\Leftarrow \\) (3): 比较经典，先通过分块矩阵的初等变换凑出 E:",
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            "content": "2.6 特殊矩阵性质总结"
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            "content": "再通过 E 消去其他部分:",
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            "content": "\\[\n\\rightarrow \\left(\\begin{array}{c c c c c}E&O&O&\\dots&O\\\\A _ {2}&A _ {2} - A _ {2} ^ {2}&- A _ {2} A _ {3}&\\dots&- A _ {2} A _ {m}\\\\A _ {3}&- A _ {3} A _ {2}&A _ {3} - A _ {3} ^ {2}&\\dots&- A _ {3} A _ {m}\\\\\\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\A _ {m}&- A _ {m} A _ {2}&- A _ {m} A _ {3}&\\dots&A _ {m} - A _ {m} ^ {2}\\end{array}\\right)\\rightarrow \\left(\\begin{array}{c c c c c}E&O&O&\\dots&O\\\\O&A _ {2} - A _ {2} ^ {2}&- A _ {2} A _ {3}&\\dots&- A _ {2} A _ {m}\\\\O&- A _ {3} A _ {2}&A _ {3} - A _ {3} ^ {2}&\\dots&- A _ {3} A _ {m}\\\\\\vdots&\\vdots&\\vdots&\\ddots&\\vdots\\\\O&- A _ {m} A _ {2}&- A _ {m} A _ {3}&\\dots&A _ {m} - A _ {m} ^ {2}\\end{array}\\right)\n\\]"
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            "content": "根据  \\( r(A_{1}) + r(A_{2}) + \\cdots + r(A_{m}) = n \\) ，得到",
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            "content": "\\[\n\\left( \\begin{array}{c c c c} A _ {2} - A _ {2} ^ {2} & - A _ {2} A _ {3} & \\dots & - A _ {2} A _ {m} \\\\ - A _ {3} A _ {2} & A _ {3} - A _ {3} ^ {2} & \\dots & - A _ {3} A _ {m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ - A _ {m} A _ {2} & - A _ {m} A _ {3} & \\dots & A _ {m} - A _ {m} ^ {2} \\end{array} \\right) = O\n\\]"
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            "content": "这说明  \\( \\forall i \\neq j \\)  有  \\( A_{i}^{2} = A_{i} \\) ，且  \\( A_{i}A_{j} = O(i, j = 2, 3, \\cdots, m) \\) ，再对  \\( A_{1} + \\cdots + A_{m} = E_{n} \\)  两边同乘  \\( A_{i} \\)  得到\n\n\\( A_{1}A_{i} + A_{i} = A_{i} \\)",
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            "content": "因此 \\(A_{1}A_{i} = O\\) ，结论成立",
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            "content": "(3) \\( \\Rightarrow(1) \\) ：直接对  \\( A_{1}+\\cdots+A_{m}=E_{n} \\)  两边右乘  \\( A_{i} \\)  得到",
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            "content": "\\[\nA _ {1} A _ {i} + A _ {2} A _ {i} + \\dots + A _ {m} A _ {i} = A _ {i} ^ {2} = A _ {i}\n\\]"
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            "content": "第3章 行列式"
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            "content": "3.1 行列式基本概念与性质"
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            "content": "定理 3.1 (行列式初等变换)"
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            "angle": 0,
            "content": "矩阵初等变换相对行列式的变换为:",
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            "angle": 0,
            "content": "- 互换两行/列：取反号",
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            "angle": 0,
            "content": "- 同行/列可同时提出公因式",
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            "angle": 0,
            "content": "- 倍加：不改变行列式大小",
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            "content": "定理 3.2 (行列式分拆)"
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            "content": "行列式可以关于一行或者一列进行分拆，即",
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            "content": "\\[\n\\left| \\begin{array}{c c c c} a _ {1 1} & a _ {1 2} & \\dots & a _ {1 n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b _ {s 1} + c _ {s 1} & b _ {s 2} + c _ {s 2} & \\dots & b _ {s n} + c _ {s n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a _ {n 1} & a _ {n 2} & \\dots & a _ {n n} \\end{array} \\right| = \\left| \\begin{array}{c c c c} a _ {1 1} & a _ {1 2} & \\dots & a _ {1 n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ b _ {s 1} & b _ {s 2} & \\dots & b _ {s n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a _ {n 1} & a _ {n 2} & \\dots & a _ {n n} \\end{array} \\right| + \\left| \\begin{array}{c c c c} a _ {1 1} & a _ {1 2} & \\dots & a _ {1 n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ c _ {s 1} & c _ {s 2} & \\dots & c _ {s n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a _ {n 1} & a _ {n 2} & \\dots & a _ {n n} \\end{array} \\right|.\n\\]"
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            "content": "3.2 行列式展开"
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            "content": "定义 3.1 (余子式与代数余子式)"
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            "content": "将矩阵 A 去除第 i 行，第 j 列后矩阵的行列式称为  \\( a_{ij} \\)  的余子式  \\( M_{ij} \\) ，代数余子式  \\( A_{ij} := (-1)^{i+j} M_{ij} \\)",
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            "content": "定理 3.3 (行列式展开)"
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            "content": "固定列 \\(j\\) ，则按列展开为 \\(|A| = \\sum_{i=1}^{n} a_{ij} A_{ij}\\) 。固定行 \\(i\\) ，则按行展开为 \\(|A| = \\sum_{j=1}^{n} a_{ij} A_{ij}\\)",
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            "content": "定理 3.4 (不同行相乘为 0)"
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            "content": "\\(A\\) 的第 \\(i\\) 行元素与 \\(k\\) 行的代数余子式乘积和为0: \\(\\sum_{j=1}^{n} a_{ij} A_{kj} = 0\\)",
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            "content": "3.3 特殊的行列式"
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            "content": "3.3.1 爪型行列式"
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            "content": "定理 3.5 (正爪型行列式)"
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            "content": "正爪型行列式通过对角消去一条直边，再按三角行列式展开",
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            "content": "3.3 特殊的行列式"
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        {
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            "content": "定理 3.6 (异爪型行列式)"
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            "content": "异爪型行列式(一横两斜)分以下几种情况:",
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            "content": "- 行和为0型：全部加到第一行，按第一列展开",
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            "angle": 0,
            "content": "- 两对角各自相等型：按斜直夹边展开，获得递推公式",
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        {
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            "content": "练习3.1异爪型行列式(1)计算 \\(|A| = \\left| \\begin{array}{cccccc}0 & -1 & -1 & \\dots & -1 & -1\\\\ 1 & 1 & 0 & \\dots & 0 & 0\\\\ 0 & 1 & 1 & \\dots & 0 & 0\\\\ \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\ 0 & 0 & 0 & \\dots & 1 & 1 \\end{array} \\right|\\)",
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            "content": "(2) 设 \\(s(x) = \\begin{cases} \\frac{x}{|x|}, & x \\neq 0 \\\\ 0, & x = 0 \\end{cases}\\)，已知 \\(A = (a_{ij})\\)，\\(a_{ij} = s(i - j)\\)，求 \\(|A|\\)",
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            "content": "解(1)按最后一列展开，得到 \\(D_{n} = (-1)^{n + 2} + D_{n - 1}\\) ，因此 \\(n\\) 为奇数时 \\(D_{n} = 0\\) ， \\(n\\) 为偶数时 \\(D_{n} = 1\\)",
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            "content": "(2) 上下相减后变为 (1)",
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            "content": "3.3.2 大对角型 (三斜线)"
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            "content": "3.3.3 行和相同型"
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            "content": "定理 3.7 (行和相同型)"
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            "content": "行和相同型行列式分以下两种情况：",
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            "angle": 0,
            "content": "- 内部大量相同：加到第一列，提出公因式，消去内部大量元素",
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            "angle": 0,
            "content": "- 循环型：上下相减 (例题见循环型一节)",
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            "content": "练习 3.2 内部大量相同型 计算行列式",
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            "content": "\\[\n\\left| \\begin{array}{c c c c c} a & b & b & \\dots & b \\\\ b & a & b & \\dots & b \\\\ b & b & a & \\dots & b \\\\ \\vdots & \\vdots & & & \\vdots \\\\ b & b & b & \\dots & a \\end{array} \\right|\n\\]"
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            "content": "解加到第一列，提出 \\([a + (n - 1)b]\\)，用第一列消去其他元素，最终得到答案 \\((a - b)^{n - 1}[a + (n - 1)b]\\)",
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            "content": "3.3.4 循环型"
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            "content": "循环型一般无论形态都是上下相减，一般只有行和相同型和反对称型两种。",
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            "content": "练习3.3循环型 计算 \\(|A|, |B|\\)",
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            "content": "\\[\n| A | = \\left| \\begin{array}{c c c c c c} 0 & 1 & 2 & \\dots & n - 2 & n - 1 \\\\ 1 & 0 & 1 & \\dots & n - 3 & n - 2 \\\\ 2 & 1 & 0 & \\dots & n - 4 & n - 3 \\\\ \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\ n - 2 & n - 3 & n - 4 & \\dots & 0 & 1 \\\\ n - 1 & n - 2 & n - 3 & \\dots & 1 & 0 \\end{array} \\right| \\quad | B | = \\left| \\begin{array}{c c c c c c} 1 & 2 & 3 & \\dots & n - 1 & n \\\\ n & 1 & 2 & \\dots & n - 2 & n - 1 \\\\ n - 1 & n & 1 & \\dots & n - 3 & n - 2 \\\\ \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\ 3 & 4 & 5 & \\dots & 1 & 2 \\\\ 2 & 3 & 4 & \\dots & n & 1 \\end{array} \\right|\n\\]"
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            "content": "解(1)依次上面一行减下面一行, 变成如下行列式, 再每一列加上第一列即可",
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            "content": "\\[\n\\left| \\begin{array}{c c c c c c} - 1 & 1 & 1 & \\dots & 1 & 1 \\\\ - 1 & - 1 & 1 & \\dots & 1 & 1 \\\\ - 1 & - 1 & - 1 & \\dots & 1 & 1 \\\\ \\vdots & \\vdots & \\vdots & & \\vdots & \\vdots \\\\ - 1 & - 1 & - 1 & \\dots & - 1 & 1 \\\\ n - 1 & n - 2 & n - 3 & \\dots & 1 & 0 \\end{array} \\right|\n\\]"
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            "content": "(2) 依次上面一行减下面一行，然后将最后一列的 -1 倍加到每一列。",
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            "content": "定理 3.8 (行列式加边法)"
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            "content": "计算矩阵 \\(A\\) 的行列式 \\(|A|\\) 时，可加上一边：",
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            "content": "\\[\n| A | = \\left| \\begin{array}{c c c c} a _ {1 1} & a _ {1 2} & \\dots & a _ {1 n} \\\\ a _ {2 1} & a _ {2 2} & \\dots & a _ {2 n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a _ {n 1} & a _ {n 2} & \\dots & a _ {n n} \\end{array} \\right| = \\left| \\begin{array}{c c c c c} 1 & 1 & 1 & 1 & 1 \\\\ 0 & a _ {1 1} & a _ {1 2} & \\dots & a _ {1 n} \\\\ 0 & a _ {2 1} & a _ {2 2} & \\dots & a _ {2 n} \\\\ \\vdots & \\vdots & \\vdots & & \\vdots \\\\ 0 & a _ {n 1} & a _ {n 2} & \\dots & a _ {n n} \\end{array} \\right|\n\\]"
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            "content": "笔记 行列式加边法往往在内部元素大量相同，但行和不同的情况。",
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            "content": "定理 3.9 (Cramer 法则)"
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            "content": "\\(n\\) 个方程的 \\(n\\) 元线性方程组 \\(AX = B\\) ，系数矩阵行列式 \\(|A|\\neq 0\\) ，则其唯一解为",
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            "content": "\\[\n\\left(\\frac {| B _ {1} |}{| A |}, \\frac {| B _ {2} |}{| A |}, \\dots , \\frac {| B _ {n} |}{| A |}\\right)\n\\]"
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            "content": "其中 \\(B_{i}\\) 为 \\(A\\) 将第 \\(i\\) 列换为 \\(B\\) 组成的矩阵。",
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            "content": "4.1 线性方程组解的判定定理"
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            "content": "4.1.1 解存在判定定理"
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            "content": "定理 4.1 (解存在充要条件)",
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            "content": "K 上线性方程组 Ax=b 有解的充要条件为  \\( r(A)=r(\\overline{A}) \\) ，这里  \\( \\overline{A} := [A, b] \\)",
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            "content": "定理 4.2 (同解定理)",
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            "content": "\\(A_{s\\times n},B_{m\\times n}\\) 为两个矩阵，则 \\(AX = 0,BX = 0\\) 同解当且仅当 \\(A,B\\) 行向量等价",
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            "content": "练习 4.1 具体应用 (1) 已知下面两个线性方程组同解，求 a, b, c 的值，且求通解",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} x _ {1} + 2 x _ {2} + 3 x _ {3} = 0 \\\\ 2 x _ {1} + 3 x _ {2} + 5 x _ {3} = 0 \\\\ x _ {1} + x _ {2} + a x _ {3} = 0 \\end{array} \\right. \\quad \\left\\{ \\begin{array}{l} x _ {1} + b x _ {2} + c x _ {3} = 0 \\\\ 2 x _ {1} + b ^ {2} x _ {2} + (c + 1) x _ {3} = 0 \\end{array} \\right.\n\\]"
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            "content": "(2) \\(A_{s \\times n}, B_{t \\times n}\\) 为两个矩阵，\\(AX = 0, BX = 0\\) 的解空间分别为 \\(V_{1}, V_{2}\\)，证明：\\(V_{1} \\subseteq V_{2}\\) 当且仅当存在矩阵 \\(C_{t \\times s}\\) 使得 \\(B = CA\\)",
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            "content": "证明 (1) 设方程组为 \\(AX = 0, BX = 0\\)，即 \\(AX = 0\\) 与 \\(\\left( \\begin{array}{l} A \\\\ B \\end{array} \\right) X = 0\\) 同解，做初等行变换即可。",
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            "content": "(2) \\( V_{1}\\subseteq V_{2} \\) 当且仅当AX=0与 \\( \\left(\\begin{array}{l}A\\\\ B\\end{array}\\right)X=0 \\) 同解。",
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            "content": "若 K 上 Ax = b 满足  \\( r(A) = r(\\overline{A}) \\) :",
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            "content": "若 \\(r(A) = n\\) ：则 \\(Ax = b\\) 有唯一解",
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            "content": "若 \\(r(A) < n\\)：则 \\(Ax = b\\) 有无穷解",
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            "content": "4.2 线性方程组的求解"
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            "content": "定义 4.1 (自由未知量)"
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            "content": "消元后多余的 0 = 0 位置对应的变量  \\( x_{r+1}, \\cdots, x_{n} \\)  称为方程组自由未知量",
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            "content": "自由变量的个数为 \\(n - r(A)\\)",
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            "content": "让自由未知量 \\(x_{r - 1 + i}\\) 取1，其余取0，构造解向量 \\(\\eta_{i}\\) 称为基础解系：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\eta_ {1} = (*, \\dots , *, 1, 0, \\dots , 0) \\\\ \\eta_ {2} = (*, \\dots , *, 0, 1, \\dots , 0) \\\\ \\vdots \\\\ \\eta_ {n - r} = (*, \\dots , *, 0, 0, \\dots , 1) \\end{array} \\right.\n\\]"
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            "content": "笔记 如果只有一个自由未知数，则一般取 1 而非 0，否则无法表示出基础解系，只是要给特解。",
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            "content": "齐次线性方程组 \\(Ax = 0\\) 的任意解是其基础解系的线性组合",
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            "content": "定理 4.6 (非齐次线性方程组的解)"
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            "content": "非齐次线性方程组 Ax = b 的任意解是其特解  \\( \\gamma_{0} \\)  加上齐次方程组 Ax = 0 基础解系线性组合，即  \\( x = \\gamma_{0} + L\\{\\eta_{1}, \\cdots, \\eta_{n-r}\\} \\)",
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            "content": "练习 4.2 求解非齐次线性方程组 (1) 讨论以下线性方程组解的情况",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} x _ {1} + x _ {2} - x _ {3} = 1 \\\\ 2 x _ {1} + 3 x _ {2} + k x _ {3} = 3 \\\\ x _ {1} + k x _ {2} + 3 x _ {3} = 2 \\end{array} \\right.\n\\]"
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            "content": "(2) 已知 \\(A = \\left( \\begin{array}{ccc}1 & -1 & -1\\\\ 4 & t & 1\\\\ -2 & 2 & t \\end{array} \\right),b = \\left( \\begin{array}{c}1\\\\ 2\\\\ -t \\end{array} \\right)\\) ，当 \\(t\\) 为何值时，\\(AX = b\\) 无解，有唯一解，有无穷多解，并给出有解时对应的解。",
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            "content": "解(1)对增广矩阵做初等变换得到",
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            "content": "4.3 最小二乘解"
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            "content": "(2) 做增广矩阵初等变换得到",
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            "content": "\\[\n\\left[ \\begin{array}{c c c c} 1 & - 1 & - 1 & 1 \\\\ 0 & t + 4 & 5 & - 2 \\\\ 0 & 0 & t - 2 & 2 - t \\end{array} \\right]\n\\]"
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            "content": "t = -4 时无解。t = 2 无穷多解，其他情况有唯一解。具体求解略去。",
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            "content": "4.2.3 线性方程组公共解的计算"
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            "content": "练习 4.3 公共解的情况 (1) 线性方程组 I 的通解为  \\( (1,1,0,0)^{T} + k_{1}(1,0,-1,0)^{T} + k_{2}(2,3,0,1)^{T} \\) ，线性方程组 II 如下，若两个方程组有无穷组公共解，求 a, b 的值并表示出所有公共解",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} 7 x _ {1} - 6 x _ {2} + 3 x _ {3} = b \\\\ 8 x _ {1} - 9 x _ {2} + a x _ {4} = 7 \\end{array} \\right.\n\\]"
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            "content": "解 由于有无穷多组解，因此有无穷组 \\(k_{1}, k_{2}\\) 使得 \\((1 + k_{1} + 2k_{2}, 1 + 3k_{2}, -k_{1}, k_{2})\\) 为方程组 II 的解。代入方程化简得到",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} 4 k _ {1} - 4 k _ {2} = b - 1 \\\\ 8 k _ {1} + (a - 1 1) k _ {2} = 8 \\end{array} \\right. \\Rightarrow \\left\\{ \\begin{array}{l} 4 k _ {1} - 4 k _ {2} = b - 1 \\\\ 0 k _ {1} + (a - 3) k _ {2} = 1 0 - 2 b \\end{array} \\right.\n\\]"
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            "content": "需要有无穷多解，得有自由位置量，因此 \\(a = 3, b = 5\\) ，\\(k_{1} = k_{2} + 1\\) ，解系为 \\((2,1, -1,0)^{T} + k_{2}(3,3, -1,1)^{T}\\) 笔记若给两个完整的线性方程组，则先求出一个的基础解系，再往另一个方程中代。",
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            "content": "4.3 最小二乘解"
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            "content": "4.3.1 \\(A^T A\\) 的研究"
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            "content": "定理 4.7 ( \\( A^{T}A \\)  与 A 的秩)"
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            "content": "设 A 是  \\( m \\times n \\)  的实矩阵，则  \\( r(A^{T}A) = r(A) \\) 。复数域有  \\( r(A\\overline{A}^{T}) = r(\\overline{A}^{T}A) = r(A) \\) ，但不一定有  \\( r(A^{T}A) = r(A) \\) 。",
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            "content": "证明 (1) 只需要考虑 AX = 0 和  \\( A^{T}AX = 0 \\)  解的情况。首先根据 AX = 0 可以推出  \\( A^{T}AX = 0 \\) 。其次  \\( A^{T}AX = 0 \\)  可推出  \\( X^{T}A^{T}AX = (AX)^{T}(AX) = 0 \\) ，而 AX 是一个实向量， \\( \\alpha^{T}\\alpha = 0 \\)  当且仅当  \\( \\alpha = 0 \\) ，因此 AX = 0。故它们同解， \\( n - r(A) = n - r(A^{T}A) \\)  得到  \\( r(A) = r(A^{T}A) \\)",
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            "content": "(2) 若 \\(AX = 0\\)，则 \\(\\overline{A}^T A X = 0\\)。若 \\(\\overline{A}^T A X = 0\\)，则 \\(\\overline{X}^T \\overline{A}^T A X = (\\overline{A X})^T A X = 0\\)，得到 \\(A X = 0\\) 笔记该定理最重要的结论是最小二乘解一定存在。",
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            "content": "练习  \\( 4.4A^{T}A \\)  与 A 性质 (1) A 是 R 上任意  \\( s \\times n \\)  矩阵，证明： \\( r(AA^{T}A) = r(A) \\)",
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            "content": "证明 (1) \\( r(AA^{T}A)\\leq r(A) \\) ，而 \\( r(AA^{T}A)\\geq r(A^{T}AA^{T}A)=r(A^{T}A)=r(A) \\)",
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            "content": "定义 4.3 (最小二乘解)"
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            "content": "对一个无解的线性方程组 (矛盾方程组) AX = b，若向量  \\( (x_{1}, \\cdots, x_{s}) \\in \\mathbb{R}^{s} \\)  使得距离",
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            "content": "\\[\nd = \\sum_ {i = 1} ^ {n} (a _ {i 1} x _ {1} + \\dots + a _ {i s} x _ {s} - b _ {i}) ^ {2}\n\\]"
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            "content": "最小，则称该向量为最小二乘解。",
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            "content": "对于任一方程组 \\(AX = b\\) ，最小二乘解一定存在，且 \\(X_0\\) 是最小二乘解当且仅当 \\(A^T A X = A^T b\\)",
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            "content": "证明 设  \\( \\alpha_{1},\\cdots,\\alpha_{s} \\)  是 A 的列向量，定义  \\( W=L(\\alpha_{1},\\cdots,\\alpha_{s}) \\) ， \\( X_{0} \\)  是最小二乘解当且仅当  \\( AX_{0} \\)  是 b 在上的投影，即  \\( (b-AX_{0})\\perp\\alpha_{i} \\) ，因此",
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            "content": "\\[\n\\alpha_ {i} ^ {T} (b - A X _ {0}) \\Rightarrow A ^ {T} (b - A X _ {0}) = 0 \\Rightarrow A ^ {T} A X _ {0} = A ^ {T} b\n\\]"
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            "content": "方程组 \\(A^T A X = A^T b\\) 一定有解。",
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            "content": "定理 4.10 (凸闭集最佳逼近元的唯一性)"
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            "content": "\\(\\mathcal{H}\\) 是Hilbert空间， \\(M\\subset \\mathcal{H}\\) 是凸闭集，则 \\(\\forall x\\in \\mathcal{H}\\) 在 \\(M\\) 中存在唯一最佳逼近。",
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            "content": "第 5 章 线性空间与线性变换"
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            "content": "\\(\\mathbb{P}\\) 是一个数域， \\(V\\) 是非空集合， \\(V\\) 上定义了二元运算 \\(+\\) 和数乘，若这两种运算满足：",
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            "content": "二元运算：(1) 交换 (2) 结合 (3) 零元 (4) 负元",
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            "content": "- 数乘：(1)一元(2)结合 \\((c_{1}c_{2})\\alpha = c_{1}(c_{2}\\alpha)(3)\\) 向量分配 \\((c_{1} + c_{2})\\alpha = c_{1}\\alpha + c_{2}\\alpha (4)\\) 数分配 \\(c(\\alpha +\\beta) = c\\alpha +c\\beta\\)",
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            "content": "则称 \\(V\\) 是 \\(\\mathbb{P}\\) 上的线性空间",
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            "content": "5.1.2 线性相关、线性无关与向量组的秩"
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            "content": "定义 5.2 (线性相关和线性无关)"
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            "content": "\\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  为向量组，若存在不全为0的 \\( k_{1},\\cdots,k_{n} \\) 使得",
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            "content": "\\[\nk _ {1} \\alpha_ {1} + k _ {2} \\alpha_ {2} + \\dots + k _ {n} \\alpha_ {n} = 0\n\\]"
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            "content": "则称它们线性相关。若该等式成立当且仅当  \\( k_{1}=k_{2}=\\cdots=k_{n}=0 \\) ，则称线性无关。",
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            "content": "给定一列向量  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) ，将它们作为矩阵的列向量拼成一个矩阵，对矩阵做初等行变换化为阶梯型矩阵，阶梯头所在列对应的向量为极大线性无关组。(其实按行也可以，按列更不容易出错，因为换行不影响结果)",
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            "content": "5.1 线性空间"
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            "content": "\\[\n\\left[\\begin{array}{c c c c c}1&0&3&1&2\\\\- 1&3&0&- 1&1\\\\2&1&7&2&5\\\\4&2&1 4&0&6\\end{array}\\right]\\rightarrow \\left[\\begin{array}{c c c c c}1&0&3&1&- 2\\\\0&1&1&0&1\\\\0&0&0&1&1\\\\0&0&0&0&0\\end{array}\\right]\n\\]"
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            "content": "因此极大线性无关组为 \\(\\alpha_{1},\\alpha_{2},\\alpha_{4}\\)",
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            "content": "\\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  为向量组，称其中极大线性无关组的向量个数为它们的秩。",
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            "content": "设  \\( \\alpha_{1},\\cdots,\\alpha_{r} \\)  可以被  \\( \\beta_{1},\\cdots,\\beta_{s} \\)  线性表出，r>s，则  \\( \\alpha_{1},\\cdots,\\alpha_{r} \\)  线性相关。",
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            "content": "定义 5.4 (同构)",
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            "content": "若两个代数系统之间存在双射，且该双射保持代数系统的运算，则称两个代数系统同构。",
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            "content": "\\(V\\) 为 \\(\\mathbb{P}\\) 上线性空间，\\(\\alpha_{1}, \\cdots, \\alpha_{n}\\) 为 \\(V\\) 的一组基，构造线性映射 \\(f\\)，将 \\(\\forall \\alpha \\in V\\) 映射到其坐标，则 \\(f\\) 为 \\(V \\to \\mathbb{P}^{n}\\) 的同构映射",
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            "content": "\\( \\alpha_{1},\\cdots,\\alpha_{n} \\) 是 V 的一组基， \\( A_{n\\times s} \\) ，向量组 \\( \\beta_{1},\\cdots,\\beta_{s} \\) 满足",
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            "content": "\\[\n(\\beta_ {1}, \\dots , \\beta_ {s}) = (\\alpha_ {1}, \\dots , \\alpha_ {n}) A\n\\]"
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            "content": "则  \\( \\dim(\\beta_{1},\\cdots,\\beta_{s})=r(A) \\)",
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            "content": "证明  \\( A = [X_{1}, \\cdots, X_{s}] \\)  即  \\( \\beta_{1}, \\cdots, \\beta_{s} \\)  的坐标，因此 A 的列向量与  \\( \\beta_{1}, \\cdots, \\beta_{s} \\)  等价，从而  \\( r(A) = \\dim(\\beta_{1}, \\cdots, \\beta_{s}) \\)",
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            "content": "推论 5.1 (线性无关情况)"
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            "content": "\\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  线性无关， \\( (\\beta_{1},\\cdots,\\beta_{n})=(\\alpha_{1},\\cdots,\\alpha_{n})A \\) ，则  \\( \\beta_{i} \\)  线性无关当且仅当 A 可逆。",
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            "content": "练习5.3同构的应用 \\((1)\\beta_{1} = \\alpha_{1} + \\alpha_{2},\\beta_{2} = \\alpha_{1} + \\alpha_{3},\\beta_{3} = \\alpha_{1} + \\alpha_{2}\\) ，证明： \\(\\alpha_{1},\\alpha_{2},\\alpha_{3}\\) 和 \\(\\beta_{1},\\beta_{2},\\beta_{3}\\) 等价",
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            "content": "(2) 证明 n 为奇数时， \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) 线性无关当且仅当 \\( \\alpha_{1}+\\alpha_{2},\\cdots,\\alpha_{n-1}+\\alpha_{n},a_{n}+\\alpha_{1} \\) 线性无关",
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            "content": "(3) 设  \\( \\beta_{1} = \\alpha_{2} + \\cdots + \\alpha_{n}, \\beta_{2} = \\alpha_{1} + \\alpha_{3} + \\cdots + \\alpha_{n}, \\beta_{n} = \\alpha_{1} + \\cdots + \\alpha_{n-1} \\) ，证明  \\( \\alpha_{1}, \\cdots, \\alpha_{n} \\)  和  \\( \\beta_{1}, \\cdots, \\beta_{n} \\)  有相同的秩",
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            "content": "证明 (1) 如下同构映射至坐标, 显然 \\(A\\) 可逆, 因此等价",
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            "content": "\\[\n(\\beta_ {1}, \\beta_ {2}, \\beta_ {3}) = (\\alpha_ {1}, \\alpha_ {2}, \\alpha_ {3}) \\left[ \\begin{array}{c c c} 1 & 1 & 1 \\\\ 1 & & 1 \\\\ 0 & 1 \\end{array} \\right] := (\\alpha_ {1}, \\alpha_ {2}, \\alpha_ {3}) A\n\\]"
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            "content": "(2) 将后者设为  \\( \\beta_{1}, \\cdots, \\beta_{n} \\) ，如下线性表出",
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            "content": "定义 5.5 (基与坐标)"
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            "content": "V 是线性空间， \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  是 V 中线性无关的向量组，若  \\( \\forall\\alpha\\in V \\)  可被该向量组线性表出，则称该向量组为 V 的一组基。且若",
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            "content": "\\[\n\\alpha = (\\epsilon_ {1}, \\dots , \\epsilon_ {n}) X\n\\]"
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            "content": "则称 X 为  \\( \\alpha \\)  在  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  基下的坐标",
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            "content": "笔记 坐标的计算：直接根据  \\( \\alpha = (\\epsilon_{1}, \\cdots, \\epsilon_{n})X \\)  解线性方程组 (或者用初等变换求逆)",
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            "content": "\\[\nX = (\\epsilon_ {1}, \\dots , \\epsilon_ {n}) ^ {- 1} \\alpha\n\\]"
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            "content": "练习 5.4 证明基 给定  \\( \\alpha_{1}=(1,1,\\cdots,1,1) \\) ,  \\( \\alpha_{2}=(1,1,\\cdots,1,0) \\) ,  \\( \\cdots \\) ,  \\( \\alpha_{n}=(1,0,\\cdots,0) \\) ，证明其是  \\( P^{n} \\)  的一组基证明 当成列向量排成一个矩阵，满秩即可",
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            "content": "练习 5.5 坐标的计算 (1) 已知  \\( \\epsilon_{1}, \\epsilon_{2}, \\epsilon_{3} \\)  是  \\( P^{3} \\)  的一组基，计算  \\( \\alpha = (1, 2, 1)^{T} \\)  在这组基下的坐标",
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            "content": "\\[\n\\epsilon_ {1} = (1, 1, 1) ^ {T}, \\epsilon_ {2} = (1, 1, - 1) ^ {T}, \\epsilon_ {3} = (1, - 1, - 1) ^ {T}\n\\]"
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            "content": "解(1)只需要解方程组 \\(\\alpha = k_{1}\\epsilon_{1} + k_{2}\\epsilon_{2} + k_{3}\\epsilon_{3}\\)，即 \\([\\epsilon_1,\\epsilon_2,\\epsilon_3]X = \\alpha\\)",
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            "content": "\\[\n\\left[ \\begin{array}{c c c} 1 & 1 & 1 \\\\ 1 & 1 & - 1 \\\\ 1 & - 1 & - 1 \\end{array} \\right] X = \\left[ \\begin{array}{c} 1 \\\\ 2 \\\\ 1 \\end{array} \\right]\n\\]"
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            "content": "解得 \\(k_{1} = 1, k_{2} = \\frac{1}{2}, k_{3} = -\\frac{1}{2}\\)",
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            "content": "练习5.6难度进阶(1)求非零向量 \\(\\alpha \\in \\mathbb{P}^4\\) ，使得其在下面两组基中有相同的坐标",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\epsilon_ {1} = (1, 0, 0, 0) ^ {T} \\\\ \\epsilon_ {2} = (0, 1, 0, 0) ^ {T} \\\\ \\epsilon_ {3} = (0, 0, 1, 0) ^ {T} \\\\ \\epsilon_ {4} = (0, 0, 0, 1) ^ {T} \\end{array} \\right., \\quad \\left\\{ \\begin{array}{l} \\eta_ {1} = (2, 1, - 1, 1) ^ {T} \\\\ \\eta_ {2} = (0, 3, 1, 0) ^ {T} \\\\ \\eta_ {3} = (5, 3, 2, 1) ^ {T} \\\\ \\eta_ {4} = (6, 6, 1, 3) ^ {T} \\end{array} \\right.\n\\]"
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            "content": "解 (1) 设 \\(\\alpha\\) 在第一组基下的坐标为 \\((x_{1}, x_{2}, x_{3}, x_{4})^{T}\\), 则由于同坐标",
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            "content": "\\[\n\\alpha = x _ {1} \\eta_ {1} + x _ {2} \\eta_ {2} + x _ {3} \\eta_ {3} + x _ {4} \\eta_ {4}\n\\]"
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            "angle": 0,
            "content": "5.1 线性空间"
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "解线性方程组  \\( \\left[\\eta_{1},\\eta_{2},\\eta_{3},\\eta_{4}\\right]X=\\alpha \\)  即可 (注意将  \\( \\alpha \\)  减到左边):",
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        },
        {
            "type": "equation",
            "bbox": [
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            "content": "\\[\n\\left[ \\begin{array}{r r r r} 1 & 0 & 5 & 6 \\\\ 1 & 2 & 3 & 6 \\\\ - 1 & 1 & 1 & 1 \\\\ 1 & 0 & 1 & 2 \\end{array} \\right] X = 0\n\\]"
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        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "解得  \\( \\alpha = k(1,1,1,-1) \\)",
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        {
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            "content": "5.1.5 坐标变换"
        },
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            "content": "定义 5.6 (过渡矩阵)"
        },
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            "type": "text",
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            "content": "给定 \\( (\\epsilon_{1},\\cdots,\\epsilon_{n}),(\\eta_{1},\\cdots,\\eta_{n}) \\) 两组基，若存在矩阵T满足：",
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            "content": "\\[\n(\\epsilon_ {1}, \\dots , \\epsilon_ {n}) = (\\eta_ {1}, \\dots , \\eta_ {n}) T\n\\]"
        },
        {
            "type": "text",
            "bbox": [
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            "content": "则称 T 为  \\( \\eta_{1},\\cdots,\\eta_{n} \\)  到  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  的过渡矩阵 (不要搞反)",
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        {
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            "content": "笔记 过渡矩阵的计算：直接根据  \\( (\\epsilon_{1},\\cdots,\\epsilon_{n})=(\\eta_{1},\\cdots,\\eta_{n})T \\)  得到",
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            "content": "\\[\nT = (\\eta_ {1}, \\dots , \\eta_ {n}) ^ {- 1} (\\epsilon_ {1}, \\dots , \\epsilon_ {n})\n\\]"
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            "content": "用初等变换求  \\( A^{-1}B \\)  的方式直接求 T 即可。",
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            "angle": 0,
            "content": "定理 5.5 (坐标变换)"
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            "content": "44"
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            "content": "\\(V\\) 是 \\(n\\) 维线性空间， \\(\\eta_1,\\dots ,\\eta_n\\) 和 \\(\\epsilon_1,\\dots ,\\epsilon_n\\) 是空间两组基，且 \\((\\eta_{1},\\dots ,\\eta_{n}) = (\\epsilon_{1},\\dots ,\\epsilon_{n})T\\) ，向量 \\(x\\in V\\) 在 \\(\\epsilon_i,\\eta_i\\) 下的坐标分别为 \\(X,Y\\) ，则",
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            "content": "\\[\nX = T Y\n\\]"
        },
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            "type": "text",
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            "content": "证明  \\( x=(\\epsilon_{1},\\cdots,\\epsilon_{n})X=(\\eta_{1},\\cdots,\\eta_{n})Y=(\\epsilon_{1},\\cdots,\\epsilon_{n})TY \\)  即可。",
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            "type": "text",
            "bbox": [
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            "content": "笔记 坐标变换用荡秋千记忆，即  \\( (\\eta_{1},\\cdots,\\eta_{n})=(\\epsilon_{1},\\cdots,\\epsilon_{n})T \\) ，分别对应 Y,X，则把 T 荡到另一边，即 TY=X。这与线性变换在不同基下的矩阵一样都需要记住。",
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            "content": "练习 5.7 过渡矩阵的计算 给定两组基，求它们之间的过渡矩阵，并求  \\( \\xi = (1, 0, 0, -1) \\)  在  \\( \\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\alpha_{4} \\)  下的坐标",
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            "content": "\\[\n\\begin{array}{l l l l} \\alpha_ {1} = (1, 1, 1, 1) & \\alpha_ {2} = (1, 1, - 1, - 1) & \\alpha_ {3} = (1, - 1, 1, - 1) & \\alpha_ {4} = (1, - 1, - 1, 1) \\\\ \\beta_ {1} = (1, 1, 0, 1) & \\beta_ {2} = (2, 1, 3, 1) & \\beta_ {3} = (1, 1, 0, 0) & \\beta_ {4} = (0, 1, - 1, - 1) \\end{array}\n\\]"
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            "content": "解(1)先取标准基  \\( \\epsilon_{1}, \\epsilon_{2}, \\epsilon_{3}, \\epsilon_{4} \\) ，设",
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            "content": "\\[\n(\\alpha_ {1}, \\alpha_ {2}, \\alpha_ {3}, \\alpha_ {4}) = (\\epsilon_ {1}, \\epsilon_ {2}, \\epsilon_ {3}, \\epsilon_ {4}) A, (\\beta_ {1}, \\beta_ {2}, \\beta_ {3}, \\beta_ {4}) = (\\epsilon_ {1}, \\epsilon_ {2}, \\epsilon_ {3}, \\epsilon_ {4}) B\n\\]"
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            "content": "\\[\nA = \\left[ \\begin{array}{r r r r} 1 & 1 & 1 & 1 \\\\ 1 & 1 & - 1 & - 1 \\\\ 1 & - 1 & 1 & - 1 \\\\ 1 & - 1 & - 1 & 1 \\end{array} \\right], B = \\left[ \\begin{array}{r r r r} 1 & 2 & 1 & 0 \\\\ 1 & 1 & 1 & 1 \\\\ 0 & 3 & 0 & - 1 \\\\ 1 & 1 & 0 & - 1 \\end{array} \\right]\n\\]"
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            "content": "5.1 线性空间"
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            "content": "综上得到过渡矩阵为 (具体计算时用行初等变换:  \\( (A|B) \\rightarrow (E|A^{-1}B) \\) )",
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            "content": "\\[\nA ^ {- 1} B = \\frac {1}{4} \\left[ \\begin{array}{r r r r} 3 & 7 & 2 & - 1 \\\\ 1 & - 1 & 2 & 3 \\\\ - 1 & 3 & 0 & - 1 \\\\ 1 & - 1 & 0 & - 1 \\end{array} \\right]\n\\]"
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            "content": "(2) \\(\\xi\\) 在 \\((\\epsilon_1, \\epsilon_2, \\epsilon_3, \\epsilon_4)\\) 下的坐标即本身，因此在 \\((\\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4)\\) 下坐标为 \\(A^{-1}\\xi\\) (具体计算时也和 \\((A|B)\\) 摆在一起)，结果为 \\((0, \\frac{1}{2}, \\frac{1}{2}, 0)^T\\)",
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            "content": "5.1.6 和与交、维数定理"
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            "content": "定义 5.7 (线性空间和与交)"
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            "content": "\\(V_{1}, V_{2}\\) 为线性空间，则",
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            "content": "\\[\nV _ {1} \\cap V _ {2} := \\{\\alpha : \\alpha \\in V _ {1}, \\alpha \\in V _ {2} \\}, V _ {1} + V _ {2} := \\{\\alpha = \\alpha_ {1} + \\alpha_ {2}: \\alpha_ {1} \\in V _ {1}, \\alpha_ {2} \\in V _ {2} \\}\n\\]"
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            "content": "且显然 \\(V_{1} \\cap V_{2}, V_{1} + V_{2}\\) 都是子空间。",
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            "content": "笔记 \\(V_{1} \\cup V_{2}\\) 由于运算不封闭，一般而言不是子空间。注意分别 \\(V_{1} \\cup V_{2}\\) 和 \\(V_{1} + V_{2}\\)",
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            "content": "练习 5.8 和与交基本计算 (1) 已知  \\( V_{1} = L(\\alpha_{1}, \\alpha_{2}, \\alpha_{3}), V_{2} = L(\\beta_{1}, \\beta_{2}) \\) ，其中",
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            "content": "\\[\n\\alpha_ {1} = \\left( \\begin{array}{c} 1 \\\\ 2 \\\\ 1 \\\\ 0 \\end{array} \\right), \\alpha_ {2} = \\left( \\begin{array}{c} - 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{array} \\right), \\alpha_ {3} = \\left( \\begin{array}{c} 0 \\\\ 3 \\\\ 2 \\\\ 1 \\end{array} \\right), \\beta_ {1} = \\left( \\begin{array}{c} 2 \\\\ - 1 \\\\ 0 \\\\ 1 \\end{array} \\right), \\beta_ {2} = \\left( \\begin{array}{c} 1 \\\\ - 1 \\\\ 3 \\\\ 7 \\end{array} \\right)\n\\]"
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            "content": "求 \\(V_{1} + V_{2}, V_{1} \\cap V_{2}\\) 的基与维数",
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            "content": "(2) 已知 \\( W_{1} = L(\\alpha_{1}, \\alpha_{2}, \\alpha_{3}), W_{2} = L(\\beta_{1}, \\beta_{2}) \\)，求 \\( W_{1} \\cap W_{2}, W_{1} + W_{2} \\) 的基与维数。",
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            "content": "\\[\n\\alpha_ {1} = \\left( \\begin{array}{r} 1 \\\\ 2 \\\\ - 1 \\\\ - 3 \\end{array} \\right), \\alpha_ {2} = \\left( \\begin{array}{r} - 1 \\\\ - 1 \\\\ 2 \\\\ 1 \\end{array} \\right), \\alpha_ {3} = \\left( \\begin{array}{r} - 1 \\\\ - 3 \\\\ 0 \\\\ 5 \\end{array} \\right), \\beta_ {1} = \\left( \\begin{array}{r} - 1 \\\\ 0 \\\\ 4 \\\\ 2 \\end{array} \\right), \\beta_ {2} = \\left( \\begin{array}{r} 0 \\\\ 5 \\\\ 9 \\\\ - 1 4 \\end{array} \\right)\n\\]"
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            "content": "解 (1) \\( V_{1}+V_{2}=L(\\alpha_{1},\\alpha_{2},\\alpha_{3},\\beta_{1},\\beta_{2}) \\) ，因此只需要排成列，做初等变换：",
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            "content": "\\[\n\\left( \\begin{array}{c c c c c} 1 & - 1 & 0 & 2 & 1 \\\\ 2 & 1 & 3 & - 1 & - 1 \\\\ 1 & 1 & 2 & 0 & 3 \\\\ 0 & 1 & 1 & 1 & 7 \\end{array} \\right) \\to \\left( \\begin{array}{c c c c c} 1 & 0 & 1 & 0 & - 1 \\\\ 0 & 1 & 1 & 0 & 4 \\\\ 0 & 0 & 0 & 1 & 3 \\\\ 0 & 0 & 0 & 0 & 0 \\end{array} \\right) := A\n\\]"
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            "angle": 0,
            "content": "因此维数为3，基为 \\(\\alpha_{1},\\alpha_{2},\\beta_{1}\\)",
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            "content": "任取 \\(\\alpha \\in V_1 \\cap V_2\\)，设 \\(\\alpha = x_1\\alpha_1 + x_2\\alpha_2 + x_3\\alpha_3 = y_1\\beta_1 + y_2\\beta_2\\)，因此系数需要满足方程组：",
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            "content": "\\[\nx _ {1} \\alpha_ {1} + x _ {2} \\alpha_ {2} + x _ {3} \\alpha_ {3} - y _ {1} \\beta_ {1} - y _ {2} \\beta_ {2} = 0\n\\]"
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            "content": "设 \\(x_{4} = -y_{1}, x_{5} = y_{2}\\)，即求解 \\(AX = [\\alpha_{1}, \\alpha_{2}, \\alpha_{3}, \\beta_{1}, \\beta_{2}]X = 0\\)，解得基础解系为",
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            "content": "\\[\n\\eta_ {1} = \\left( \\begin{array}{c} - 1 \\\\ - 1 \\\\ 1 \\\\ 0 \\\\ 0 \\end{array} \\right), \\eta_ {2} = \\left( \\begin{array}{c} 1 \\\\ - 4 \\\\ 0 \\\\ - 3 \\\\ 1 \\end{array} \\right)\n\\]"
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    [
        {
            "type": "header",
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            "angle": 0,
            "content": "5.1 线性空间"
        },
        {
            "type": "text",
            "bbox": [
                0.091,
                0.086,
                0.301,
                0.1
            ],
            "angle": 0,
            "content": "对应 \\(V_{1} \\cap V_{2}\\) 的基础解系为：",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
                0.284,
                0.109,
                0.714,
                0.124
            ],
            "angle": 0,
            "content": "\\[\n- \\alpha_ {1} - \\alpha_ {2} + \\alpha_ {3} = (0, 0, 0, 0) ^ {T}, \\alpha_ {1} - 4 \\alpha_ {2} = (5, - 2, - 3, - 4) ^ {T}\n\\]"
        },
        {
            "type": "text",
            "bbox": [
                0.091,
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                0.593,
                0.147
            ],
            "angle": 0,
            "content": "只取非零的，故只能为 \\(k(5, -2, -3, -4)\\)，维数为1，基即 \\(\\alpha_{1} - 4\\alpha_{2}\\)。",
            "merge_prev": false
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "练习 5.9 和与并的区别 (1) 设  \\( V_{1}, V_{2} \\)  是 P 上线性空间 V 的两个非平凡子空间，证明： \\( \\exists \\alpha \\in V \\)  使得  \\( \\alpha \\notin V_{1}, \\alpha \\notin V_{2} \\) \n\n(2)  \\( V_{1}, \\cdots, V_{s} \\)  是 V 的 s 个非平凡子空间，证明：V 中至少有一个向量不属于  \\( V_{1}, \\cdots, V_{s} \\)  中的任意一个。",
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        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "(3) \\( V_{1}, V_{2} \\) 是两个子空间，证明： \\( V_{1} + V_{2} = V_{1} \\cup V_{2} \\) 当且仅当 \\( V_{1} \\subseteq V_{2} \\) 或者 \\( V_{2} \\subseteq V_{1} \\)",
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        },
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            "type": "text",
            "bbox": [
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            "content": "证明 (1) 由于 \\(V_{1}\\) 非平凡，\\(\\exists \\alpha \\notin V_{1}\\)，若 \\(\\alpha \\notin V_{2}\\)，则结论成立；否则 \\(\\alpha \\in V_{2}\\)，由于 \\(V_{2}\\) 非平凡，因此 \\(\\exists \\beta \\notin V_{2}\\)，若 \\(\\beta \\notin V_{1}\\)，则结论成立；否则 \\(\\beta \\in V_{1}\\)，此时有",
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        },
        {
            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n\\alpha \\notin V _ {1}, \\beta \\in V _ {1}, \\alpha \\in V _ {2}, \\beta \\notin V _ {2}\n\\]"
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "因此 \\(\\gamma = \\alpha +\\beta \\notin V_1,\\gamma \\notin V_2\\)",
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        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "(2) 用归纳法",
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            "type": "text",
            "bbox": [
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            "content": "(3) 右推左显然。左推右：设 \\(V_{1} + V_{2} = V_{1} \\cup V_{2}\\)，但 \\(V_{1} \\notin V_{2}\\)，则 \\(\\exists \\alpha_{1} \\in V_{1}, \\alpha_{1} \\notin V_{2}\\)，但",
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            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n\\forall \\alpha_ {2} \\in V _ {2}, \\alpha = \\alpha_ {1} + \\alpha_ {2} \\in V _ {1} + V _ {2}\n\\]"
        },
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "由于 \\(V_{1} + V_{2} = V_{1}\\cup V_{2}\\) ，因此 \\(\\alpha\\) 至少属于 \\(V_{1},V_{2}\\) 中的一个，显然 \\(\\alpha \\notin V_2\\) (根据加法封闭)，因此 \\(\\alpha \\in V_{1}\\) ，故",
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        },
        {
            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n\\alpha_ {2} = \\alpha - \\alpha_ {1} \\in V _ {1}\n\\]"
        },
        {
            "type": "page_number",
            "bbox": [
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            "angle": 0,
            "content": "46"
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "综上得到 \\(V_{2} \\subseteq V_{1}\\)",
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        },
        {
            "type": "paragraph_title",
            "bbox": [
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            "angle": 0,
            "content": "定理 5.6 (维数公式)"
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "设 \\(V_{1}, V_{2}\\) 是有限维线性空间 \\(V\\) 的两个子空间，则",
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        },
        {
            "type": "equation",
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            "angle": 0,
            "content": "\\[\n\\dim V _ {1} + \\dim V _ {2} = \\dim (V _ {1} + V _ {2}) + \\dim (V _ {1} \\cap V _ {2})\n\\]"
        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "证明 设小扩大。设  \\( V_{1} \\cap V_{2} \\)  的基为  \\( \\alpha_{1}, \\cdots, \\alpha_{s} \\) ， \\( V_{1}, V_{2} \\)  的基分别为：",
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        },
        {
            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n\\left\\{ \\begin{array}{l} \\alpha_ {1}, \\dots , \\alpha_ {s}, \\beta_ {s + 1}, \\dots , \\beta_ {m} \\\\ \\alpha_ {1}, \\dots , \\alpha_ {s}, \\gamma_ {s + 1}, \\dots , \\gamma_ {n} \\end{array} \\right.\n\\]"
        },
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "则  \\( V_{1} + V_{2} = L(\\alpha_{1}, \\cdots, \\alpha_{s}, \\beta_{s+1}, \\cdots, \\beta_{m}, \\gamma_{s+1}, \\cdots, \\gamma_{n}) \\) ，下面证明上面的向量全部线性无关。不妨设",
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        },
        {
            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\nk _ {1} \\alpha_ {1} + \\dots + k _ {s} \\alpha_ {s} + k _ {s + 1} \\beta_ {s + 1} + \\dots + k _ {n} \\beta_ {n} + l _ {s + 1} \\gamma_ {s + 1} + \\dots + l _ {n} \\gamma_ {n} = 0\n\\]"
        },
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            "bbox": [
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            "angle": 0,
            "content": "将 \\(l_{i}\\gamma_{i}\\) 的部分移到右侧，此时左侧线性组合属于 \\(V_{1}\\)，右侧属于 \\(V_{2}\\)，因此都属于 \\(V_{1} \\cap V_{2}\\)。根据左侧得到：",
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            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\nk _ {1} \\alpha_ {1} + \\dots + k _ {s} \\alpha_ {s} + k _ {s + 1} \\beta_ {s + 1} + \\dots + k _ {n} \\beta_ {n} \\in V _ {1} \\cap V _ {2} \\Rightarrow k _ {s + 1} = k _ {s + 2} = \\dots = k _ {n} = 0 \\Rightarrow k _ {1} = \\dots = k _ {s} = 0\n\\]"
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            "content": "5.1 线性空间"
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            "content": "5.1.7 直和理论"
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            "content": "定义 5.8 (直和)"
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            "content": "\\( V_{1}, V_{2} \\) 是线性空间 \\( V \\) 的子空间，若 \\( V_{1} \\cap V_{2} = \\{0\\} \\)，则记 \\( V_{1} \\oplus V_{2} := V_{1} + V_{2} \\)",
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            "content": "定义 5.9 (补空间)"
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            "content": "若 \\(V = V_{1} \\oplus V_{2}\\)，则称 \\(V_{2}\\) 是 \\(V_{1}\\) 的补空间。",
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            "content": "笔记 补空间一般是不唯一的，因为  \\( \\alpha_{1},\\cdots,\\alpha_{m} \\)  扩充为  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  的方式不唯一。",
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            "content": "定理 5.7 (直和等价条件)"
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            "content": "\\(W = V_{1} \\oplus V_{2}\\) 当且仅当",
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            "content": "- \\(W\\) 中任意向量 \\(\\alpha = \\alpha_{1} + \\alpha_{2},\\alpha_{1}\\in V_{1},\\alpha_{2}\\in V_{2}\\) 唯一表出",
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            "content": "- \\(W\\) 中零向量被 \\(V_{1}, V_{2}\\) 中向量唯一表出: \\(0 = 0 + 0\\)",
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            "content": "- \\(\\dim W = \\dim V_1 + \\dim V_2\\)",
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            "content": "练习 5.11 判断直和 (1) 数域 P 上所有 n 阶矩阵组成线性空间  \\( V = M_{n}(P) \\) ， \\( V_{1} \\)  表示所有对称矩阵组成的集合， \\( V_{2} \\)  表示所有反对称矩阵组成的集合，证明  \\( V_{1}, V_{2} \\)  是子空间且  \\( V = V_{1} \\oplus V_{2} \\)",
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            "content": "证明 (1) 显然关于加法和数乘封闭，因此是子空间。首先证明 \\( V = V_{1} + V_{2} \\)，由于矩阵 \\( A \\) 都可以写为：",
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            "content": "\\[\nA = \\frac {A + A ^ {T}}{2} + \\frac {A ^ {T} - A}{2}\n\\]"
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            "content": "从而  \\( V = V_{1} + V_{2} \\) 。",
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            "content": "证明 \\(V = V_{1} \\oplus V_{2}\\) ：可用 \\(V_{1} \\cap V_{2} = \\{\\}\\)，此时取 \\(A \\in V_{1} \\cap V_{2}\\)，则 \\(A = A^{T} = -A^{T}\\)，得到 \\(A^{T} = A = O\\)，因此为直和。也可以用 \\(O\\) 分解唯一，设 \\(O = B + C = B^{T} - C^{T}\\)，而 \\(O^{T} = B^{T} + C^{T}\\)，因此 \\(B = C = O\\)。",
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            "content": "定理 5.8 (多空间直和等价条件)"
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            "content": "\\( V_{1},\\cdots,V_{s} \\) 是 W 的子空间，且 \\( W=V_{1}+V_{2}+\\cdots+V_{s} \\)，则 \\( W=V_{1}\\oplus V_{2}\\oplus\\cdots\\oplus V_{s} \\) 当且仅当",
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            "content": "- \\(W\\) 中零向量分解唯一",
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            "content": "- \\(W\\) 中任意向量 \\(\\alpha\\) 分解唯一",
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            "content": "\\[\nV _ {i} \\cap \\sum_ {j \\neq i} V _ {j} = \\{0 \\}\n\\]"
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            "content": "\\[\n\\bullet \\dim W = \\dim V _ {1} + \\dots + \\dim V _ {s}\n\\]"
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            "content": "\\[\n0 = \\alpha_ {1} + \\dots + \\alpha_ {n} \\Rightarrow \\alpha_ {i} = \\alpha_ {1} + \\dots + \\alpha_ {i - 1} + \\alpha_ {i + 1} + \\dots + \\alpha_ {n}\n\\]"
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            "content": "5.2 线性映射"
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            "content": "定理 5.9 (直和合并定理)"
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            "content": "若  \\( V = V_{1} \\oplus V_{2}, V_{1} = V_{11} \\oplus V_{12} \\oplus \\cdots \\oplus V_{1s}, V_{2} = V_{21} \\oplus \\cdots \\oplus V_{2t} \\) ，则",
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            "content": "\\[\nV = V _ {1 1} \\oplus \\dots V _ {1 s} \\oplus V _ {2 1} \\oplus \\dots \\oplus V _ {2 t}\n\\]"
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            "content": "证明 只证明  \\( V = V_{1} \\oplus V_{2} \\) ， \\( V_{1} = V_{11} \\oplus V_{12} \\)  时， \\( V = V_{11} \\oplus V_{12} \\oplus V_{2} \\) 。",
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            "content": "首先显然有 \\(V = V_{11} + V_{12} + V_2\\) 。设",
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            "content": "\\[\n0 = \\alpha_ {1 1} + \\alpha_ {1 2} + \\alpha_ {2} = (\\alpha_ {1 1} + \\alpha_ {1 2}) + \\alpha_ {2}\n\\]"
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            "content": "由于 \\(V = V_{1} \\oplus V_{2}\\)，因此得到 \\(\\alpha_{11} + \\alpha_{12} = \\alpha_{2} = 0\\)，同理根据 \\(V_{1} = V_{11} \\oplus V_{12}\\) 得到 \\(\\alpha_{11} = \\alpha_{12} = 0\\)，综上0元素分解唯一，为直和。",
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            "content": "5.1.8 商空间"
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            "content": "定义 5.10 (商空间的元素)"
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            "content": "W 是线性空间 V 的子空间， \\( \\alpha \\in V \\) ，定义",
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            "content": "\\[\n\\alpha + W := \\{\\alpha + \\omega : \\omega \\in W \\}\n\\]"
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            "content": "定义 \\(V / W\\) 为 \\(\\{\\alpha + W : \\alpha \\in V\\}\\)，即线性空间子集组成的集合。可以证明其是一个线性空间",
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            "content": "定义 \\(\\eta : V \\to V / W\\) 为自然映射",
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            "content": "5.2 线性映射"
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            "content": "\\(V, W\\) 是 \\(\\mathbb{P}\\) 上的线性空间，\\(\\varphi: V \\to W\\) 若满足：",
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            "content": "\\[\n\\bullet \\varphi (\\alpha + \\beta) = \\varphi (\\alpha) + \\varphi (\\beta)\n\\]"
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            "content": "则称 \\(\\varphi\\) 为 \\(V\\) 到 \\(W\\) 的线性映射，将 \\(V\\) 到 \\(W\\) 的线性映射全体记为集合 \\(\\mathrm{Hom}_{\\mathbb{P}}(V,W)\\)",
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            "content": "核与像是最常见的不变子空间，因此在进一步研究线性映射之前，要先研究这两个特殊的集合。",
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            "content": "\\(U, V\\) 是 \\(\\mathbb{P}\\) 上的线性空间，\\(f: U \\to V\\)，定义 \\(\\operatorname{Ker}(f) = \\{\\alpha \\in U: f(\\alpha) = 0\\}\\)，\\(\\operatorname{Im}(f) = \\{f(\\alpha): \\alpha \\in U\\}\\)",
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            "content": "笔记像的维度对应着 \\( (\\alpha_{1},\\cdots,\\alpha_{n})A \\) 的秩，核的维度对应着线性方程组AX=0的解空间维数。",
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            "content": "5.2 线性映射"
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            "content": "定理 5.10 (核像维数公式)"
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            "content": "\\( f \\in \\text{Hom}(U, V) \\) 其核与像满足：",
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            "content": "\\[\n\\dim \\operatorname{Ker} (f) + \\dim \\operatorname{Im} (f) = \\dim (U)\n\\]"
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            "bbox": [
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            "content": "证明 设  \\( \\alpha_{1},\\cdots,\\alpha_{r} \\)  是 Kerf 的基,  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  是 U 的基, 而  \\( \\mathrm{Im}f=L(f(\\alpha_{1}),\\cdots,f(\\alpha_{n}))=L(f(\\alpha_{r+1}),\\cdots,f(\\alpha_{n})) \\) , 下面证明线性无关。假设  \\( k_{r+1}f(\\alpha_{r+1})+\\cdots+k_{n}f(\\alpha_{n})=0 \\) , 根据线性性可知  \\( f(k_{r+1}\\alpha_{r+1}+\\cdots+k_{n}\\alpha_{n})=0 \\) , 即  \\( k_{r+1}\\alpha_{r+1}+\\cdots+k_{n}\\alpha_{n}\\in\\mathrm{Ker}(f) \\) , 根据基的线性无关性, 显然  \\( k_{r+1},\\cdots,k_{n} \\)  均为 0",
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            "content": "笔记 该定理的证明会在题目中出现，主要方式是先设原空间的子空间的一组基，扩展成原空间的一组基，再作用 f 后与像空间进行联系。",
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            "content": "练习 5.12 核像维数公式推广证明 证明以下核像命题"
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            "content": "(1) 定义域为子空间: \\(\\mathcal{A}\\) 是有限维线性空间 \\(V\\) 的线性变换, \\(W\\) 是 \\(V\\) 的子空间, 证明:",
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            "content": "\\[\n\\dim (\\mathcal {A} W) + \\dim (\\operatorname{Ker} \\mathcal {A} \\cap W) = \\dim (W)\n\\]"
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            "content": "(2) 复合映射： \\( \\sigma,\\tau \\)  是有限维空间  \\( U\\to V,V\\to W \\)  的线性映射，证明：",
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            "content": "\\[\n\\dim (\\operatorname{Ker} (\\sigma)) + \\dim (\\operatorname{Im} (\\sigma) \\cap \\operatorname{Ker} (\\tau)) = \\dim (\\operatorname{Ker} (\\tau \\sigma))\n\\]"
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            "content": "证明 (1) 设  \\( \\alpha_{1},\\cdots,\\alpha_{r} \\)  是  \\( \\operatorname{Ker}\\mathcal{A}\\cap W \\)  的一组基， \\( \\alpha_{1},\\cdots,\\alpha_{n} \\)  是 W 的一组基，则  \\( \\mathcal{A}W=L(\\mathcal{A}\\alpha_{r+1},\\cdots,\\mathcal{A}\\alpha_{n}) \\) 。考虑 AW 的维数，设  \\( k_{r+1}\\mathcal{A}\\alpha_{r+1}+\\cdots+k_{n}\\mathcal{A}\\alpha_{n}=0 \\) ，则有  \\( k_{r+1}\\alpha_{r+1}+\\cdots+k_{n}\\alpha_{n}\\in\\operatorname{Ker}\\mathcal{A}\\cap W \\) ，因此得到  \\( k_{r+1}=\\cdots=k_{n}=0 \\) 。",
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            "content": "(2) 先得找原空间最小的空间，可知  \\( \\mathrm{Ker}(\\sigma) \\subseteq \\mathrm{Ker}(\\tau\\sigma) \\) ，因此设  \\( \\alpha_{1}, \\cdots, \\alpha_{r} \\)  是  \\( \\mathrm{Ker}(\\sigma) \\)  的基， \\( \\alpha_{1}, \\cdots, \\alpha_{n} \\)  是  \\( \\mathrm{Ker}(\\tau\\sigma) \\)  的基。设  \\( V_{1} = L(\\sigma\\alpha_{r+1}, \\cdots, \\sigma\\alpha_{n}) \\) 。显然  \\( V_{1} \\subseteq \\mathrm{Im}(\\sigma) \\cap \\mathrm{Ker}(\\tau) \\) 。",
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            "content": "另一侧 \\(\\forall \\alpha \\in \\mathrm{Im}(\\sigma) \\cap \\mathrm{Ker}(\\tau)\\)，\\(\\exists \\beta\\) 使得 \\(\\sigma(\\beta) = \\alpha, \\tau(\\alpha) = 0\\)，这说明 \\(\\tau(\\sigma(\\beta)) = 0\\)，因此 \\(\\beta \\in \\mathrm{Ker}(\\tau\\sigma)\\)，即 \\(\\alpha \\in V_1\\)，最终得出 \\(V_1 = \\mathrm{Im}(\\sigma) \\cap \\mathrm{Ker}(\\tau)\\)。",
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            "content": "定理 5.11 (核像与单射、满射)"
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            "content": "\\( \\sigma \\) 是 \\( U\\to W \\) 的线性映射，则：",
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            "content": "- \\(\\sigma\\) 是满射当且仅当 \\(\\operatorname{Im}(\\sigma) = W\\)",
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            "content": "- \\(\\sigma\\) 是单射当且仅当 \\(\\operatorname{Ker}(\\sigma) = \\{\\mathbf{0}\\}\\)",
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            "angle": 0,
            "content": "49"
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            "content": "- 推论：有限维线性空间上线性变换(非映射)是单射当且仅当其是满射(一般单射更好证)",
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            "content": "定理 5.12 (核、像任意性)"
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            "content": "\\(V\\) 是 \\(P\\) 上 \\(n\\) 维线性空间，\\(V_{1}, V_{2}\\) 是其子空间，\\(\\dim V_{1} + \\dim V_{2} = n\\)，则存在 \\(V\\) 上的线性变换 \\(\\mathcal{A}\\) 使得 \\(\\operatorname{Ker} \\mathcal{A} = V_{1}, \\operatorname{Im} \\mathcal{A} = V_{2}\\)",
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            "content": "证明 设  \\( V_{1} \\)  的一组基为  \\( \\alpha_{1},\\cdots,\\alpha_{r} \\) ，扩充为 V 的一组基  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) ，设  \\( \\beta_{r+1},\\cdots,\\beta_{n} \\)  是  \\( V_{2} \\)  的一组基，构造映射  \\( \\mathcal{A}(k_{1}\\alpha_{1}+\\cdots+k_{n}\\alpha_{n})=k_{r+1}\\beta_{r+1}+\\cdots+k_{n}\\beta_{n} \\) 。可证明 A 是线性变换，且核、像满足条件。",
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            "content": "练习 5.13 计算核与像 (1) \\( A: P^{4} \\rightarrow P^{3} \\) ，求 KerA, ImA",
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            "content": "\\[\n\\mathcal {A} \\left( \\begin{array}{c} x _ {1} \\\\ x _ {2} \\\\ x _ {3} \\\\ x _ {4} \\end{array} \\right) = \\left( \\begin{array}{c} x _ {1} - 3 x _ {2} + x _ {3} - 2 x _ {4} \\\\ - x _ {1} - 1 1 x _ {2} + 2 x _ {3} - 5 x _ {4} \\\\ 3 x _ {1} + 5 x _ {2} + x _ {4} \\end{array} \\right)\n\\]"
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            "content": "解 (1) \\(\\operatorname{Ker} \\mathcal{A}\\) 相当于 \\(AX = 0\\) 的解空间，\\(\\operatorname{Im} \\mathcal{A}\\) 相当于 \\(A\\) 的列空间",
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            "content": "5.3 线性变换"
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            "content": "5.2.3 幂等变换的核与像"
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            "content": "一般而言核与像虽然有维数关系，但不具备直和关系(可能相交也可能和不为全空间)，但是幂等变换是一个特例。",
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            "content": "定义 5.15 (幂等变换)"
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            "content": "若线性变换 \\(\\mathcal{A}\\) 满足 \\(\\mathcal{A}^2 = \\mathcal{A}\\)，则称其为幂等变换。",
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            "content": "定理 5.13 (幂等变换核像空间分解)"
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            "content": "\\(\\mathcal{A}\\) 是 \\(V\\) 上的幂等变换，则",
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            "content": "\\[\nV = \\operatorname{Ker} \\mathcal {A} \\oplus \\operatorname{Im} \\mathcal {A}\n\\]"
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            "content": "证明 (1) 先证明 \\(V\\) 是核与像的和: \\(\\forall \\alpha \\in V\\), 有 \\(\\alpha = \\alpha - \\mathcal{A}\\alpha + \\mathcal{A}\\alpha\\), 其中 \\(\\mathcal{A}\\alpha \\in \\operatorname{Im}\\mathcal{A}\\)。而 \\(\\mathcal{A}(\\alpha - \\mathcal{A}\\alpha) = \\mathcal{A}\\alpha - \\mathcal{A}^2\\alpha = 0\\), 因此 \\(\\alpha - \\mathcal{A}\\alpha \\in \\operatorname{Ker}\\mathcal{A}\\)",
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            "content": "(2) 证明直和: \\(\\forall \\alpha \\in \\operatorname{Ker} \\mathcal{A} \\cap \\operatorname{Im} \\mathcal{A}\\), 则 \\(\\mathcal{A} \\alpha = 0\\) 且 \\(\\exists \\beta, \\mathcal{A} \\beta = \\alpha\\), 两侧同时作用 \\(\\mathcal{A}\\), 左侧得到 \\(\\mathcal{A}^2 \\beta = \\mathcal{A} \\beta = \\alpha\\), 右侧 \\(\\mathcal{A} \\alpha = 0\\), 因此 \\(\\alpha = 0\\)",
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            "content": "练习 5.14 经典幂等变换核像题 A, B 为 P 上线性空间中的幂等变换，证明：",
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            "content": "(1) \\(\\mathcal{A},\\mathcal{B}\\) 有相同的像当且仅当 \\(\\mathcal{AB} = \\mathcal{B},\\mathcal{BA} = \\mathcal{A}\\)",
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            "content": "(2) \\(\\mathcal{A},\\mathcal{B}\\) 有相同的核当且仅当 \\(\\mathcal{AB} = \\mathcal{A},\\mathcal{BA} = \\mathcal{B}\\)",
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            "content": "证明 (1) 左推右: \\(\\forall \\alpha \\in V\\) 有 \\(\\mathcal{B} \\alpha = \\in \\operatorname{Im}(\\mathcal{B})\\), 根据条件即 \\(\\mathcal{B} \\alpha \\in \\operatorname{Im} \\mathcal{A}\\), 因此根据幂等变换投影的性质 (\\(\\mathcal{A}\\) 不改变 \\(\\operatorname{Im} \\mathcal{A}\\) 中的向量) 得到",
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            "content": "\\[\n\\mathcal {A} (\\mathcal {B} \\alpha) = \\mathcal {B} \\alpha \\Rightarrow \\mathcal {A B} = \\mathcal {B}\n\\]"
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            "content": "同理得到 \\(\\mathcal{BA} = \\mathcal{A}\\)",
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            "content": "右推左：\\(\\forall \\gamma \\in \\operatorname{Im}(\\mathcal{A})\\)，有 \\(\\mathcal{A}\\gamma = \\gamma\\)，而 \\(\\mathcal{B}\\mathcal{A}\\gamma = \\mathcal{A}\\gamma = \\gamma\\)，因此 \\(\\mathcal{B}\\gamma = \\gamma\\)，即 \\(\\gamma \\in \\operatorname{Im}(\\mathcal{B})\\)，另一侧同理",
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            "content": "(2) 左推右：由于  \\( V = \\operatorname{Im}(\\mathcal{A}) \\oplus \\operatorname{Ker}(\\mathcal{A}) \\) ，因此  \\( \\forall \\alpha \\in V \\)  有",
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            "content": "\\[\n\\alpha = \\alpha_ {1} + \\alpha_ {2}, \\alpha_ {1} \\in \\operatorname{Im} (\\mathcal {A}), \\alpha_ {2} \\in \\operatorname{Ker} \\mathcal {A}\n\\]"
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            "content": "\\(\\alpha - \\alpha_{1} \\in \\operatorname{Ker} \\mathcal{A}\\)，因此 \\(\\alpha - \\alpha_{1} \\in \\operatorname{Ker} \\mathcal{B}\\)，故 \\(\\mathcal{B}(\\alpha - \\alpha_{1}) = 0\\)，即 \\(\\mathcal{B} \\alpha = \\mathcal{B} \\alpha_{1}\\)。而 \\(\\mathcal{A} \\alpha = \\alpha_{1}\\)，得到",
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            "content": "\\[\n(\\mathcal {B A}) \\alpha = \\mathcal {B} \\alpha_ {1} = \\mathcal {B} \\alpha\n\\]"
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            "content": "因此得出 \\(\\mathcal{BA} = \\mathcal{B}\\)",
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            "content": "右推左",
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            "content": "5.2.4 一般线性变换的核像分解"
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            "content": "定理 5.14 (线性变换核像空间分解)"
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            "content": "- \\(T\\) 是单射",
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            "content": "- \\(T\\) 是满射",
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            "content": "证明 (1) 从矩阵角度看即可",
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            "content": "(2) 左推右：若 \\(T\\alpha_{1} = T\\alpha_{2}\\)，则两侧作用 \\(T^{-1}\\) 得到 \\(\\alpha_{1} = \\alpha_{2}\\)",
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            "content": "右推左：下证  \\( T\\epsilon_{1},\\cdots,T\\epsilon_{n} \\)  线性无关，若  \\( k_{1}T\\epsilon_{1}+\\cdots+k_{n}T\\epsilon_{n}=0 \\) ，则",
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            "content": "\\[\nT (k _ {1} \\epsilon_ {1} + \\dots + k _ {n} \\epsilon_ {n}) = 0\n\\]"
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            "content": "由于 T 是单射，得到  \\( k_{1}\\epsilon_{1}+\\cdots+k_{n}\\epsilon_{n}=0 \\) ，因此  \\( k_{1}=\\cdots=k_{n}=0 \\) ，根据命题 (1) 知可逆",
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            "content": "(3) 左推右：设 \\(T\\) 可逆，则 \\(\\forall \\alpha \\in V\\)，存在 \\(T^{-1}\\alpha\\)，而 \\(\\alpha = T(T^{-1}\\alpha)\\)，因此是满射",
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            "content": "右推左：由于满射，对任一一组基  \\( \\eta_{1},\\cdots,\\eta_{n} \\) ，总存在  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\) ，使得",
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            "content": "\\[\nT \\epsilon_ {1} = \\eta_ {1}, \\dots , T \\epsilon_ {n} = \\eta_ {n}\n\\]"
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            "content": "\\[\nT (k _ {1} \\epsilon_ {1} + \\dots + k _ {n} \\epsilon_ {n}) = k _ {1} \\eta_ {1} + \\dots + k _ {n} \\eta_ {n} = 0\n\\]"
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            "content": "定义 5.17 (线性变换的矩阵)"
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            "content": "5.3 线性变换"
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            "content": "练习5.16线性变换在基下的矩阵已知 \\(A = (a_{ij}),i,j = 1,2,3\\) 是 \\(\\mathcal{A}\\) 在 \\(\\epsilon_1,\\epsilon_2,\\epsilon_3\\) 下的矩阵，计算",
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            "content": "(1) \\(\\mathcal{A}\\) 在 \\(\\epsilon_3, \\epsilon_2, \\epsilon_1\\) 下的矩阵",
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            "content": "(2) \\(\\mathcal{A}\\) 在 \\(\\epsilon_1, k\\epsilon_2, \\epsilon_3\\) 下的矩阵",
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            "content": "(3) \\(\\mathcal{A}\\) 在 \\(\\epsilon_1 + \\epsilon_2, \\epsilon_2, \\epsilon_3\\) 下的矩阵",
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            "content": "解 根据 \\(T(\\epsilon_1, \\epsilon_2, \\epsilon_3) = (\\epsilon_1, \\epsilon_2, \\epsilon_3)A\\)，得到",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} T \\epsilon_ {1} = a _ {1 1} \\epsilon_ {1} + a _ {2 1} \\epsilon_ {2} + a _ {3 1} \\epsilon_ {3} \\\\ T \\epsilon_ {2} = a _ {1 2} \\epsilon_ {1} + a _ {2 2} \\epsilon_ {2} + a _ {3 2} \\epsilon_ {3} \\\\ T \\epsilon_ {3} = a _ {1 3} \\epsilon_ {1} + a _ {2 3} \\epsilon_ {2} + a _ {3 3} \\epsilon_ {3} \\end{array} \\right.\n\\]"
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            "content": "(1) 重新排列得到:",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} T \\epsilon_ {3} = a _ {3 3} \\epsilon_ {3} + a _ {2 3} \\epsilon_ {2} + a _ {1 3} \\epsilon_ {1} \\\\ T \\epsilon_ {2} = a _ {3 2} \\epsilon_ {3} + a _ {2 2} \\epsilon_ {2} + a _ {1 2} \\epsilon_ {1} \\\\ T \\epsilon_ {1} = a _ {3 1} \\epsilon_ {3} + a _ {2 1} \\epsilon_ {2} + a _ {1 1} \\epsilon_ {1} \\end{array} \\right.\n\\]"
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            "content": "得到矩阵为",
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            "content": "\\[\n\\left[ \\begin{array}{l l l} a _ {3 3} & a _ {3 2} & a _ {3 1} \\\\ a _ {2 3} & a _ {2 2} & a _ {2 1} \\\\ a _ {1 3} & a _ {1 2} & a _ {1 1} \\end{array} \\right]\n\\]"
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            "content": "(2)(3) 同理列开即可",
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            "content": "练习 5.17 幂零变换的矩阵 T 是 n 维线性空间 V 上的线性变换， \\( T^{n-1} \\neq O, T^n = O \\) ，证明：T 在某组基下的矩阵为",
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            "content": "\\[\n\\left[ \\begin{array}{c c c c c} 0 & 0 & \\dots & 0 & 0 \\\\ 1 & 0 & \\dots & 0 & 0 \\\\ \\vdots & \\vdots & & \\vdots & \\vdots \\\\ 0 & 0 & \\dots & 1 & 0 \\end{array} \\right]\n\\]"
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            "content": "证明 首先  \\( \\alpha, T\\alpha, \\cdots, T^{n-1}\\alpha \\)  均线性无关，因此它们是 V 的一组基，T 在这组基下的矩阵即目标。",
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            "content": "定理 5.16 (线性变换后的坐标)"
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            "content": "若  \\( \\alpha \\)  在  \\( (\\epsilon_{1},\\cdots,\\epsilon_{n}) \\)  基下的坐标为 X，则  \\( A\\alpha \\)  在这组基下的坐标为 AX。",
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            "content": "证明  \\( \\mathcal{A}\\alpha=\\mathcal{A}(x_{1}\\epsilon_{1}+\\cdots+x_{n}\\epsilon_{n})=\\mathcal{A}(\\epsilon_{1},\\cdots,\\epsilon_{n})X=(\\epsilon_{1},\\cdots,\\epsilon_{n})AX, \\)",
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            "content": "5.3.3 线性变换在不同基下的矩阵：相似关系"
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            "content": "定理 5.17 (线性变换在不同基下的矩阵)"
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            "content": "\\( \\epsilon_{i},\\eta_{i}(i=1,\\cdots,n) \\) 是 V 两组基，且 \\( (\\eta_{1},\\cdots,\\eta_{n})=(\\epsilon_{1},\\cdots,\\epsilon_{n})T \\) 。若 A 在 \\( \\epsilon_{i},\\eta_{i} \\) 基下的矩阵分别为 A,B，则有 \\( B=T^{-1}AT \\) 。",
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            "content": "证明 由线性变换矩阵定义， \\( \\mathcal{A}(\\epsilon_{1},\\cdots,\\epsilon_{n})=(\\epsilon_{1},\\cdots,\\epsilon_{n})A,\\mathcal{A}(\\eta_{1},\\cdots,\\eta_{n})=(\\eta_{1},\\cdots,\\eta_{n})B \\) 。而  \\( \\mathcal{A}(\\eta_{1},\\cdots,\\eta_{n})=\\mathcal{A}(\\epsilon_{1},\\cdots,\\epsilon_{n})T=(\\epsilon_{1},\\cdots,\\epsilon_{n})AT \\) ，且  \\( \\mathcal{A}(\\eta_{1},\\cdots,\\eta_{n})=(\\epsilon_{1},\\cdots,\\epsilon_{n})TB \\)  上面两式取等号即可。",
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            "content": "5.4 特征值、相似对角化"
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            "content": "练习 5.18 线性变换在不同基下的矩阵 给定下述两组基，线性变换 T 满足  \\( T \\epsilon_{i} = \\eta_{i} \\) ,",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\epsilon_ {1} = (1, 0, 1) ^ {T}, \\epsilon_ {2} = (2, 1, 0) ^ {T}, \\epsilon_ {3} = (1, 1, 1) ^ {T} \\\\ \\eta_ {1} = (1, 2, - 1) ^ {T}, \\eta_ {2} = (2, 2, - 1) ^ {T}, \\eta_ {3} = (2, - 1, - 1) ^ {T} \\end{array} \\right.\n\\]"
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            "content": "(1) 写出 \\(\\epsilon_{1}, \\epsilon_{2}, \\epsilon_{3}\\) 到 \\(\\eta_{1}, \\eta_{2}, \\eta_{3}\\) 的过渡矩阵",
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            "content": "(2) 写出 \\(T\\) 在 \\(\\epsilon_1, \\epsilon_2, \\epsilon_3\\) 下的矩阵",
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            "content": "(3) 写出 \\(T\\) 在 \\(\\eta_1, \\eta_2, \\eta_3\\) 下的矩阵",
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            "content": "解 (1) \\( e_{1}, e_{2}, e_{3} \\) 下的矩阵为",
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            "content": "\\[\n(\\eta_ {1}, \\eta_ {2}, \\eta_ {3}) =\n\\]"
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            "content": "定理 5.18 (线性变换构造的等价类)"
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            "content": "矩阵 \\(A, B\\) 相似的充要条件为它们是一个线性变换在两组基下的矩阵。",
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            "content": "5.3.4 相似矩阵的性质"
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            "content": "定理 5.19 (相似矩阵的基本性质)"
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            "content": "\\(K\\) 上矩阵 \\(A, B\\) 相似，且满足 \\(B = P^{-1}AP\\) ，则",
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            "content": "•  \\( r(A) = r(B) \\)",
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            "content": "- \\(f(x)\\) 为多项式，则 \\(f(A) \\sim f(B)\\)",
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            "content": "若 \\(A_{1} \\sim B_{1}, A_{2} \\sim B_{2}\\)，不一定有 \\(A_{1} + A_{2} \\sim B_{1} + B_{2}\\)，要求过渡矩阵相同才行",
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            "content": "证明 (1) \\( r(A)=r(B) \\) ：因为  \\( B=P^{-1}AP \\) ，而 P 满秩，等价于一系列初等变换，故不改变秩。",
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            "content": "5.4 特征值、相似对角化"
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            "content": "5.4.1 特征值与特征向量"
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            "content": "\\(\\mathcal{A}\\) 是数域 \\(\\mathbb{P}\\) 上线性空间 \\(V\\) 内的一个线性变换，若对于 \\(\\lambda \\in \\mathbb{P}\\)，\\(\\exists \\xi \\neq 0\\)，使得 \\(\\mathcal{A}\\xi = \\lambda \\xi\\)，则称 \\(\\lambda\\) 是 \\(\\mathcal{A}\\) 的一个特征值，\\(\\xi\\) 是 \\(\\mathcal{A}\\) 属于 \\(\\lambda\\) 的特征向量。",
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            "content": "对于  \\( \\lambda\\in P \\) ，定义其特征子空间为所有满足  \\( A\\xi=\\lambda\\xi \\)  的特征向量  \\( \\xi \\)  所组成的集合  \\( V_{\\lambda} \\) 。",
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            "content": "引理 5.1 (特征子空间的性质)"
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            "content": "特征子空间 \\(V_{\\lambda}\\) 有如下性质：",
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            "content": "- 对于 \\(\\lambda \\in \\mathbb{P}\\), \\(V_{\\lambda} \\neq \\{\\mathbf{0}\\}\\) 的充要条件为 \\(|\\lambda E - A| = 0\\), 其中 \\(A\\) 是 \\(\\mathcal{A}\\) 在任意基下的矩阵。",
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            "content": "5.4 特征值、相似对角化"
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            "content": "定义 5.20 (特征多项式)"
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            "content": "对于 P 上的矩阵 A，特征多项式定义为  \\( f(\\lambda) = |\\lambda E - A| \\) 。",
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            "content": "定理 5.20 (相似矩阵的特征多项式)"
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            "content": "相似矩阵有相同特征多项式。即对于同一个线性变换，特征多项式是不变的。",
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            "content": "证明 若 \\(B = T^{-1}AT\\)，则 \\(|\\lambda E - B| = |\\lambda E - T^{-1}AT| = |T^{-1}||\\lambda E - A||T| = |\\lambda E - A|\\)。",
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            "content": "定理 5.21 (特征值基本性质)"
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            "content": "\\( \\lambda_{1},\\cdots,\\lambda_{n} \\) 是矩阵 A 的特征值。则 \\( \\sum\\lambda_{i}=\\mathrm{tr}(A),\\quad\\prod\\lambda_{i}=|A| \\)",
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            "content": "证明 用行列式展开与韦达定理证明。",
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            "content": "练习 5.19 特征值与特征向量的基本练习 (1) \\( \\lambda_{1}, \\lambda_{2} \\) 是 T 两个不同的特征值，\\( \\alpha_{1}, \\alpha_{2} \\) 分别是对应的特征向量，证明：\\( \\alpha_{1} + \\alpha_{2} \\) 不是 T 的特征向量",
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            "content": "(2) \\(T\\) 是可逆线性变换，若 \\(\\lambda\\) 是 \\(T\\) 的特征值，则 \\(\\lambda^{-1}\\) 是 \\(T^{-1}\\) 的特征值证明 (1) \\(T\\alpha_{1} = \\lambda_{1}\\alpha_{1}, T\\alpha_{2} = \\lambda_{2}\\alpha_{2}\\)，因此",
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            "content": "\\[\nT (\\alpha_ {1} + \\alpha_ {2}) = \\lambda_ {1} \\alpha_ {1} + \\lambda_ {2} \\alpha_ {2}\n\\]"
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            "content": "若 \\(\\alpha_{1} + \\alpha_{2}\\) 是特征向量，\\(\\lambda\\) 是特征值，则 \\(T(\\alpha_{1} + \\alpha_{2}) = \\lambda (\\alpha_{1} + \\alpha_{2})\\) ，得到",
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            "content": "\\[\n(\\lambda - \\lambda_ {1}) \\alpha_ {1} + (\\lambda - \\lambda_ {2}) \\alpha_ {2} = 0\n\\]"
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            "content": "因此  \\( \\lambda - \\lambda_{1} = \\lambda - \\lambda_{2} = 0 \\) ，这与假设矛盾。",
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            "content": "(2) 由于 \\(T\\alpha = \\lambda \\alpha\\)，因此 \\(\\alpha = \\lambda T^{-1}(\\alpha)\\)，即 \\(T^{-1}(\\alpha) = \\lambda^{-1}\\alpha\\)，因此 \\(\\lambda^{-1}\\) 是 \\(T^{-1}\\) 的特征值",
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            "content": "5.4.2 特征值与特征向量进一步结论"
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            "content": "定理 5.22 (AB 与 BA 的特征值)"
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            "content": "54"
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            "content": "\\( V, V' \\) 是 K 上的线性空间， \\( \\mathcal{A} \\in \\mathrm{Hom}(V, V') \\) ， \\( \\mathcal{B} \\in \\mathrm{Hom}(V', V) \\) （矩阵同理），则\n- AB 与 BA 有相同非零特征值",
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            "content": "- 若 \\(\\xi\\) 是 \\(AB\\) 属于 \\(\\lambda_0\\) 的特征向量，则 \\(B\\xi\\) 是 \\(BA\\) 属于 \\(\\lambda_0\\) 的一个特征向量",
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            "content": "证明 设  \\( \\lambda_{0} \\)  是 AB 非零特征值， \\( \\exists\\xi\\in V^{\\prime} \\)  使得  \\( (\\mathcal{AB})\\xi=\\lambda_{0}\\xi \\) ，两侧同时作用 B 得到：",
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            "content": "\\[\n\\mathcal {B} (\\mathcal {A B}) \\xi = \\mathcal {B} (\\lambda_ {0} \\xi) \\Rightarrow (\\mathcal {B A}) (\\mathcal {B} \\xi) = \\lambda_ {0} (\\mathcal {B} \\xi)\n\\]"
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            "content": "\\(A_{m\\times n},B_{n\\times m}\\) ， \\(f_{AB}(\\lambda),f_{BA}(\\lambda)\\) 是 \\(AB,BA\\) 的特征多项式，则",
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            "content": "\\[\n\\lambda^ {n} f _ {A B} (\\lambda) = \\lambda^ {m} f _ {B A} (\\lambda)\n\\]"
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            "content": "证明 证明：设 \\(\\lambda \\neq 0\\) ，根据",
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            "content": "5.4 特征值、相似对角化"
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            "content": "同理根据",
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            "content": "\\[\n\\left[ \\begin{array}{c c} E _ {n} & \\frac {1}{\\lambda} B \\\\ A & E _ {m} \\end{array} \\right] \\cdot \\left[ \\begin{array}{c c} E _ {n} & 0 \\\\ - A & E _ {m} \\end{array} \\right] = \\left[ \\begin{array}{c c} E _ {n} - \\frac {1}{\\lambda} A B & \\frac {1}{\\lambda} B \\\\ 0 & E _ {m} \\end{array} \\right] \\Rightarrow | H | = \\left(\\frac {1}{\\lambda}\\right) ^ {n} | \\lambda E _ {n} - B A |\n\\]"
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            "content": "根据  \\( |H| \\)  的值即可推出结论。",
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            "content": "练习5.20相关结论(1) \\(A, B\\) 是两个 \\(n\\) 阶方阵，证明：\\(AB + A\\) 和 \\(BA + A\\) 有相同的特征多项式",
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            "content": "(2) 计算下面矩阵 \\(C\\) 的特征值",
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            "content": "\\[\nC = \\left[ \\begin{array}{c c c c} a _ {1} b _ {1} & a _ {1} b _ {2} & \\dots & a _ {1} b _ {n} \\\\ a _ {2} b _ {1} & a _ {2} b _ {2} & \\dots & a _ {2} b _ {n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ a _ {n} b _ {1} & a _ {n} b _ {2} & \\dots & a _ {n} b _ {n} \\end{array} \\right]\n\\]"
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            "content": "(3) 特例：已知  \\( A = (a_{1}, \\cdots, a_{n}) \\) ，证明  \\( A^{T}A \\)  的特征值为 0 和  \\( a_{1}^{2} + \\cdots + a_{n}^{2} \\) ，这里 0 是 n - 1 重根证明 (1) 由于  \\( AB + A = A(B + E) \\) ， \\( BA + A = (B + E)A \\) ，因此根据前面定理可知。",
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            "content": "(2) 令 \\(A = (a_{1}, \\cdots, a_{n})^{T}\\), \\(B = (b_{1}, \\cdots, b_{n})\\), 则 \\(C = AB\\), 而 \\(BA = \\sum_{i=1}^{n} a_{i} b_{i}\\) 为一阶矩阵, 其特征值为 \\(\\sum_{i=1}^{n} a_{i} b_{i}\\), \\(AB\\) 和 \\(BA\\) 特征多项式相差 \\(\\lambda^{n-1}\\), 因此 \\(C = AB\\) 的特征多项式为",
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            "content": "\\[\n\\lambda^ {n - 1} (\\lambda - \\sum_ {i = 1} ^ {n} a _ {i} b _ {i})\n\\]"
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            "content": "(3) 即 (2) 的特例",
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            "content": "定理 5.24 (可逆的特征值表达)"
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            "content": "\\(n\\) 阶方阵 \\(A\\) 可逆当且仅当 \\(A\\) 的特征值全不为0",
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            "content": "证明 用  \\( \\left|A\\right|=\\lambda_{1}\\cdots\\lambda_{n} \\)  即可",
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            "content": "定理 5.25 (逆矩阵的特征值)"
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            "content": "n 阶可逆矩阵 A 的特征值为  \\( \\lambda_{1},\\cdots,\\lambda_{n} \\) ，则  \\( A^{-1} \\)  的全部特征值为  \\( \\lambda_{1}^{-1},\\cdots,\\lambda_{n}^{-1} \\)",
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            "content": "证明 用 A 相似于 Jordan 阵 J， \\( J^{-1} \\)  为上三角且对角为  \\( \\lambda_{1}^{-1},\\cdots,\\lambda_{n}^{-1} \\) ，因此  \\( A^{-1} \\)  的特征值也为  \\( \\lambda_{1}^{-1},\\cdots,\\lambda_{n}^{-1} \\)",
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            "content": "5.4.3 相似的进一步研究"
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            "content": "定理 5.26 (相似的矩阵分解)"
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            "content": "\\(A, B\\) 相似当且仅当存在方阵 \\(P, Q\\)，使得",
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            "content": "\\[\nA = P Q, B = Q P\n\\]"
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            "content": "这里 \\(P, Q\\) 至少有一个可逆",
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            "content": "证明 由于 \\(A, B\\) 相似，故存在可逆 \\(P\\) 使得 \\(P^{-1}AP = B\\) ，令 \\(Q = P^{-1}A\\) ，则得到 \\(A = PQ, B = QP\\)",
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            "content": "定理 5.27 (一些相似结论)"
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            "content": "\\(A, B\\) 为 \\(n\\) 阶方阵，则",
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            "content": "- \\(AB\\) 与 \\(BA\\): 若方阵 \\(A\\) 可逆, 则 \\(AB\\) 与 \\(BA\\) 相似",
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            "content": "- \\(A^{-1}\\) 与 \\(B^{-1}\\): 若 \\(A, B\\) 可逆, 则 \\(A^{-1}\\) 与 \\(B^{-1}\\) 相似",
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            "content": "- 多项式：任一多项式 \\(f(x)\\)，有 \\(f(A), f(B)\\) 相似",
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            "content": "5.4 特征值、相似对角化"
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            "angle": 0,
            "content": "- 伴随矩阵：若 \\(A, B\\) 相似，则 \\(A^{*}\\) 与 \\(B^{*}\\) 相似",
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            "content": "证明 (1) \\(BA = (A^{-1}A)(BA) = A^{-1}(AB)A\\)",
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            "angle": 0,
            "content": "(2) \\((P^{-1}AP)^{-1} = B^{-1}\\), 得到 \\(P^{-1}A^{-1}P = B^{-1}\\)",
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            "content": "(3) 显然",
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            "content": "\\[\n(4) B ^ {*} = (P ^ {- 1} A P) ^ {*} = P ^ {*} A ^ {*} (P ^ {- 1}) ^ {*} = P ^ {*} A ^ {*} (P ^ {*}) ^ {- 1}\n\\]"
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            "content": "推论5.2(伴随矩阵的特征值)"
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            "content": "若矩阵 A 的特征值为  \\( \\lambda_{1},\\cdots,\\lambda_{m} \\) ，则  \\( A^{*} \\)  的特征值为  \\( \\prod_{i\\neq j}\\lambda_{i} \\)",
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            "content": "证明 考虑 \\(A\\) 在 \\(\\mathbb{C}\\) 上相似于 Jordan 阵 \\(J\\)，而 \\(J^{*}\\) 的特征值易算，相似时特征值不变，因此可算出 \\(A^{*}\\) 的特征值。",
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            "content": "5.4.4 相似对角化"
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            "content": "定义 5.21 (相似对角化)"
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            "content": "\\(\\mathcal{A}\\) 是 \\(V\\) 内的线性变换，如果存在 \\(V\\) 的一组基使得 \\(\\mathcal{A}\\) 在该基下的矩阵为对角阵，则称 \\(\\mathcal{A}\\) 可对角化。",
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            "content": "定义 5.22 (代数重数、几何重数)",
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            "content": "代数重数指特征值 \\(\\lambda_{i}\\) 在特征多项式 \\(f(\\lambda)\\) 中的重数。几何重数指 \\(\\lambda_{i}\\) 对应特征子空间的维数。",
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            "content": "引理 5.2 (代数重数与几何重数关系)"
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            "content": "对每个特征值 \\(\\lambda_{i}\\)，其代数重数大于等于几何重数。",
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            "content": "56"
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            "content": "引理 5.3 (不同特征值的特征向量)"
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            "angle": 0,
            "content": "A 属于不同特征值的特征向量线性无关。",
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            "content": "证明 设  \\( \\lambda_{1},\\cdots,\\lambda_{s} \\)  是 A 的 s 个不同的特征值，对应特征向量  \\( \\xi_{1},\\cdots,\\xi_{s} \\) 。用归纳法，假设  \\( \\xi_{1},\\cdots,\\xi_{k-1} \\)  线性无关，若",
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            "angle": 0,
            "content": "c _ {1} \\xi_ {1} + \\dots + c _ {k} \\xi_ {k} = 0\\tag{5.1}"
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            "content": "两侧作用 A 得到  \\( c_{1}\\lambda_{1}\\xi_{1}+\\cdots+c_{k}\\lambda_{k}\\xi_{k}=0 \\) ，用该式子减去  \\( \\lambda_{k} \\)  倍的 (5.1) 得到",
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            "content": "\\[\nc _ {1} \\left(\\lambda_ {1} - \\lambda_ {k}\\right) \\xi_ {1} + \\dots + c _ {k - 1} \\left(\\lambda_ {k - 1} - \\lambda_ {k}\\right) \\xi_ {k - 1} = 0\n\\]"
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            "content": "由于  \\( \\lambda_{i} \\)  两两不相等，因此  \\( c_{1}=\\cdots=c_{k-1}=0 \\) ，即特征向量彼此线性无关。",
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            "content": "- 充要条件3：所有特征值的代数重数等于几何重数",
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            "content": "- 充要条件4： \\(\\dim V = \\sum_{i=1}^{m} \\dim(V_i)\\)",
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            "content": "5.5 最小多项式"
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            "content": "证明 (1) 左推右：由对角化形式，右侧乘开 \\(n\\) 个基向量均是特征向量，显然基向量都线性无关。",
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            "content": "\\[\n(A \\eta_ {1}, A \\eta_ {2}, \\dots , A \\eta_ {n}) = (\\eta_ {1}, \\eta_ {2}, \\dots , \\eta_ {n}) \\left[ \\begin{array}{c c c} \\lambda_ {1} & & \\\\ & \\lambda_ {2} & \\\\ & \\ddots & \\lambda_ {n} \\end{array} \\right]\n\\]"
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            "content": "右推左：n 个线性无关特征向量当一组基即可。",
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            "content": "练习5.21对角化计算(1)计算 \\(A = \\left( \\begin{array}{ll}3 & 4\\\\ 5 & 2 \\end{array} \\right)\\) 的特征值，并计算 \\(Q\\) 使得 \\(Q^{-1}AQ\\) 是对角矩阵",
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            "content": "定理 5.29 (相关矩阵的可对角化性质)"
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            "content": "证明 (1) 设  \\( P^{-1}AP = \\text{diag}\\{\\lambda_1, \\cdots, \\lambda_n\\} \\) ，两侧取伴随得到  \\( P^*A^*(P^*)^{-1} = P^{-1}A^*P = \\text{diag}\\{\\prod_{i \\neq 1} \\lambda_i, \\prod_{i \\neq 2} \\lambda_i, \\cdots, \\prod_{i \\neq n} \\lambda_i\\} \\) 。其中伴随转逆用了  \\( A^{-1} = \\frac{1}{|A|}A^* \\)",
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            "content": "证明 n=1 时显然成立。假设 n-1 时成立，则取  \\( x_{1} \\)  为 A 对应  \\( \\lambda_{1} \\)  的特征向量，构造矩阵  \\( P_{1}=[x_{1},\\cdots,x_{n}](x_{2}\\sim x_{n}) \\)  无要求)，可以得到：",
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            "content": "\\[\nA P _ {1} = (x _ {1}, \\dots , x _ {n}) \\left[ \\begin{array}{c c c c} \\lambda_ {1} & b _ {1 2} & \\dots & b _ {1 n} \\\\ 0 & b _ {2 2} & \\dots & b _ {2 n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ 0 & b _ {n 1} & \\dots & b _ {n n} \\end{array} \\right] \\Rightarrow P _ {1} ^ {- 1} A P _ {1} = \\left[ \\begin{array}{c c c c} \\lambda_ {1} & b _ {1 2} & \\dots & b _ {1 n} \\\\ 0 & b _ {2 2} & \\dots & b _ {2 n} \\\\ \\vdots & \\vdots & & \\vdots \\\\ 0 & b _ {n 1} & \\dots & b _ {n n} \\end{array} \\right]. = \\left[ \\begin{array}{c c} \\lambda_ {1} & * \\\\ 0 & A _ {1} \\end{array} \\right]\n\\]"
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            "content": "5.5 最小多项式"
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            "content": "5.5 最小多项式"
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            "content": "5.5.1 最小多项式与基本性质"
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            "content": "定义 5.23 (零化多项式与最小多项式)"
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            "content": "\\(A\\) 是数域 \\(K\\) 上的 \\(n\\) 阶方阵，\\(g(x)\\) 是 \\(K\\) 上的多项式，",
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            "angle": 0,
            "content": "- 零化多项式：若 \\(g(A) = 0\\) ，则称 \\(g(x)\\) 是 \\(A\\) 的一个零化多项式。",
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            "angle": 0,
            "content": "- 最小多项式：将 \\(A\\) 的次数最低首一零化多项式称为 \\(A\\) 的最小多项式",
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            "content": "定理 5.31 (最小多项式的性质)"
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            "content": "最小多项式有以下性质：",
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            "content": "- 唯一性：最小多项式是唯一的",
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            "angle": 0,
            "content": "- 整除性：最小多项式整除任意零化多项式",
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            "content": "- 分块对角： \\( A = \\text{diag}\\{A_1, \\cdots, A_s\\} \\)，则 \\( m_A(\\lambda) = [m_1(\\lambda), \\cdots, m_s(\\lambda)] \\) (最小公倍数)",
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            "content": "相似：相似矩阵有相同最小多项式",
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            "content": "证明 (1) 假设有两个不同最小多项式 \\(m_{1}(\\lambda), m_{2}(\\lambda)\\)，则令 \\(h(\\lambda) = m_{1}(\\lambda) - m_{2}(\\lambda)\\)，有 \\(h(A) = m_{1}(A) - m_{2}(A) = 0\\)，显然 \\(h(\\lambda)\\) 是最小多项式，矛盾。",
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            "content": "(2) 对任意零化多项式 \\( f(x) \\)，其可以表示为 \\( f(x) = q(x)m(x) + r(x) \\)，代入 \\( x = A \\) 得到 \\( r(A) = 0 \\)，其只能为 \\( r(x) = 0 \\)，否则与最小多项式矛盾，因此整除。",
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            "content": "(3) 由于  \\( m(A)=\\mathrm{diag}\\{m(A_{1}),\\cdots,m(A_{s})\\} \\) ，最大公因式即可满足所有分块均为 0。",
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            "content": "(4) 若 \\(B = P^{-1}AP\\)，则 \\(f(B) = f(P^{-1}AP) = P^{-1}f(A)P\\)，因此可知最小多项式相同。",
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            "content": "5.5.2 Hamilton-Cayley 定理"
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            "content": "定理 5.32 (Hamilton-Cayley 定理)"
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            "content": "\\(A\\) 是 \\(K\\) 上的 \\(n\\) 阶矩阵，\\(f(\\lambda) = |\\lambda E - A|\\) 即 \\(A\\) 的特征多项式，则 \\(f(x)\\) 是 \\(A\\) 的零化多项式。",
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            "content": "证明 设  \\( B(\\lambda) \\)  是  \\( \\lambda E - A \\)  的伴随矩阵，则  \\( B(\\lambda)(\\lambda E - A) = f(\\lambda)E \\) ，由于  \\( B(\\lambda) \\)  元素含  \\( 1 \\sim \\lambda^{n-1} \\) ，因此可以写为  \\( B(\\lambda) = \\lambda^{n-1}B_{0} + \\cdots + B_{n-1} \\) ，从而得到：",
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            "content": "\\[\nB (\\lambda) (\\lambda E - A) = \\lambda^ {n} B _ {0} + \\lambda^ {n - 1} (B _ {1} - B _ {0} A) + \\dots + \\lambda (B _ {n - 1} - B _ {n - 2} A) - B _ {n - 1} A\n\\]"
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            "content": "根据  \\( B(\\lambda)(\\lambda E - A) = f(\\lambda)E = \\lambda^{n}E + a_{1}\\lambda^{n-1}E + \\cdots + a_{n-1}\\lambda E + a_{n}E \\)  进行比较得到：",
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            "content": "- 最小多项式与特征多项式同根(重数可以不同)。",
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            "content": "定理 5.33 (最小多项式与相似对角化)",
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            "content": "\\(\\mathbb{P}\\) 上的 \\(n\\) 阶矩阵 \\(A\\) 可对角化当且仅当 \\(A\\) 的最小多项式在 \\(\\mathbb{P}\\) 上可分解为互素一次因子乘积。",
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            "content": "第 6 章 Jordan 标准型"
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            "content": "6.1 Jordan 标准形：空间分解法"
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            "content": "定义 6.1 (幂零线性变换)"
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            "content": "A 是 V 内的线性变换，若  \\( \\exists m \\)  使得  \\( A^{m}=0 \\) ，则 A 是幂零线性变换。",
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            "content": "引理6.1(幂零基)"
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            "content": "A 是幂零变换， \\( \\forall\\alpha\\in V \\)  若  \\( A^{k-1}\\alpha\\neq0,A^{k}\\alpha=0 \\) ，则  \\( \\alpha,A\\alpha,\\cdots,A^{k-1}\\alpha \\)  线性无关。",
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            "content": "证明 设  \\( a_{0}\\alpha + a_{1}A\\alpha + \\cdots + a_{k-1}A^{k-1}\\alpha = 0 \\) ，若从左至右第一个非零的是  \\( a_{i} \\) ，即  \\( a_{i}A^{i}\\alpha + \\cdots + a_{k}A^{k-1}\\alpha = 0 \\) ，两侧同时作用  \\( A^{k-1-i} \\) ，上式变为  \\( a_{i}A^{k-1}\\alpha = 0 \\) ，而  \\( A^{k-1}\\alpha \\neq 0 \\) ，故  \\( a_{i} = 0 \\) ，与假设矛盾。",
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            "content": "对  \\( \\forall\\alpha\\in V,\\exists k \\)  使得  \\( A^{k}=0,A^{k-1}\\neq0 \\)  ，则  \\( I(\\alpha):=L(\\alpha,A\\alpha,\\cdots,A^{k-1}\\alpha) \\)  称为  \\( \\alpha \\)  生成的 A 的循环不变子空间。",
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            "content": "下面说明下面的向量组线性无关:",
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            "content": "6.1 Jordan 标准形：空间分解法"
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            "content": "(2) 对于 \\(\\alpha \\in N_i\\)，由定义 \\(\\exists \\beta \\in V\\) 使得 \\(\\alpha = \\mathcal{A}^i\\beta\\)，从而 \\(\\alpha = \\mathcal{A}^{i-1}(\\mathcal{A}\\beta) \\in N_{i-1}\\)。由 \\(\\alpha\\) 任意性可知 \\(N_i \\subseteq N_{i-1}\\)。",
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            "content": "引理 6.3 (核空间增长)"
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            "content": "核空间  \\( M_{i} \\)  的增长满足以下性质：",
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            "bbox": [
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            "angle": 0,
            "content": "若 \\(M_{k} = M_{k + 1}\\) ，则 \\(M_{k} = M_{k + 1} = M_{k + 2} = \\dots\\)",
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            "bbox": [
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            "content": "若 dimV = n，则  \\( M_{n} = M_{n+1} = M_{n+2} = \\cdots \\) 。",
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            "content": "证明 (1) 显然有 \\(M_{k+1} \\subseteq M_{k+2}\\), 只需要证明 \\(M_{k+2} \\subseteq M_{k+1}\\)。对于 \\(\\forall \\alpha \\in M_{k+2}\\), 有 \\(\\mathcal{A}^{k+2} \\alpha = 0\\), 即 \\(\\mathcal{A}^{k+1}(\\mathcal{A} \\alpha) = 0\\), 因此 \\(\\mathcal{A} \\alpha \\in M_{k+1} = M_k\\), 即 \\(\\mathcal{A}^{k+1} \\alpha = 0 \\to \\alpha \\in M_{k+1}\\), 由任意性可知 \\(M_{k+2} \\subseteq M_{k+1}\\), 故 \\(M_{k+1} = M_{k+2}\\), 后续类似。",
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            "content": "(2) 只需要证明  \\( M_{n} = M_{n+1} \\) ，假设不然，则  \\( \\{0\\} = M_{0} \\subset M_{1} \\subset \\cdots \\subset M_{n} \\subset M_{n+1}, dimM_{n+1} > n \\) ，与  \\( M_{n+1} \\subseteq V \\)  矛盾。",
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            "content": "定理 6.2 (幂次变换核像分解)"
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            "content": "\\(V = M_{n} \\oplus N_{n}\\), 其中 \\(n = dimV\\)。",
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            "content": "证明 显然由核像性质有  \\( dimM_{n} + dimN_{n} = n \\) ，只需证  \\( M_{n} \\cap N_{n} = \\{0\\} \\) 。 \\( \\forall v \\in M_{n} \\cap N_{n} \\) ，有  \\( A^{n}v = 0 \\) ，且  \\( \\exists u \\in V \\)  使得  \\( v = A^{n}u \\) ，从而  \\( A^{2n}u = 0 \\) ，而  \\( M_{2n} = M_{2n-1} = \\cdots = M_{n} \\) ，故  \\( A^{n}u = 0 \\) ，从而 v = 0。",
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            "content": "定义 6.3 (广义特征向量与根子空间 (广义特征子空间))"
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            "content": "- 广义特征向量：线性变换 \\(\\mathcal{A}\\) 关于特征值 \\(\\lambda\\) 的广义特征向量定义为满足 \\(\\exists j \\in \\mathbb{N}^{+}, (\\mathcal{A} - \\lambda E)^{j}v = 0\\) 的所有向量。",
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            "content": "- 根子空间：线性变换 \\(\\mathcal{A}\\) 关于 \\(\\lambda\\) 的所有广义特征向量加上零向量组成的空间称为 \\(\\mathcal{A}\\) 关于特征值 \\(\\lambda\\) 的根子空间，记作 \\(G(\\lambda, \\mathcal{A})\\)。",
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            "content": "定理 6.3 (根子空间的具体表达)"
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            "content": "根子空间  \\( G(\\lambda, \\mathcal{A}) \\)  满足  \\( G(\\lambda, \\mathcal{A}) = \\operatorname{Ker}(\\mathcal{A} - \\lambda E)^{dimV} \\) 。",
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            "content": "证明 右推左：若 \\(v \\in Ker(\\mathcal{A} - \\lambda E)^{dimV}\\)，则由 \\(G(\\lambda, \\mathcal{A})\\) 定义可知 \\(v \\in G(\\lambda, \\mathcal{A})\\)。左推右：\\(v \\in G(\\lambda, \\mathcal{A})\\) 即 \\(\\exists j \\in \\mathbb{N}\\) s.t. \\((\\mathcal{A} - \\lambda E)^j v = 0\\)，而前面已证明这样的 \\(v\\) 必然满足 \\(v \\in Ker(\\mathcal{A} - \\lambda E)^{dimV}\\)。",
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            "content": "笔记 为了知识体系的完整，这里补充一下： \\( \\lambda_{i} \\)  根子空间的次数最低压缩为  \\( \\lambda_{i} \\)  对应 Jordan 块的最高阶数，即最小多项式中的  \\( \\lambda_{i} \\)  的阶数。其他书也常用特征多项式的代数重数表示 (题目中会遇到)，上面引理的 dimV 明显放缩过大。",
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            "content": "6.1.3 空间第一分解定理：根子空间分解"
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            "content": "定理 6.4 (根子空间的不变性)"
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            "angle": 0,
            "content": "根子空间  \\( G(\\lambda, \\mathcal{A}) \\)  是 A 不变的。",
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            "content": "证明 由定理6.3可知 \\(G(\\lambda, \\mathcal{A}) = Ker(\\mathcal{A} - \\lambda E)^n\\)，\\(\\forall \\alpha \\in G(\\lambda, \\mathcal{A})\\) 有 \\((\\mathcal{A} - \\lambda E)^n \\alpha = 0\\)，显然 \\((\\mathcal{A} - \\lambda E)^n \\mathcal{A} \\alpha = \\mathcal{A}((\\mathcal{A} - \\lambda E)^n \\alpha) = 0\\)，故 \\(\\mathcal{A} \\alpha \\in G(\\lambda, \\mathcal{A})\\)。",
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            "content": "定理 6.5 (空间第一分解定理：根子空间分解)"
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            "content": "若 V 是复线性空间，A 是线性变换， \\( \\lambda_{1},\\cdots,\\lambda_{m} \\)  是其特征值，则  \\( V=G(\\lambda_{1},\\mathcal{A})\\oplus\\cdots\\oplus G(\\lambda_{m},\\mathcal{A}) \\) 。",
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            "content": "6.2 空间分解法：零化多项式角度"
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            "content": "去除  \\( \\lambda_{1} \\)  所有对应特征向量后， \\( A|_{U} \\)  在 U 有  \\( \\lambda_{2},\\cdots,\\lambda_{m} \\)  特征值，对 U 用归纳法， \\( U=G(\\lambda_{2},\\mathcal{A}|_{U})\\oplus\\cdots\\oplus G(\\lambda_{m},\\mathcal{A}|_{U}) \\) ，由于  \\( \\forall j\\geq2,G(\\lambda_{j},\\mathcal{A}|_{U})=G(\\lambda_{j},\\mathcal{A}) \\) ，从而得证。",
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            "content": "6.1.4 Jordan 标准形：空间分解法"
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            "content": "定理 6.6 (幂零变换的 Jordan 标准型)"
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            "content": "幂零变换 A 在循环不变子空间分解基下的矩阵如下。称为幂零变换的 Jordan 标准形。",
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            "content": "\\[\nJ = \\left[ \\begin{array}{c c c c} J _ {1} & & & \\\\ & J _ {2} & & 0 \\\\ 0 & & \\ddots & \\\\ & & & J _ {s} \\end{array} \\right], \\quad J _ {i} = \\left[ \\begin{array}{c c c c} 0 & 1 & & 0 \\\\ & 0 & \\ddots & \\\\ & & \\ddots & 1 \\\\ 0 & & & 0 \\end{array} \\right] _ {n _ {i} \\times n _ {i}}\n\\]"
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            "content": "证明 代入计算显然可得。",
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            "content": "定理 6.7 (核限制变换是幂零变换)"
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            "content": "证明 由于 \\(M_{k} = \\mathrm{Ker}\\mathcal{B}^{k}\\)，因此 \\(\\forall \\alpha \\in M_{k}\\) 有 \\(\\mathcal{B}^{k}\\alpha = 0\\)，故 \\((\\mathcal{B}|_{M_k})^k = 0\\)。",
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            "content": "定理 6.8 (Jordan 标准型：空间分解法)"
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            "content": "对于 \\(\\mathbb{C}\\) 上线性变换 \\(\\mathcal{A}\\) 而言，其有在先根子空间分解，根子空间再循环不变子空间分解的基下有以下形式。称为线性变换的 Jordan 标准形。",
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            "content": "\\[\nJ = \\left[ \\begin{array}{c c c c} J _ {1} & & & 0 \\\\ & J _ {2} & & \\\\ & & \\ddots & \\\\ 0 & & & J _ {s} \\end{array} \\right], \\quad J _ {i} = \\left[ \\begin{array}{c c c c} \\lambda_ {i} & 1 & & 0 \\\\ & \\lambda_ {i} & \\ddots & \\\\ & & \\ddots & 1 \\\\ 0 & & & \\lambda_ {i} \\end{array} \\right] _ {n _ {i} \\times n _ {i}}\n\\]"
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            "content": "63"
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            "content": "6.2 空间分解法：零化多项式角度"
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            "content": "且满足 \\(f(A)\\alpha = f_{1}(A)f_{2}(A)\\alpha = 0\\) 。故考察 \\(f_{2}(A)\\alpha_{1}\\)",
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            "bbox": [
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            "content": "\\[\nf _ {2} (A) \\alpha_ {1} = f _ {2} (A) u (A) f _ {1} (A) \\alpha = u (A) f _ {1} (A) f _ {2} (A) = 0 \\Rightarrow \\alpha_ {1} \\in V _ {2}\n\\]"
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            "content": "同理得到 \\(f_{1}(A)\\alpha_{2} = 0, \\alpha_{2} \\in V_{1}\\)，从而得到 \\(V = V_{1} + V_{2}\\)。",
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            "content": "下证直和， \\(\\forall \\alpha \\in V_1\\cap V_2\\) ，则 \\(f_{1}(A)\\alpha = f_{2}(A)\\alpha = 0\\) ，因此",
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            "content": "\\[\n\\alpha = u (A) f _ {1} (A) \\alpha + v (A) f _ {2} (A) \\alpha = 0\n\\]"
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            "content": "因此结论成立。",
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            "content": "定理 6.10 (多空间分解)"
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            "content": "A 是数域 P 上 n 阶方阵， \\( f(x), f_{1}(x), \\cdots, f_{s}(x) \\in P[x] \\) ， \\( f(x) = f_{1}(x)f_{2}(x)\\cdots f_{s}(x) \\) ，且  \\( f_{1}(x), \\cdots, f_{s}(x) \\)  两两互素，设  \\( f(A)X = 0, f_{1}(A)X = 0, \\cdots, f_{s}(A)X = 0 \\)  的解空间为  \\( V, V_{1}, \\cdots, V_{s} \\) ，则",
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            "content": "\\[\nV = V _ {1} \\oplus V _ {2} \\oplus \\dots \\oplus V _ {s}\n\\]"
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            "content": "证明 用数学归纳法即可。",
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            "content": "定理 6.11 (空间分解研究矩阵秩)"
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            "content": "考虑  \\( f(A)=f_{1}(A)\\cdots f_{s}(A) \\) ，则",
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            "content": "\\[\nn - \\mathrm{r} (f (A)) = n - \\mathrm{r} (f _ {1} (A)) + \\dots + n - \\mathrm{r} (f _ {s} (A))\n\\]"
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            "content": "证明 设  \\( f(A)X = 0, f_{1}(A)X = 0, \\cdots, f_{s}(A)X = 0 \\)  的解空间为  \\( V, V_{1}, \\cdots, V_{s} \\) ，根据解空间分解得到  \\( V = V_{1} \\oplus \\cdots \\oplus V_{s} \\) ，因此根据解空间的维数可得到结论。",
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            "content": "练习 6.1 空间分解研究矩阵的秩 证明以下命题",
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            "content": "(1) \\( A^{2}=A \\) 当且仅当 \\( r(A)+r(A-E)=n \\)",
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            "content": "(2) \\(A^2 = E\\) 当且仅当 \\(r(A + E) + r(A - E) = n\\)",
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            "content": "(3) \\( A^{3}=E \\) 当且仅当 \\( r(A-E)+r(A^{2}+A+E)=n \\)",
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            "content": "证明 (1) \\( f(A)=A^{2}-A=A(A-E) \\) ，因此 \\( n-r(A^{2}-A)=n-r(A)+n-r(A-E) \\) ，左侧等于n，右侧等于 \\( 2n-r(A)-r(A-E) \\) ，因此 \\( r(A)+r(A-E)=n \\)",
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            "content": "6.2.2 根子空间分解：线性变换角度"
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            "content": "V 是 P 上的线性空间，O 是零变换，则其核空间为整个空间：",
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            "content": "\\[\n\\operatorname{Ker} (\\mathcal {O}) = V\n\\]"
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            "content": "\\(\\mathcal{A}\\) 为线性变换， \\(f(x) = f_{1}(x)f_{2}(x)\\) 为多项式，且 \\((f_{1}(x),f_{2}(x)) = 1\\) ，则有如下核空间分解：",
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            "content": "\\[\n\\operatorname{Ker} (f (\\mathcal {A})) = \\operatorname{Ker} (f _ {1} (\\mathcal {A})) \\oplus \\operatorname{Ker} (f _ {2} (\\mathcal {A}))\n\\]"
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            "content": "证明 根据 \\((f_{1}(x), f_{2}(x)) = 1\\) 得到 \\(\\exists u(x), v(x)\\) 使得 \\(u(x)f_{1}(x) + v(x)f_{2}(x) = 1\\)，因此",
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            "content": "对  \\( \\forall\\alpha\\in\\mathrm{Ker}(f(\\mathcal{A})) \\)",
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            "content": "6.2 空间分解法：零化多项式角度"
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            "content": "显然 \\(f_{2}(\\mathcal{A})\\alpha_{1} = u(\\mathcal{A})f(\\mathcal{A})\\alpha = 0\\) ，得到 \\(\\alpha_{1}\\in \\mathrm{Ker}(f_{1}(\\mathcal{A}))\\) ，同理有 \\(\\alpha_{2}\\in \\mathrm{Ker}(f_{2}(\\mathcal{A}))\\) ，综上得到",
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            "content": "\\[\n\\operatorname{Ker} (f (\\mathcal {A})) = \\operatorname{Ker} (f _ {1} (\\mathcal {A})) + \\operatorname{Ker} (f _ {2} (\\mathcal {A}))\n\\]"
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            "content": "下证直和：对 \\(\\forall \\alpha \\in \\mathrm{Ker}(f_1(\\mathcal{A}))\\cap \\mathrm{Ker}(f_2(\\mathcal{A}))\\) ，有",
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            "content": "\\[\n\\alpha = u (\\mathcal {A}) f _ {1} (\\mathcal {A}) \\alpha + v (\\mathcal {A}) f _ {2} (\\mathcal {A}) \\alpha = 0\n\\]"
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            "content": "因此为直和",
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            "content": "定理 6.13 (全空间分解：零化多项式角度)"
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            "content": "V 是 P 上的线性空间, A 是 V 上的线性变换,  \\( f(x) = f_{1}(x)f_{2}(x)\\cdots f_{s}(x) \\)  为零化多项式, 且  \\( f_{1}(x), \\cdots, f_{s}(x) \\)  两两互素, 则",
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            "angle": 0,
            "content": "\\[\nV = \\operatorname{Ker} (f _ {1} (\\mathcal {A})) \\oplus \\dots \\oplus \\operatorname{Ker} (f _ {s} (\\mathcal {A}))\n\\]"
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            "content": "证明 由于 \\(f(A) = \\mathcal{O}\\)，根据零变换的核空间以及核空间分解定理可知。",
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            "content": "引理 6.5 (Hamilton-Cayley)"
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            "content": "若 \\(f(\\lambda)\\) 是线性变换 \\(\\mathcal{A}\\) 的特征多项式，则 \\(f(\\lambda)\\) 是 \\(\\mathcal{A}\\) 的零化多项式",
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            "content": "证明 该定理在最小多项式一章已经证明过了",
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            "content": "定理 6.14 (根子空间分解：零化多项式角度)"
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            "content": "若 \\(\\mathbb{C}\\) 上线性变换 \\(\\mathcal{A}\\) 的特征多项式为 \\(f(\\lambda) = (\\lambda - \\lambda_1)^{k_1} \\cdots (\\lambda - \\lambda_s)^{k_s}\\)，最小多项式为 \\(m(\\lambda) = (\\lambda - \\lambda_1)^{l_1} \\cdots (\\lambda - \\lambda_s)^{l_s}\\)，则有以下根子空间分解",
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            "content": "\\[\nV = \\operatorname{Ker} (\\mathcal {A} - \\lambda_ {1} \\mathcal {E}) ^ {k _ {1}} \\oplus \\dots \\oplus \\operatorname{Ker} (\\mathcal {A} - \\lambda_ {s} \\mathcal {E}) ^ {k _ {s}} = \\operatorname{Ker} (\\mathcal {A} - \\lambda_ {1} \\mathcal {E}) ^ {l _ {1}} \\oplus \\dots \\oplus \\operatorname{Ker} (\\mathcal {A} - \\lambda_ {s} \\mathcal {E}) ^ {l _ {s}}\n\\]"
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            "content": "证明 根据 Hamilton-Cayley 定理以及零化多项式全空间分解可知。",
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            "content": "练习 6.2 根子空间分解：零化多项式角度 (1) ZJU2021.8: V 是线性空间， \\( \\varphi \\) 是 V 上线性变换，特征多项式 \\( f(\\lambda)=\\left(\\lambda-2\\right)^{6}(\\lambda+2)^{4} \\) ，将 V 分解为两个非平凡不变子空间的直和，并证明结论。",
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            "content": "定理 6.15 (根子空间分解与 Jordan 标准型)"
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            "content": "根子空间分解将Jordan阵按照特征值 \\(\\lambda_{i}\\) 分为若干份，每份由 \\(\\lambda_{i}\\) 为对角的若干个Jordan块组成，对角 \\(\\lambda_{i}\\) 在Jordan阵中的出现次数等于其代数重数。",
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            "content": "6.2.3 Jordan 阵与循环不变子空间"
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            "content": "引理 6.6 (核限制变换是幂零变换)"
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            "content": "\\(\\lambda_{i}\\) 为 \\(\\mathcal{A}\\) 的特征值，则 \\((\\mathcal{A} - \\lambda_i\\mathcal{E})\\big|_{\\mathrm{Ker}(\\mathcal{A} - \\lambda_i\\mathcal{E})^k}\\) 在 \\(\\mathrm{Ker}(\\mathcal{A} - \\lambda_i\\mathcal{E})^k\\) 中为幂零变换",
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            "content": "证明 由于 \\(\\forall \\alpha \\in \\mathrm{Ker}(\\mathcal{A} - \\lambda_i\\mathcal{E})^k\\) 满足 \\((\\mathcal{A} - \\lambda_i\\mathcal{E})^k\\alpha = 0\\) ，因此限制变换满足 \\((\\mathcal{A} - \\lambda_i\\mathcal{E})^k\\big|_{\\mathrm{Ker}(\\mathcal{A} - \\lambda_i\\mathcal{E})^k} = \\mathcal{O}\\)",
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            "content": "定理 6.16 (循环不变子空间分解)"
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            "content": "对于 V 内的幂零变换 A， \\( \\exists v_{1},\\cdots,v_{n}\\in V,\\ m_{1},\\cdots,m_{n}\\in N \\)  使得  \\( V=I(v_{1})\\oplus I(v_{2})\\oplus\\cdots\\oplus I(v_{n}) \\) 。其中  \\( I(v_{i})=L(v_{i},\\mathcal{A}v_{i}\\cdots,\\mathcal{A}^{m_{i}}v_{i}) \\) ，且  \\( A^{m_{i}+1}v_{i}=0 \\) 。",
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            "content": "证明 在空间第二分解定理中已经证明过了。",
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            "content": "6.3 Jordan 标准形： \\( \\lambda \\)  矩阵法"
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            "content": "定理 6.17 (Jordan 阵与循环不变子空间)"
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            "content": "线性变换 A 在根子空间  \\( \\mathrm{Ker}(\\mathcal{A}-\\lambda_{i}\\mathcal{E})^{k} \\)  下的限制变换  \\( A|_{\\mathrm{Ker}(\\mathcal{A}-\\lambda_{i}\\mathcal{E})^{k}} \\)  在一组循环不变基  \\( ((\\mathcal{A}-\\lambda_{i}\\mathcal{E})^{s}\\alpha, (\\mathcal{A}-\\lambda_{i}\\mathcal{E})^{s-1}\\alpha,\\cdots,\\alpha) \\)  下的矩阵为一个 Jordan 块：",
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            "content": "\\[\nJ _ {s} (\\lambda_ {i}) = \\left[ \\begin{array}{c c c c c} \\lambda_ {i} & 1 & & & \\\\ & \\lambda_ {i} & 1 & & \\\\ & & \\ddots & \\ddots & \\\\ & & & \\lambda_ {i} & 1 \\\\ & & & & \\lambda_ {i} \\end{array} \\right]\n\\]"
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            "content": "证明 直接验证即可。",
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            "content": "定理 6.18 (Jordan 阵与几何重数)"
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            "content": "特征值 \\(\\lambda_{i}\\) 的代数重数是特征多项式 \\(f(\\lambda)\\) 的 \\(\\lambda = \\lambda_{i}\\) 根的重数，几何重数 \\((\\lambda_{i}E - A = 0\\) 解空间维数) 是 \\(\\lambda_{i}\\) 对应以 \\(\\lambda_{i}\\) 为对角的 Jordan 块个数。",
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            "content": "证明 根据 \\(\\lambda_{i}E - A = 0\\) 的解空间维数等于 \\(\\lambda_{i}E - J = 0\\) 解空间维数，而每有一个以 \\(\\lambda_{i}\\) 为对角的Jordan块，\\(\\lambda_{i}E - J\\) 上就有一行全为0，因此秩减1，从而块数等于几何重数。",
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            "content": "笔记 即以  \\( \\lambda_{i} \\)  为对角的 Jordan 块个数为  \\( n - r(\\lambda E - A) \\) ，反过来写也要分得清。",
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            "content": "6.3 Jordan 标准形：\\(\\lambda\\) 矩阵法"
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            "content": "6.3.1 \\(\\lambda\\) 矩阵与相抵"
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            "content": "定义 6.4 ( \\( \\lambda \\) 矩阵)",
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            "content": "\\(A(\\lambda)\\) 是元素为 \\(\\lambda\\) 多项式的矩阵，则将 \\(A(\\lambda)\\) 称为 \\(\\lambda\\) 矩阵。",
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            "content": "定义 6.5 ( \\( \\lambda \\) 矩阵初等变换)"
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            "content": "\\(\\lambda\\) 矩阵有三种初等变换：",
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            "content": "- 互换两行/列",
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            "content": "- 用常数 \\(c\\) 乘一行/列 (不能乘 \\(\\lambda\\) 多项式！)",
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            "content": "- 将一行/列乘以一个 \\(\\lambda\\) 多项式加到另一行/列上",
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            "content": "定义 6.6 (相抵)"
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            "content": "若可以用初等变换将一个 \\(\\lambda\\) 矩阵转换为另一个 \\(\\lambda\\) 矩阵，则称两个 \\(\\lambda\\) 矩阵相抵。",
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            "content": "定理 6.19 (相抵标准形：法式)"
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            "content": "对于数值矩阵 A， \\( \\lambda E - A \\)  总是相抵于  \\( \\mathrm{diag}\\{1,\\cdots,1,d_{1}(\\lambda),\\cdots,d_{m}(\\lambda)\\} \\) ，其中  \\( d_{i}(\\lambda) \\)  是首一多项式，且  \\( d_{i}(\\lambda)|d_{i+1}(\\lambda) \\) 。",
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            "content": "引理 6.7 (对角整除化)"
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            "content": "\\(A(\\lambda)\\) 为非零 \\(\\lambda\\) 矩阵，则 \\(A(\\lambda)\\) 一定相抵于 \\(B(\\lambda)\\)，满足 \\(b_{11}(\\lambda) \\neq 0\\) 且 \\(b_{11}(\\lambda)\\) 可整除 \\(B(\\lambda)\\) 中的所有元素",
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            "content": "证明 先将 \\(A(\\lambda)\\) 中次数最低的元素放置于对角 \\(a_{11}(\\lambda)\\)",
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            "content": "6.3 Jordan 标准形：\\(\\lambda\\) 矩阵法"
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            "content": "(1) 消去第一行、第一列：不妨设 \\(a_{i1}(\\lambda)\\) 不被 \\(a_{11}(\\lambda)\\) 整除，则做带余除法 \\(a_{i1}(\\lambda) = q(\\lambda)a_{11}(\\lambda) + r(\\lambda)\\)，再将 \\(r(\\lambda)\\) 放置于 \\(a_{11}\\)，直至全部整除，可消去第一列和第一行。",
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            "content": "(2) 考虑第一列、第一行外的其他元素，若存在 \\(a_{ij}(\\lambda)\\) 不被 \\(a_{11}\\) 整除，则把第 \\(j\\) 列加到第一列上，做带余除法，把余项放到 \\(a_{11}\\)，以此类推。",
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            "content": "定理 6.20 (法式的计算：初等变换法)"
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            "content": "每次把最低次移动到角上，尝试消除，消除过程按照以下步骤",
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            "content": "- 通过初等变换将当前对角元素变为所有非零元素中次数最低的 \\(\\lambda\\) 多项式",
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            "content": "- 将矩阵对角整除化：具体见引理的过程",
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            "content": "- 消去一列，再消去一行，重复对角整除化和消去",
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            "content": "定理 6.21 (相抵与相似)"
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            "content": "A 与 B 相似当且仅当  \\( \\lambda E - A \\)  与  \\( \\lambda E - B \\)  相抵。",
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            "content": "证明 (1) 左推右：若 \\(B = P^{-1}AP\\)，则 \\(\\lambda E - B = P^{-1}(\\lambda E - A)P\\)，显然右侧两个 \\(\\lambda\\) 矩阵相抵。",
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            "content": "(2) 右推左：若  \\( \\lambda E - B = P(\\lambda)(\\lambda E - A)Q(\\lambda) \\) ，设  \\( Q(\\lambda) = Q_k \\lambda^k + \\cdots + Q_1 \\lambda + Q_0 \\) ， \\( Q(\\lambda)^{-1} = R_m \\lambda^m + \\cdots + R_1 \\lambda + R_0 \\) ，下记  \\( W = Q(B) \\) 。由  \\( Q(\\lambda)Q^{-1}(\\lambda) = E \\)  得  \\( R_m Q(\\lambda) \\lambda^m + \\cdots + R_1 Q(\\lambda) \\lambda + R_0 Q(\\lambda) = E \\) ，代入 B 得到  \\( R_m WB^m + \\cdots + R_1 WB + R_0 W = E \\) 。",
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            "content": "由于 \\(P(\\lambda)^{-1}(\\lambda E - B) = (\\lambda E - A)Q(\\lambda) = Q(\\lambda)\\lambda - AQ(\\lambda)\\)，代入 \\(\\lambda = B\\) 得到 \\(Q(B)B = AQ(B)\\) (前面一式子左 \\(= 0\\))，即 \\(WB = AW\\)，从而 \\(WB^{l} = A^{l}W\\) (多作用一个 \\(A\\) 相当于多右乘一个 \\(B\\))。",
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            "content": "再由前面的式子得到 \\( (R_{m}A^{m}+\\cdots+R_{0})W=E \\) ，则表明W可逆，从而 \\( B=W^{-1}AW \\) 。",
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            "content": "6.3.2 行列式因子、不变因子"
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            "content": "定义 6.7 (行列式 k 阶子式)",
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            "content": "\\( |A| \\) 的 k 阶子式是其任选 k 行、k 列组成的矩阵的行列式。",
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            "content": "定义 6.8 (行列式因子)"
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            "content": "\\(A(\\lambda)\\) 是 \\(\\lambda\\) 矩阵，若其所有 \\(k\\) 阶子式(行列式)均为0，则定义其 \\(k\\) 阶行列式因子 \\(D_{k}(\\lambda) = 0\\) ，否则定义为所有非零 \\(k\\) 阶子式的首一最大公因式。",
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            "content": "笔记一般行列式因子比初等变换方便，不过一定注意这里的子式是所有子式，而非主子式，甚至可以不连着，0的情况也得分开考虑。",
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            "content": "引理 6.8 (行列式因子的相除性)"
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            "content": "设  \\( D_{1}(\\lambda),\\cdots,D_{r}(\\lambda) \\)  是  \\( A(\\lambda) \\)  非零行列式因子，则  \\( D_{i}(\\lambda)|D_{i+1}(\\lambda) \\) 。",
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            "content": "证明 设 \\(A_{i+1}\\) 是 \\(A(\\lambda)\\) 任意 \\(i + 1\\) 阶子式，这个行列式按某一行展开，则其每一展开项都是一个多项式与一个 \\(i\\) 阶子式乘积，而 \\(D_i(\\lambda)\\) 是所有 \\(i\\) 阶子式公因子，因此 \\(D_i(\\lambda)|_{A_{i+1}(\\forall i)}\\) ，而 \\(D_{i+1}(\\lambda)\\) 是所有 \\(A_{i+1}\\) 的最大公因式，因此证毕。",
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            "content": "6.4 \\(\\lambda\\) 矩阵的进一步研究"
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            "content": "定理 6.22 (行列式因子、不变因子的性质)"
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            "angle": 0,
            "content": "- 相抵矩阵的因子：相抵的 \\(\\lambda\\) 矩阵有相同的行列式因子、不变因子 (初等变换不改变行列式、不变因子)",
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            "angle": 0,
            "content": "- 法式与不变因子：\\(\\lambda E - A\\) 法式的所有对角元素即 \\(\\lambda E - A\\) 的不变因子。",
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            "content": "证明 (1) 只需证明行列式因子在三种初等变换不变即可。而不变因子由行列式因子唯一确定。",
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            "content": "(2) 由于不变因子在初等变化中不变，求出法式的不变因子 (即对角)，因此 \\(\\lambda E - A\\) 的不变因子即其法式对角元素。",
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            "content": "定理 6.23 (不变因子与最小多项式)"
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            "content": "\\(A\\) 的最小多项式即 \\(\\lambda E - A\\) 的最后一个不变因子",
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            "content": "证明 设 A 的互异特征值为  \\( \\lambda_{1},\\cdots,\\lambda_{s} \\) ，因此可以列出所有的初等因子 (具体见下一节)，最小多项式即初等因子的最小公倍式，最后一个不变因子也是，因此结论成立",
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            "content": "6.3.3 初等因子与 Jordan 标准型"
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            "content": "定义 6.10 (初等因子)"
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            "content": "将所有不变因子分解为不可约因子之积，这些不可约因子全体称为初等因子组。",
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            "content": "定理 6.24 (对角化计算初等因子)"
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            "content": "\\( \\lambda E-A \\)  经过初等变化变为对角阵 (不一定要法式)",
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            "content": "\\[\n\\left( \\begin{array}{c c c c} f _ {1} (\\lambda) & & & \\\\ & f _ {2} (\\lambda) & & \\\\ & & \\ddots & \\\\ & & & f _ {n} (\\lambda) \\end{array} \\right)\n\\]"
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            "content": "其中 \\(f_{i}(\\lambda)\\) 为非零首一多项式，则 \\(A\\) 的初等因子组是所有 \\(f_{i}(\\lambda)\\) 做不可约分解得到的因式组合。",
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            "content": "引理 6.9 (Jordan 块、Jordan 阵的初等因子)"
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            "content": "r 阶  \\( \\lambda_{0} \\)  对角的 Jordan 块的初等因子组为  \\( (\\lambda - \\lambda_{0})^{r} \\) 。对角为  \\( J_{1}, J_{2}, \\cdots, J_{k} \\)  的 Jordan 阵的初等因子组为  \\( (\\lambda - \\lambda_{1})^{r_{1}}, \\cdots, (\\lambda - \\lambda_{k})^{r_{k}} \\) 。",
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            "content": "证明 (1) Jordan 块的初等因子组直接用行列式因子计算即可 (只有 \\(r\\) 阶非 1，因此不变因子好算)",
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            "content": "(2)Jordan 阵的初等因子进行分块初等变换，将各个 Jordan 块分块变成法式，再用引理6.24即可。",
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            "content": "定理 6.25 (使用初等因子计算 Jordan 标准型)"
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            "content": "若 K 上的矩阵 A 有初等因子组  \\( (\\lambda - \\lambda_{1})^{r_{1}}, \\cdots, (\\lambda - \\lambda_{k})^{r_{k}} \\) ，则 A 相似于 Jordan 阵  \\( J = \\text{diag}\\{J_{1}, \\cdots, J_{k}\\} \\) ，其中  \\( J_{i} \\)  是以  \\( \\lambda_{i} \\)  为对角的  \\( r_{i} \\)  阶 Jordan 块。",
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            "content": "定理 6.26 (Jordan 标准型存在充要条件)"
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            "content": "K 上矩阵 A 有 Jordan 标准型当且仅当其初等因子都是一次项的幂次乘积。",
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            "content": "一、初等变换方法：只需要将 \\(\\lambda E - A\\) 转换为对角矩阵，对角所有分解即初等因子组，直接构造Jordan标准型",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c} - 1 & 1 & 3 \\\\ 3 & 0 & - 4 \\\\ - 2 & 1 & 4 \\end{array} \\right]\n\\]"
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            "content": "\\[\n\\left[\\begin{array}{c c c}1&0&0\\\\0&\\lambda - 1&- \\lambda + 1\\\\0&\\lambda^ {2} + \\lambda - 3&- 3 \\lambda + 4\\end{array}\\right]\\rightarrow \\left[\\begin{array}{c c c}1&0&0\\\\0&\\lambda - 1&- \\lambda + 1\\\\0&(\\lambda^ {2} + \\lambda - 3) - (\\lambda - 1) (\\lambda + 2)&(- 3 \\lambda + 4) - (\\lambda + 2) (- \\lambda + 1)\\end{array}\\right] = \\left[\\begin{array}{c c c}1&0&\\\\0&\\lambda - 1&- \\lambda\\\\0&- 1&\\lambda^ {2} - \\lambda^ {2}\\end{array}\\right]\n\\]"
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            "content": "因此最终可以化简为：",
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            "content": "\\[\n\\left[ \\begin{array}{c c c c} 1 & & & \\\\ & 1 & & \\\\ & & (\\lambda - 1) ^ {3} \\end{array} \\right]\n\\]"
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            "content": "练习6.4初等变换方法计算Jordan标准型 计算 \\(A = \\left[ \\begin{array}{ccc}1 & -1 & 2\\\\ 3 & -3 & 6\\\\ 2 & -2 & 4 \\end{array} \\right]\\) 的Jordan标准型解通过初等变换得到 \\(\\lambda E - A\\) 的法式为 \\(\\left[ \\begin{array}{lll}1 & 0 & 0\\\\ 0 & \\lambda & 0\\\\ 0 & 0 & -\\lambda^2 +2\\lambda \\end{array} \\right]\\) ，因此初等因子为 \\(\\lambda ,\\lambda ,( \\lambda -2)\\) ，对应Jordan阵为：\\(J = \\left[ \\begin{array}{ccccc}0 &  &  &  \\\\  & 0 &  &  \\\\  &  & 2 \\end{array} \\right]\\)",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c c} a & b & \\dots & b & b \\\\ & a & \\dots & b & b \\\\ & & \\ddots & \\vdots & \\vdots \\\\ & & & a & b \\\\ & & & & a \\end{array} \\right]\n\\]"
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            "content": "6.4 \\(\\lambda\\) 矩阵的进一步研究"
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            "content": "解 若 b=0，则 Jordan 标准型即 aE。若  \\( b\\neq0 \\) ，则考虑  \\( \\lambda E-A \\) ：",
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            "content": "\\[\n\\lambda E - A = \\left[ \\begin{array}{c c c c c} \\lambda - a & - b & \\dots & - b & - b \\\\ & \\lambda - a & \\dots & - b & - b \\\\ & & \\ddots & \\vdots & \\vdots \\\\ & & & \\lambda - a & - b \\\\ & & & & \\lambda - a \\end{array} \\right]\n\\]"
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            "content": "考虑 \\(n - 1\\) 阶子式，右下子式行列式为 \\((\\lambda - a)^{n - 1}\\)，右上子式在 \\(\\lambda = a\\) 时显然不是0(根据余数定理)，因此互素，即初等因子为 \\((\\lambda - a)^n\\)，因此Jordan阵为一个 \\(n\\) 阶的以 \\(a\\) 为对角的Jordan块",
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            "content": "练习6.6行列式因子应用相关练习 求不变因子:(1) \\( A=\\left[\\begin{array}{cccc} & & 1 & \\lambda+2 \\\\ & 1 & \\lambda+2 \\\\ 1 & \\lambda+2 & & \\\\ \\lambda+2 & & & \\end{array}\\right] \\) (2) \\( A=\\left[\\begin{array}{cccc} \\lambda & -1 & & \\\\ & \\lambda & -1 & \\\\ & & \\lambda & -1 \\\\ 5 & 4 & 3 & \\lambda+2 \\end{array}\\right] \\)",
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            "content": "解(1)左上行列式因子为-1，因此不变因子为 \\( 1,1,1,(\\lambda+2)^{4} \\)",
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            "content": "(2) 右上行列式因子为 -1，因此不变因子 1,1,1, \\( |A| \\)",
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            "content": "定理 6.27 (重要  \\( \\lambda \\)  矩阵)"
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            "content": "下面  \\( \\lambda \\)  矩阵 A 的不变因子为  \\( 1,1,\\cdots,\\lambda^{n}+a_{1}\\lambda^{n-1}+a_{2}\\lambda^{n-2}+\\cdots+a_{n-1}\\lambda+a_{n} \\)",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c c c} \\lambda & & & & & a _ {n} \\\\ - 1 & \\lambda & & & & a _ {n - 1} \\\\ & - 1 & \\lambda & & & a _ {n - 2} \\\\ & & \\ddots & \\ddots & & \\vdots \\\\ & & & - 1 & \\lambda & a _ {2} \\\\ & & & & - 1 & \\lambda + a _ {1} \\end{array} \\right]\n\\]"
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            "content": "证明 考虑左下方行列式因子，其值为 1，因此不变因子为  \\( 1, 1, \\cdots, 1, |A| \\) ，而计算出  \\( |A| = \\lambda^{n} + a_{1}\\lambda^{n-1} + \\cdots + a_{n} \\) ，因此结论成立",
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            "content": "推论6.1"
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c c c} 0 & & & & & a _ {n} \\\\ - 1 & 0 & & & & a _ {n - 1} \\\\ & - 1 & 0 & & & a _ {n - 2} \\\\ & & \\ddots & \\ddots & & \\vdots \\\\ & & & - 1 & 0 & a _ {2} \\\\ & & & & - 1 & a _ {1} \\end{array} \\right]\n\\]"
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            "content": "练习6.7一些用行列式因子计算方便的特殊矩阵 计算Jordan标准型(1) \\(A = \\left[ \\begin{array}{cccc}1 & 2 & 3 & 4\\\\ & 1 & 2 & 3\\\\ & & 1 & 2\\\\ & & & 1 \\end{array} \\right]\\) (2) \\(A = \\left[ \\begin{array}{cccc}a & & & \\\\ a & a & & \\\\ \\vdots & \\vdots & \\ddots & \\\\ a & a & \\cdots & a \\end{array} \\right]\\)",
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            "content": "(2) 左上行列式  \\( (\\lambda - a)^{n-1} \\) ，左下在  \\( \\lambda = a \\)  显然非零，因此不变因子  \\( 1, 1, \\cdots, 1, |\\lambda E - A| \\)",
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            "content": "(3)(4) 显然",
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            "content": "笔记 如果对角相等，则用行列式因子计算可能会比较方便",
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            "content": "6.4.2 \\(k\\) 阶Jordan块的个数"
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            "content": "定理 6.28 (k 阶 Jordan 块个数)"
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            "content": "\\(n\\) 阶复矩阵 \\(A\\) 的Jordan标准型为 \\(J\\)，则 \\(J\\) 中对角元 \\(\\lambda\\) 的 \\(k\\) 阶Jordan块 \\(J_{k}(\\lambda)\\) 的个数如下。这里约定 \\(r((A - \\lambda E_n)^0) = n\\)",
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            "content": "\\[\nr ((A - \\lambda E _ {n}) ^ {k - 1}) - 2 r ((A - \\lambda E _ {n}) ^ {k}) + r ((A - \\lambda E _ {n}) ^ {k + 1})\n\\]"
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            "content": "证明 考虑零对角 Jordan 块的幂次:",
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            "content": "\\[\nr (J _ {r} (0) ^ {k}) = \\left\\{ \\begin{array}{l l} r - k, & 0 \\leq k <   r \\\\ 0, & k \\geq r \\end{array} \\right. \\quad r (J _ {r} (0) ^ {k}) - r (J _ {r} (0) ^ {k + 1}) = \\left\\{ \\begin{array}{l l} 1, & 0 \\leq k <   r \\\\ 0, & k \\geq r \\end{array} \\right.\n\\]"
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            "content": "而非零对角 Jordan 块的幂次满足  \\( r(J_{r}(a)^{k}) - r(J_{r}(a)^{k+1}) = 0 \\) 。",
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            "content": "设 A 的 Jordan 标准型为  \\( J = \\text{diag}\\{J_{r_1}(\\lambda_1), \\cdots, J_{r_s}(\\lambda_s)\\} \\) ， \\( (A - \\lambda E)^k, (J - \\lambda E)^k \\)  相似。根据 J 是分块对角的性质可知  \\( r((A - \\lambda E)^k) = r((J - \\lambda E)^k) = \\sum_{i=1}^{s} r(J_{r_i}(\\lambda_i - \\lambda)^k) \\) ，因此得到：",
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            "content": "\\[\nr ((A - \\lambda E) ^ {k}) - r ((A - \\lambda E) ^ {k + 1}) = \\sum_ {i = 1} ^ {s} [ r (J _ {r _ {i}} (\\lambda_ {i} - \\lambda) ^ {k}) - r (J _ {r _ {i}} (\\lambda_ {i} - \\lambda) ^ {k + 1}) ]\n\\]"
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            "content": "根据前面的分析， \\( r((A-\\lambda E)^{k})-r((A-\\lambda E)^{k+1}) \\) 等于 \\( \\lambda_{i}=\\lambda \\) 的且阶数大于k那些块个数，同理 \\( r((A-\\lambda E)^{k-1})-r((A-\\lambda E)^{k}) \\) 等于对角为 \\( \\lambda \\) 的且阶数大于k-1的块个数，相减即可得到阶数为k的块个数。",
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            "content": "练习 6.8 上述定理的应用 计算 A 的 Jordan 标准型",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c c} a & 0 & 1 & & \\\\ & a & 0 & \\ddots & \\\\ & & a & \\ddots & 1 \\\\ & & & \\ddots & 0 \\\\ & & & & a \\end{array} \\right]\n\\]"
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            "content": "练习 6.9 A, B 是两个 n 阶矩阵，证明 A, B 相似的充要条件是  \\( \\forall \\lambda \\in C, k = 1, 2, \\cdots, n \\) ，均有  \\( r((A - \\lambda E)^{k}) = r((B - \\lambda E)^{k}) \\)",
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            "content": "70"
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            "content": "证明 直接用结论即可",
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            "content": "6.4 \\(\\lambda\\) 矩阵的进一步研究"
        },
        {
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            "content": "6.4.3 分块矩阵的 Jordan 阵"
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            "content": "定理 6.29 (分块矩阵的 Jordan 标准型)"
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            "content": "\\(P\\) 上 \\(n\\) 阶矩阵 \\(A, B\\) 无公共复特征值，且 \\(A, B\\) 的 Jordan 标准型分别为 \\(J_{1}, J_{2}\\)，则对任意 \\(C \\in P^{m \\times n}\\)，则：",
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            "content": "\\left[ \\begin{array}{c c} A & C \\\\ O & B \\end{array} \\right] \\sim \\operatorname{diag} \\{J _ {1}, J _ {2} \\}\\tag{#}"
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            "content": "证明 当 \\(A, B\\) 无公共特征值时，根据 \\(AX - XB\\) 的矩阵方程结论可知，对任意矩阵 \\(C\\)，\\(AX - XB = C\\) 都有唯一解 \\(X_0\\) 使得 \\(AX_0 - X_0B = C\\)，因此：",
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            "content": "\\[\n\\left[ \\begin{array}{c c} E _ {m} & X _ {0} \\\\ O & E _ {n} \\end{array} \\right] \\left[ \\begin{array}{c c} A & C \\\\ O & B \\end{array} \\right] \\left[ \\begin{array}{c c} E _ {m} & - X _ {0} \\\\ O & E _ {n} \\end{array} \\right] = \\left[ \\begin{array}{c c} A & C - (A X _ {0} - X _ {0} B) \\\\ O & B \\end{array} \\right] = \\left[ \\begin{array}{c c} A & O \\\\ O & B \\end{array} \\right]\n\\]"
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            "content": "因此得到：",
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            "content": "\\[\n\\left[ \\begin{array}{c c} A & C \\\\ O & B \\end{array} \\right] \\sim \\left[ \\begin{array}{c c} A & O \\\\ O & B \\end{array} \\right] \\sim \\left[ \\begin{array}{c c} J _ {1} & O \\\\ O & J _ {2} \\end{array} \\right]\n\\]"
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            "content": "证明 第二种证明：直接看 Jordan 标准型， \\( \\lambda_{i} \\)  为对角的 k 阶 Jordan 块结论直接往上套即可。",
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        {
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            "content": "推论6.2"
        },
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            "content": "对于分块上三角矩阵  \\( A, A_{11}, A_{22}, \\cdots, A_{ss} \\)  两两无公共特征值， \\( A_{ii} \\)  的 Jordan 标准型为  \\( J_i \\) ，则 A 的 Jordan 标准型为  \\( \\mathrm{diag}\\{J_1, \\cdots, J_s\\} \\)",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c} A _ {1 1} & A _ {1 2} & \\dots & A _ {1 s} \\\\ & A _ {2 2} & \\dots & A _ {2 s} \\\\ & & \\ddots & \\vdots \\\\ & & & A _ {s s} \\end{array} \\right]\n\\]"
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            "content": "练习 6.10 分块矩阵应用 求 A 的 Jordan 标准型:",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c} 1 & 5 & - 1 & 2 \\\\ 0 & 2 & 4 & 3 \\\\ 0 & 0 & - 3 & 0 \\\\ 0 & 0 & 4 & - 3 \\end{array} \\right]\n\\]"
        },
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            "content": "解 \\(\\mathrm{diag}\\{1,2,J_2(-3)\\}\\)",
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            "content": "6.4.4 过渡矩阵的计算"
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            "content": "定理 6.30 (过渡矩阵计算总结)"
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            "content": "若已知 A 的 Jordan 标准型为 J，要计算 P 使得  \\( P^{-1}AP = J \\) ，则可以按照以下步骤：",
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            "content": "1. 对于非一阶，以  \\( \\lambda \\)  为对角的 Jordan 块，设对应的 P 位置为  \\( [X_{1},\\cdots,X_{k}] \\) ，则满足",
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            "content": "\\[\nA X _ {1} = \\lambda X _ {1}, A X _ {2} = \\lambda X _ {2} + X _ {1}, \\dots , A X _ {k} = \\lambda X _ {k} + X _ {k - 1}\n\\]"
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            "content": "设右端项为  \\( b = [b_{1}, \\cdots, b_{n}]^{T} \\)  (未知量)，统一对增广矩阵做初等变换  \\( [A - \\lambda E, b] \\to [A', b'] \\) ，选取顺序为  \\( X_{1}, \\cdots, X_{k} \\) 。 \\( X_{1} \\)  要保证不能选  \\( [0, \\cdots, 0]^{T} \\) ，且选的  \\( X_{k} \\)  要保证  \\( X_{k+1} \\)  有解。",
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            "content": "2. 对于一阶：即  \\( AX = \\lambda X \\)  特征向量，先求完非一阶再求，不要选  \\( [0, \\cdots, 0]^{T} \\) ，保证与其他列向量线性无关。",
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            "content": "6.4 \\(\\lambda\\) 矩阵的进一步研究"
        },
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            "content": "练习 6.11 已知矩阵 A 及其 Jordan 标准型 J，求过渡矩阵 P 使得  \\( P^{-1}AP = J \\)",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c} 2 & 3 & 2 \\\\ 1 & 8 & 2 \\\\ - 2 & - 1 4 & - 3 \\end{array} \\right] \\quad J = \\left[ \\begin{array}{c c c} 1 & 0 & 0 \\\\ 0 & 3 & 1 \\\\ 0 & 0 & 3 \\end{array} \\right]\n\\]"
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            "content": "解设 \\(P = (X_{1}, X_{2}, X_{3})\\)，根据 \\(P^{-1}AP = J\\) 得到 \\((AX_{1}, AX_{2}, AX_{3}) = (X_{1}, X_{2}, X_{3})J\\)，因此 \\(AX_{1} = X_{1}, AX_{2} = 3X_{2}, AX_{3} = X_{2} + 3X_{3}\\)，即 \\((A - E)X_{1} = 0, (A - 3E)X_{2} = 0, (A - 3E)X_{3} = X_{2}\\)。先算后面非一阶，做初等变换得到：",
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            "content": "\\[\n[ A - 3 E | b ] \\rightarrow \\left[\\begin{array}{c c c c}{- 1}&{3}&{2}&{b _ {1}}\\\\{0}&{8}&{4}&{b _ {1} + b _ {2}}\\\\{0}&{8}&{4}&{\\frac {1}{2} b _ {1} + \\frac {5}{2} b _ {2} + b _ {3}}\\end{array}\\right]\n\\]"
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            "content": "有解必须满足 \\(\\frac{1}{2} b_1 + \\frac{5}{2} b_2 + b_3 = 0\\) ，因此 \\(AX_{2} = 0\\) 等价于：",
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            "content": "\\[\n\\left[ \\begin{array}{c c c} - 1 & 3 & 2 \\\\ 0 & 8 & 4 \\\\ \\frac {1}{2} & \\frac {5}{2} & 1 \\end{array} \\right] X _ {2} = 0\n\\]"
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            "content": "而最后一个条件显然可以被前两行消掉，因此可以无视，随便取 \\(X_{2} = [1, - 1,2]^{T}\\) ，因此 \\(X_{3}\\) 满足",
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            "content": "\\[\n\\left[ \\begin{array}{c c c} - 1 & 3 & 2 \\\\ 0 & 8 & 4 \\\\ 0 & 0 & 0 \\end{array} \\right] X _ {3} = \\left[ \\begin{array}{c} 1 \\\\ 0 \\\\ 0 \\end{array} \\right]\n\\]"
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            "content": "\\(X_{3}\\) 取法无所谓，因此取 \\(X_{3} = [-1,0,0]^{T}\\) 。而 \\(X_{1}\\) 取线性无关即可，如 \\(X_{1} = [-2,0,1]^{T}\\)",
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            "content": "练习 6.12 已知矩阵 A 及其 Jordan 标准型 J，求过渡矩阵 P 使得  \\( P^{-1}AP = J \\)",
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            "content": "\\[\nA = \\left[ \\begin{array}{r r r} 0 & 1 & 0 \\\\ - 4 & 4 & 0 \\\\ - 2 & 1 & 2 \\end{array} \\right], J = \\left[ \\begin{array}{c c c} 2 & & \\\\ & 2 & 1 \\\\ & & 2 \\end{array} \\right]\n\\]"
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            "content": "解设 \\(P = [X_1, X_2, X_3]\\)，则 \\((AX_1, AX_2, AX_3) = (2X_1, 2X_2, X_2 + 2X_3)\\)，得到方程组 \\((A - 2E)X_1 = 0, (A - 2E)X_2 = 0, (A - 2E)X_3 = X_2\\)。做初等变换后：",
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            "content": "\\[\n[ A | b ] \\to \\left[ \\begin{array}{c c c c} - 2 & 1 & 0 & b _ {1} \\\\ 0 & 0 & 0 & b _ {2} - 2 b _ {1} \\\\ 0 & 0 & 0 & b _ {3} - b _ {1} \\end{array} \\right]\n\\]"
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            "content": "因此要使 \\(AX_{3} = X_{2}\\) 有解，\\(X_{2}\\) 必须满足 \\(b_{3} = b_{1}\\)，先选取 \\(X_{2} = [1,2,1]^{T}\\)，解出 \\(X_{3} = \\left[-\\frac{1}{2},0,0\\right]^{T}\\)，\\(X_{1}\\) 要线性无关，取 \\([0,0,1]^{T}\\) 即可。",
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            "content": "\\(V\\) 是 \\(K\\) 上的线性空间，\\(f: V \\to K\\) 且满足 \\(f(\\alpha + \\beta) = f(\\alpha) + f(\\beta), f(k\\alpha) = kf(\\alpha)\\)，则称 \\(f\\) 是 \\(V\\) 上的线性函数。",
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            "content": "\\(V\\) 是 \\(K\\) 上的线性空间，\\(f: V \\times V \\to K\\)，且满足 \\(f(k_1\\alpha_1 + k_2\\alpha_2, \\beta) = k_1f(\\alpha_1, \\beta) + k_2f(\\alpha_2, \\beta), f(\\alpha, l_1\\beta_1 + l_2\\beta_2) = l_1f(\\alpha, \\beta_1) + l_2f(\\alpha, \\beta_2)\\)，则称 \\(f\\) 是 \\(V\\) 上的双线性函数",
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            "content": "7.1.2 双线性函数度量矩阵"
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            "content": "V 是 K 上的 n 维线性空间， \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  是 V 的一组基，f 为 V 上的双线性函数，则 f 的度量矩阵 G 定义如下：",
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            "content": "\\[\nG = \\left[ \\begin{array}{c c c c} f (\\epsilon_ {1}, \\epsilon_ {1}) & f (\\epsilon_ {1}, \\epsilon_ {2}) & \\dots & f (\\epsilon_ {1}, \\epsilon_ {n}) \\\\ f (\\epsilon_ {2}, \\epsilon_ {1}) & f (\\epsilon_ {2}, \\epsilon_ {2}) & \\dots & f (\\epsilon_ {2}, \\epsilon_ {n}) \\\\ \\vdots & \\vdots & & \\vdots \\\\ f (\\epsilon_ {n}, \\epsilon_ {1}) & f (\\epsilon_ {n}, \\epsilon_ {2}) & \\dots & f (\\epsilon_ {n}, \\epsilon_ {n}) \\end{array} \\right]\n\\]"
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            "content": "若  \\( x, y \\in V \\)  在  \\( \\epsilon_{1}, \\cdots, \\epsilon_{n} \\)  下的坐标为 X, Y，则  \\( f(\\alpha, \\beta) = X^{T}GY \\)",
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            "content": "\\(A, B\\) 是 \\(K\\) 上的 \\(n\\) 阶方阵，若存在 \\(K\\) 上的可逆 \\(n\\) 阶方阵 \\(T\\) 使得 \\(B = T^T A T\\) ，则称 \\(A, B\\) 合同。",
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            "content": "笔记 从矩阵角度来看，合同相当于 A 通过一系列相同的行与列变换变为 B",
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            "content": "定理 7.2 (不同基下度量矩阵关系)"
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            "content": "矩阵 \\(A, B\\) 合同当且仅当它们是双线性函数两组基下的矩阵。",
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            "content": "7.2 二次型"
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            "content": "- \\(A = \\left( \\begin{array}{cc}A_{1} & \\alpha \\\\ \\alpha^{T} & a_{nn} \\end{array} \\right)\\)，若 \\(A_{1}\\) 可逆，则取 \\(P = \\left( \\begin{array}{cc}E_{n - 1} & -A_{1}^{-1}\\alpha \\\\ 0 & 1 \\end{array} \\right)\\) (用列变换记忆)，得到",
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            "content": "\\[\nP ^ {T} A P = \\left( \\begin{array}{c c} E _ {n - 1} & 0 \\\\ 0 & a _ {n n} - \\alpha^ {T} A _ {1} ^ {- 1} \\alpha \\end{array} \\right)\n\\]"
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            "content": "\\(\\bullet A = \\left( \\begin{array}{cc}a_{11} & \\alpha^T\\\\ \\alpha & A_{n - 1} \\end{array} \\right)\\) ，若 \\(a_{11}\\neq 0\\) ，取 \\(P = \\left( \\begin{array}{cc}1 & -\\frac{1}{a_{11}}\\alpha^{T}\\\\ 0 & E_{n - 1} \\end{array} \\right)\\) ，则",
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            "content": "\\[\nP ^ {T} A P = \\left( \\begin{array}{c c} a _ {1 1} & 0 \\\\ 0 & A _ {n - 1} - \\frac {1}{a _ {1 1}} \\alpha \\alpha^ {T} \\end{array} \\right)\n\\]"
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            "content": "- 若想对上面的右下角分块 \\(A_{n - 1} - \\frac{1}{a_{11}}\\alpha \\alpha^T\\) 做合同变换，变为 \\(P_{1}^{T}\\left(A_{n - 1} - \\frac{1}{a_{11}}\\alpha \\alpha^{T}\\right)P_{1}\\)，则 \\(Q = P\\left( \\begin{array}{ll}1 & 0\\\\ 0 & P_{1} \\end{array} \\right)\\) (用列变换记忆)，得到",
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            "content": "\\[\nQ ^ {T} A Q = \\left( \\begin{array}{c c} a _ {1 1} & 0 \\\\ 0 & P _ {1} ^ {T} \\left(A _ {n - 1} - \\frac {1}{a _ {1 1}} \\alpha \\alpha^ {T}\\right) P _ {1} \\end{array} \\right)\n\\]"
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            "content": "7.2 二次型"
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            "content": "7.2.1 二次型概念及其矩阵"
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            "content": "定义 7.5 (二次型)"
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            "content": "定义多元二次多项式  \\( f(x_{1},x_{2},\\cdots,x_{n})=\\sum_{i=1}^{n}a_{ii}x_{i}^{2}+\\sum_{1\\leq i<j\\leq n}2a_{ij}x_{i}x_{j} \\)  为二次型",
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            "content": "定义 7.6 (二次型的矩阵)"
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            "content": "对于任意二次型  \\( f(x_{1},x_{2},\\cdots,x_{n}) \\) ，总能找到一个对称矩阵 A，使得  \\( f(x_{1},\\cdots,x_{n})=x^{T}Ax \\) ，称矩阵 A 为二次型的矩阵",
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            "content": "笔记 对于一个二次型  \\( f(x_{1},\\cdots,x_{n}) \\) ，存在不止一个 A 使得  \\( f(x_{1},\\cdots,x_{n})=x^{T}Ax \\) ，但是二次型的矩阵特指那个对称矩阵。",
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            "content": "练习7.1二次型的矩阵(1)给定二次型 \\(f(x_{1},x_{2},x_{3}) = 2x_{1}x_{2} + 2x_{1}x_{3} - 6x_{2}x_{3}\\) ，计算二次型的矩阵",
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            "content": "(2) 给定二次型  \\( f(X) \\) ，计算二次型的矩阵：",
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            "content": "\\[\nf (x _ {1}, x _ {2}) = \\left( \\begin{array}{c c} x _ {1} & x _ {2} \\end{array} \\right) \\left( \\begin{array}{c c} 2 & 1 \\\\ 3 & 1 \\end{array} \\right) \\binom{x _ {1}}{x _ {2}}\n\\]"
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            "content": "(3) 重点：求 \\( f(x_{1}, \\cdots, x_{n}) = \\sum_{i=1}^{m} (a_{i1}x_{1} + \\cdots + a_{in}x_{n})^{2} \\) 的矩阵",
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            "content": "解(1) \\(A = \\left( \\begin{array}{ccc}0 & 1 & 1\\\\ 1 & 0 & -3\\\\ 1 & -3 & 0 \\end{array} \\right)\\)",
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            "content": "7.2 二次型"
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            "content": "(2) 展开得到 \\( f(x_{1}, x_{2}) = 2x_{1}^{2} + 4x_{1}x_{2} + x_{2}^{2} \\)，因此矩阵为",
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            "content": "\\[\n\\left( \\begin{array}{c c} 2 & 2 \\\\ 2 & 1 \\end{array} \\right)\n\\]"
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            "content": "(3) 令 \\(A = (a_{ij})\\)，则",
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            "content": "\\[\nf = \\sum_ {i = 1} ^ {m} (a _ {i 1} x _ {1} + \\dots + a _ {i n} x _ {n}) ^ {2} = \\sum_ {i = 1} ^ {m} \\left[ \\left( \\begin{array}{c c c} x _ {1} & \\dots & x _ {n} \\end{array} \\right) \\left( \\begin{array}{c} a _ {i 1} \\\\ \\vdots \\\\ a _ {i n} \\end{array} \\right) \\right] ^ {2}\n\\]"
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            "content": "\\[\n= \\sum_ {i = 1} ^ {m} \\left[ \\left( \\begin{array}{c c c} x _ {1} & \\dots & x _ {n} \\end{array} \\right) \\left( \\begin{array}{c} a _ {i 1} \\\\ \\vdots \\\\ a _ {i n} \\end{array} \\right) \\left( \\begin{array}{c c c} a _ {i 1} & \\dots & a _ {i n} \\end{array} \\right) \\left( \\begin{array}{c} x _ {1} \\\\ \\vdots \\\\ x _ {n} \\end{array} \\right) \\right] = X ^ {T} (A ^ {T} A) X\n\\]"
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            "content": "其中计算出 \\(A^T A\\) 的过程为",
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            "content": "\\[\nA = \\left[ \\begin{array}{c} A _ {1} \\\\ A _ {2} \\\\ \\vdots \\\\ A _ {m} \\end{array} \\right] \\Rightarrow A ^ {T} A = \\left[ \\begin{array}{c c c c} A _ {1} ^ {T} & A _ {2} ^ {T} & \\dots & A _ {m} ^ {T} \\end{array} \\right] \\left[ \\begin{array}{c} A _ {1} \\\\ A _ {2} \\\\ \\vdots \\\\ A _ {m} \\end{array} \\right] = \\sum_ {i = 1} ^ {m} A _ {i} ^ {T} A _ {i}\n\\]"
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            "content": "给定二次型 \\(f = x^{T}Ax, g = y^{T}By\\)，若存在可逆矩阵 \\(T\\)，使得 \\(x = Ty\\)，且 \\(f = y^{T}(T^{T}AT)y = y^{T}By = g\\)，则称二次型 \\(f, g\\) 等价（即 \\(A, B\\) 合同）",
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            "content": "若 \\(A = C^T BC\\) ，则",
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            "content": "- 秩不变： \\( r(A)=r(B) \\)",
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            "content": "- 分块合同：若 \\(A_{1}, A_{2}\\) 分别与 \\(B_{1}, B_{2}\\) 合同，则 \\(\\operatorname{diag}\\{A_{1}, A_{2}\\}\\) 与 \\(\\operatorname{diag}\\{B_{1}, B_{2}\\}\\) 合同",
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            "content": "\\[\n\\left( \\begin{array}{c c} 1 & 0 \\\\ 0 & - 1 \\end{array} \\right) = \\left( \\begin{array}{c c} 1 & 0 \\\\ 0 & i \\end{array} \\right) \\left( \\begin{array}{c c} 1 & 0 \\\\ 0 & 1 \\end{array} \\right) \\left( \\begin{array}{c c} 1 & 0 \\\\ 0 & i \\end{array} \\right) = \\left( \\begin{array}{c c} - 1 & 0 \\\\ 0 & i \\end{array} \\right) \\left( \\begin{array}{c c} 1 & 0 \\\\ 0 & 1 \\end{array} \\right) \\left( \\begin{array}{c c} - 1 & 0 \\\\ 0 & i \\end{array} \\right)\n\\]"
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            "content": "\\[\nQ ^ {T} P ^ {T} A P Q = \\left( \\begin{array}{c c} a _ {1 1} & \\\\ & Q _ {1} ^ {T} B _ {1} Q _ {1} \\end{array} \\right)\n\\]"
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            "content": "若 \\(a_{11} = 0\\) ，但 \\(a_{kk}\\neq 0\\) ，则取 \\(P(1,k)\\) 为 \\(E_{n}\\) 第一列与第 \\(k\\) 列互换的矩阵，则 \\(P^T (1,k)AP(1,k)\\) 的 \\(a_{11}^{\\prime}\\neq 0\\) 且两个矩阵合同，转换为第一种情况。",
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            "content": "\\[\nA = \\left( \\begin{array}{c c c} 0 & a _ {1 2} & * \\\\ a _ {1 2} & 0 & * \\\\ * & * & * \\end{array} \\right) \\to \\left( \\begin{array}{c c c} 2 a _ {1 2} & a _ {1 2} & * \\\\ a _ {1 2} & a _ {2 2} & * \\\\ * & * & * \\end{array} \\right)\n\\]"
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            "content": "转换为第一种类型",
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            "content": "对于二次型 \\( f = x^{T}Ax \\)，若 \\( A \\) 与对角阵 \\( D \\) 合同，则称 \\( g = z^{T}Dz \\) 为 \\( f \\) 的标准型",
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            "content": "1. 若无 \\(x_{1}\\) 的平方项，但有 \\(x_{1}x_{2}\\) 交叉项，则用 \\(x_{1} = y_{1} + y_{2}, x_{2} = y_{1} - y_{2}\\) 换元。",
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            "content": "2. 把 \\(x_{1}\\) 的项放在一起，全部配方。",
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            "content": "3. 把凑成的  \\( (x_{1}+\\cdots)^{2} \\)  设为新变量",
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            "content": "4. 以此类推",
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            "content": "练习 7.3 配方法 (1) 用配方法计算  \\( f(x_{1}, x_{2}, x_{3}) = 2x_{1}x_{2} + 2x_{1}x_{3} - 6x_{2}x_{3} \\)  的标准型解 (1) 无  \\( x_{1}^{2} \\)  项，因此令  \\( x_{1} = y_{1} + y_{2}, x_{2} = y_{1} - y_{2}, x_{3} = y_{3} \\) ，得到",
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            "content": "\\[\nf = 2 (y _ {1} - y _ {3}) ^ {2} - 2 y _ {2} ^ {2} - 2 y _ {3} ^ {2} + 8 y _ {2} y _ {3}\n\\]"
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            "content": "令 \\(x_{1} = y_{1} - y_{3}, x_{2} = y_{2}, x_{3} = y_{3}\\) 得到 \\(f = 2x_{1}^{2} - 2(x_{2}^{2} - 4x_{2}x_{3} + 4x_{3}^{2}) + 6x_{3}^{2}\\), 再做换元 \\(y_{1} = x_{1}, y_{2} = x_{2} - 2x_{3}, y_{3} = x_{3}\\), 得到",
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            "content": "对于对称矩阵 A 使用以下格式计算初等变换 (注意这不一定得出正交矩阵):",
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            "content": "\\[\n\\binom{{{{A}}}}{{{{E}}}} \\rightarrow \\binom{{{{C ^ {T} A C}}}}{{{{C}}}} \\quad \\text {对} \\mathbf {A} \\text {行变换和对应列变换，对} \\mathbf {I} \\text {只列变换}\n\\]"
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            "content": "具体的步骤就是证明标准型存在性中的过程：若 \\(A\\) 首对角为零，存在非零对角，则互换位置。若对角均为零，则找到第一行非零的元素加到首对角位置(下面的题目马上会用)",
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            "content": "笔记 做同样的行列变换一定要小心，两种变换并非 “同时” 做的，而是先后做的，即做行变换时的矩阵已经是列变换后的，而非原本的矩阵！",
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            "content": "练习7.4标准型计算：初等变换法(1)简单：计算 \\( f = x_{1}^{2} + 2x_{1}x_{2} + 2x_{2}^{2} + 4x_{2}x_{3} + 4x_{3}^{2} \\) 的标准型",
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            "content": "(2) 对角均零：计算 \\( f(x_{1}, x_{2}, x_{3}) = 2x_{1}x_{2} + 2x_{1}x_{3} - 6x_{2}x_{3} \\)，计算其标准型，并求出过渡矩阵 \\( C \\)",
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            "content": "解(1)结果为 \\(f = y_{1}^{2} + y_{2}^{2}\\) ，千万小心别行列变换错了",
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            "content": "(2) 首先显然 \\(A = \\left( \\begin{array}{ccc}0 & 1 & 1\\\\ 1 & 0 & -3\\\\ 1 & -3 & 0 \\end{array} \\right)\\) ，列出 \\(\\left( \\begin{array}{c}A\\\\ E \\end{array} \\right)\\) 并做行列初等变换，由于 \\(A\\) 对角均为0，因此将第二列和第二行加到首对角位置：",
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            "content": "\\[\n\\left(\\begin{array}{c c c}0&1&1\\\\1&0&- 3\\\\1&- 3&0\\\\1&0&0\\\\0&1&0\\\\0&0&1\\end{array}\\right)\\rightarrow \\left(\\begin{array}{c c c}2&1&- 2\\\\1&0&- 3\\\\- 2&- 3&0\\\\1&0&0\\\\1&1&0\\\\0&0&1\\end{array}\\right)\\rightarrow \\left(\\begin{array}{c c c}2&1&0\\\\1&0&- 2\\\\0&- 2&- 2\\\\1&0&1\\\\1&1&1\\\\0&0&1\\end{array}\\right)\\rightarrow \\dots\n\\]"
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            "content": "最终变为结果为:",
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            "content": "\\[\n\\left( \\begin{array}{c c c} 2 & 0 & 0 \\\\ 0 & - \\frac {1}{2} & 0 \\\\ 0 & 0 & 6 \\\\ 1 & - \\frac {1}{2} & 3 \\\\ 1 & \\frac {1}{2} & - 1 \\\\ 0 & 0 & 1 \\end{array} \\right) = \\binom{C ^ {T} A C}{C}\n\\]"
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            "content": "定理 7.8 (正交对角化的可行性)"
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            "content": "任何实对称矩阵 A 一定正交相似于对角阵。",
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            "content": "证明 做数学归纳法，n=1 显然，假设 n-1 成立，任取特征值  \\( \\lambda \\) ，对应单位特征向量  \\( \\alpha_{1} \\) ，将  \\( \\alpha_{1} \\)  扩充为  \\( R^{n} \\)  的一组标准正交基  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) ，令  \\( T=(\\alpha_{1},\\cdots,\\alpha_{n}) \\) ，则",
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            "content": "\\[\nA T = T \\left( \\begin{array}{c c} \\lambda & \\alpha^ {T} \\\\ 0 & A _ {n - 1} \\end{array} \\right) \\Rightarrow T ^ {T} A T = \\left( \\begin{array}{c c} \\lambda & \\alpha^ {T} \\\\ 0 & A _ {n - 1} \\end{array} \\right)\n\\]"
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            "content": "由于 \\(T^T A T\\) 对称，因此 \\(\\alpha^T = 0\\) ，再根据归纳假设，存在正交矩阵 \\(P_{1}\\) 使得 \\(P_{1}^{T}A_{n - 1}P_{1}\\) 为对角阵，因此令 \\(P =\\)",
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            "content": "7.2 二次型"
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            "content": "\\(\\mathrm{diag}(1, P_1)\\)，此时",
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            "content": "\\[\nP ^ {T} T ^ {T} A T P = \\left( \\begin{array}{c c} \\lambda & 0 \\\\ 0 & P ^ {T} A _ {n - 1} P \\end{array} \\right)\n\\]"
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            "content": "显然 \\(TP = Q\\) ，则 \\(Q\\) 也是正交阵。",
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            "content": "定理 7.9 (标准型计算：正交变换)"
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            "content": "正交变换计算标准型过程如下：",
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            "content": "- 计算 \\(A\\) 的特征值 \\(\\lambda_1, \\cdots, \\lambda_n\\)",
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            "content": "- 计算 \\(\\lambda_{i}\\) 对应的特征向量 \\(\\mathbf{v}_{i}\\)，不同特征值的特征向量必正交，同一特征值的特征向量需正交化",
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            "content": "- 将特征向量单位化，组成 \\(T = [\\mathbf{v}_1, \\dots, \\mathbf{v}_n]\\)",
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            "content": "练习 7.5 正交变换法 (1) 已知矩阵 A 如下，(a) 若  \\( tE + A \\)  为正定矩阵，求 t 的取值范围 (b) 求正交矩阵 T 使得  \\( T^{T}AT \\)  为对角矩阵",
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            "content": "\\[\nA = \\left[ \\begin{array}{r r r} 2 & 2 & - 1 \\\\ 2 & 5 & - 4 \\\\ - 2 & - 4 & 5 \\end{array} \\right]\n\\]"
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            "content": "解 (1) \\( |\\lambda E - A| = (\\lambda - 1)^{2} (\\lambda - 10) \\) ，特征值为 1, 10，正定要求 t > -1。",
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            "content": "先求 \\((E - A)X = 0\\) ，得到：",
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            "content": "\\[\n\\left[ \\begin{array}{c c c} 1 & 2 & - 2 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{array} \\right] X = 0 \\Rightarrow X = k _ {1} \\left( \\begin{array}{c} - 2 \\\\ 1 \\\\ 0 \\end{array} \\right) + k _ {2} \\left( \\begin{array}{c} 2 \\\\ 0 \\\\ 1 \\end{array} \\right)\n\\]"
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            "content": "单位化得到 \\(\\frac{1}{\\sqrt{5}} (-2,1,0)^T,\\frac{1}{\\sqrt{5}} (2,0,1)^T\\)",
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            "content": "再求 \\((10E - A)X = 0\\)",
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            "content": "\\[\n\\left[ \\begin{array}{c c c} 2 & 4 & 5 \\\\ 0 & 1 & 1 \\\\ 0 & 0 & 0 \\end{array} \\right] X = 0 \\Rightarrow X = k _ {1} \\left( \\begin{array}{c} 1 \\\\ 2 \\\\ - 2 \\end{array} \\right)\n\\]"
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            "content": "单位化得到 \\(\\frac{1}{3} (1,2, - 2)\\) 。",
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            "content": "最终结果为:",
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            "content": "\\[\nT = \\left[ \\begin{array}{c c c} - \\frac {2}{\\sqrt {5}} & \\frac {2}{3 \\sqrt {5}} & \\frac {1}{3} \\\\ - \\frac {1}{\\sqrt {5}} & \\frac {4}{3 \\sqrt {5}} & \\frac {2}{3} \\\\ 0 & \\frac {5}{3 \\sqrt {5}} & - \\frac {2}{3} \\end{array} \\right], T ^ {T} A T = \\left[ \\begin{array}{c c c} 1 & & \\\\ & 1 & \\\\ & & 1 0 \\end{array} \\right]\n\\]"
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            "content": "定理 7.10 (二次型与特征值)"
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            "content": "设 n 阶实对称矩阵 A 的全部特征值排序为： \\( \\lambda_{1} \\geq \\lambda_{2} \\geq \\cdots \\geq \\lambda_{n} \\) ，则",
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            "content": "\\[\n\\max _ {X \\in \\mathbb {R} ^ {n}} \\frac {X ^ {T} A X}{X ^ {T} X} = \\lambda_ {1}, \\min _ {X \\in \\mathbb {R} ^ {n}} \\frac {X ^ {T} A X}{X ^ {T} X} = \\lambda_ {n}\n\\]"
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            "content": "证明 由于 \\(\\lambda_1E - A\\) 的特征值均非负，故 \\(\\lambda_1E - A\\) 半正定，因此 \\(\\forall X\\in \\mathbb{R}^n\\) 有 \\(X^{T}(\\lambda E - A)X\\geq 0\\) ，得到\\(X^{T}AX\\leq \\lambda_{1}X^{T}X\\) ，且 \\(\\lambda_{1}\\) 是 \\(\\lambda_1E - A\\) 的特征值，故 \\(\\exists X_0\\neq 0\\) 使得 \\(X_0^T (\\lambda_1E - A)X_0 = 0\\)",
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            "content": "练习 7.6 二次型的特征值研究 (1) 重点: A, B 是 n 阶实对称矩阵, A 的特征值都大于 a, B 的特征值都大于 b,",
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            "content": "79"
        },
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            "content": "7.3 (半) 正定二次型的研究"
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            "type": "text",
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            "angle": 0,
            "content": "证明 \\(A + B\\) 的特征值都大于 \\(a + b\\)",
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        },
        {
            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "(2) B 是 n 阶实矩阵， \\( B^{T}B \\)  的全部特征值按照大小顺序排为  \\( \\lambda_{1} \\geq \\lambda_{2} \\geq \\cdots \\geq \\lambda_{n} \\) ，证明：B 的任意实特征值  \\( \\mu \\)  满足  \\( \\sqrt{\\lambda_{n}} \\leq |\\mu| \\leq \\sqrt{\\lambda_{1}} \\)",
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        },
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            "content": "证明 (1) 根据前面定理可知 \\(X^{T}AX > aX^{T}X, X^{T}BX > bX^{T}X\\)，从而 \\(X^{T}(A + B)X > (a + b)X^{T}X\\)，因此 \\(A + B\\) 的特征值大于 \\(a + b\\)",
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            "content": "(2) 由于 \\(B\\alpha = \\mu \\alpha\\)，因此 \\(\\alpha^T B^T = \\mu \\alpha^T\\)，得到",
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            "content": "\\[\n\\alpha^ {T} B ^ {T} B \\alpha = \\mu^ {2} \\alpha^ {T} \\alpha \\in [ \\lambda_ {n} \\alpha^ {T} \\alpha , \\lambda_ {1} \\alpha^ {T} \\alpha ] \\Rightarrow \\lambda_ {n} \\leq \\mu^ {2} \\leq \\lambda_ {1}\n\\]"
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            "content": "7.2.5 R, C 上的规范型"
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            "content": "前面根据标准型对二次型进行分类，但是可以发现二次型的标准型是不唯一的，因此在 \\(\\mathbb{R},\\mathbb{C}\\) 上可以进一步划分，即每个规范型表示一个类。",
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            "content": "定理 7.11 (C 上的规范型)"
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            "content": "\\(\\mathbb{C}\\) 上任一二次型 \\(f\\) 都等价于",
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            "content": "\\[\ng = u _ {1} ^ {2} + \\dots + u _ {r} ^ {2}\n\\]"
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            "content": "推论 7.1 (C 上合同充要条件)"
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            "content": "C 上两个对称矩阵合同的充要条件为它们的秩相同。",
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            "content": "定理 7.12 (R 上的规范型)"
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            "content": "\\(\\mathbb{R}\\) 上每个二次型 \\(f\\) 都等价于一个如下二次型。将 \\(p\\) 称为正惯性指数，\\(r - p\\) 称为负惯性指数",
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            "content": "7.3 (半) 正定二次型的研究"
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            "content": "给定实数域上的二次型  \\( f(x_{1},x_{2},\\cdots,x_{n}) \\) ，若对任意不全为 0 的  \\( x_{1},\\cdots,x_{n} \\)  有  \\( f(x_{1},\\cdots,x_{n})>0 \\) ，则称  \\( f(x_{1},\\cdots,x_{n}) \\)  为正定矩阵。",
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            "content": "- 存在可逆矩阵 \\(C\\)，使得 \\(A = C^T C\\)，也可以分解为非方阵 \\(A = \\left( \\begin{array}{cc} C^T & O^T \\end{array} \\right) \\left( \\begin{array}{c} C \\\\ O \\end{array} \\right)\\).",
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            "content": "所有特征值都大于0",
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            "content": "- \\(A\\) 的各阶顺序主子式/所有主子式行列式大于0",
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            "content": "证明 等价于顺序主子式大于 0 需要证明：对 A 做如下分块：",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c} A _ {1} & \\alpha \\\\ \\alpha^ {T} & a _ {n n} \\end{array} \\right]\n\\]"
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            "content": "正定推顺序主子式：若 \\(A\\) 正定，则 \\(A_{1}\\) 正定，因此 \\(A_{1}\\) 特征值均为正，\\(|A_1| > 0, |A| > 0\\)，根据归纳法成立。顺序主子式推正定：\\(n = 1\\) 时显然成立，\\(A\\) 是 \\(n\\) 阶矩阵时，\\(A_{1}\\) 是 \\(n - 1\\) 阶矩阵且顺序主子式均正，根据归纳假设 \\(A_{1}\\) 正定，故存在 \\(P_{1}\\) 使得 \\(P_{1}^{T} A_{1} P_{1} = E_{n - 1}\\)，因此令：",
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            "content": "\\[\nP _ {2} = \\left[ \\begin{array}{c c} P _ {1} & \\\\ & 1 \\end{array} \\right], P _ {3} = \\left[ \\begin{array}{c c} E _ {n - 1} & - P _ {1} ^ {T} \\alpha \\\\ & 1 \\end{array} \\right]\n\\]"
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            "content": "得到：",
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            "content": "\\[\nP _ {3} ^ {T} P _ {2} ^ {T} A P _ {2} P _ {3} = \\left[ \\begin{array}{c c} E _ {n - 1} & \\\\ & a _ {n n} - \\alpha^ {T} P _ {1} P _ {1} ^ {T} \\alpha \\end{array} \\right]\n\\]"
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            "content": "而根据乘积矩阵的行列式性质得到",
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            "content": "\\[\na _ {n n} - \\alpha^ {T} P _ {1} P _ {1} ^ {T} \\alpha = | A | | P _ {2} | ^ {2} | P _ {3} | ^ {2} > 0\n\\]"
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            "content": "令 \\(P = P_{2}P_{3}\\mathrm{diag}\\{E_{n - 1},\\frac{1}{\\sqrt{a_{nn} - \\alpha^{T}P_{1}P_{1}^{T}\\alpha}}\\}\\) 得到",
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            "content": "\\[\nP ^ {T} A P = E\n\\]"
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            "content": "7.3 (半) 正定二次型的研究"
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            "content": "解(1)算顺序主子式即可，得到 \\( 1-t^{2}>0,-5t^{2}-4t>0 \\) ，因此 \\( -\\frac{4}{5}<t<0 \\)",
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            "content": "(2) 看特征值即可",
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            "content": "练习 7.9 正定矩阵性质的简单应用 (1) A 为 n 阶正定矩阵，证明：任意 n 阶实矩阵 B 有  \\( r(B^{T}AB) = r(B) \\)",
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            "content": "(2) \\(A\\) 为 \\(n\\) 阶正定矩阵，\\(B\\) 为 \\(n \\times m\\) 实矩阵，证明：若 \\(r(B) = m\\)，则 \\(C = B^T AB\\) 为正定矩阵证明 (1) 由于 \\(A\\) 正定，存在满秩矩阵 \\(C\\) 使得 \\(A = C^T C\\)，因此 \\(B^T AB = (CB)^T (CB)\\)，而根据 \\(A^T A\\) 的结论可知 \\(r((CB)^T (CB)) = r(CB) = r(B)\\)，因此 \\(r(B^T AB) = r(B)\\)",
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            "content": "(2) 由于 \\(A\\) 正定，因此 \\(B^T AB\\) 实对称，\\(\\forall X, X^T CX = (BX)^T ABX\\)，根据 \\(A\\) 的正定性可知其大于0，因此为正定矩阵",
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            "content": "定理 7.15 (正定矩阵对角性质)"
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            "content": "\\(A\\) 为 \\(n\\) 阶正定矩阵，则",
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            "content": "1.  \\( a_{ii} > 0, \\forall i = 1, 2, \\cdots, n \\)",
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            "content": "2.  \\( 2|a_{ij}| < a_{ii} + a_{jj} (i \\neq j) \\)",
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            "content": "3. \\(A\\) 的所有元素中，绝对值最大的元素一定是在主对角线上，且一定是正数",
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            "content": "证明 (1) 不妨取 \\(X = e_{i}\\)，则 \\(X^{T}AX = a_{ii} > 0\\)",
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            "content": "若对任意  \\( x_{1},\\cdots,x_{n} \\)  均有  \\( f(x_{1},\\cdots,x_{n})\\geq0 \\) ，则称  \\( f(x_{1},\\cdots,x_{n}) \\)  为半正定二次型，对应矩阵 A 为半正定矩阵。",
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            "content": "\\( f(x_{1},\\cdots,x_{n})=X^{T}AX \\)  为半正定二次型当且仅当：",
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            "content": "7.3 (半) 正定二次型的研究"
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            "content": "证明 根据 \\(A = C^{T}\\left( \\begin{array}{cc}E_{r} & O\\\\ O & O \\end{array} \\right)C\\) 以及",
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            "content": "\\[\n\\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) = \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) = \\binom{E _ {r}}{O} \\left( \\begin{array}{c c} E _ {r} & O \\end{array} \\right)\n\\]"
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            "content": "可以得到两种分解：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} C _ {1} = \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) C \\Rightarrow A = C _ {1} ^ {T} C _ {1} \\\\ C _ {2} = \\left( \\begin{array}{c c} E _ {r} & O \\end{array} \\right) C \\Rightarrow A = C _ {2} ^ {T} C _ {2} \\end{array} \\right.\n\\]"
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            "content": "无论哪种分解列均不满秩。",
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            "content": "定理 7.18 (半正定矩阵的  \\( CC^{T} \\)  分解)"
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            "content": "\\( f(x_{1},\\cdots,x_{n})=X^{T}AX \\)  为半正定二次型等价于下面两个命题",
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            "content": "- 存在不行满秩的 \\(n \\times n\\) 矩阵 \\(C\\)，使得 \\(A = CC^T\\)",
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            "content": "- 存在列满秩 (行不满秩) 的 \\(n \\times r (r < n)\\) 矩阵 \\(C\\)，使得 \\(A = CC^T\\)",
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            "content": "证明 右推左：直接 \\(X^{T}CC^{T}X = (C^{T}X)^{T}(C^{T}X)\\)，该多项式是 \\(C^{T}X\\) 列向量中元素的平方和，因此大于等于0。左推右：合同的 \\(A = B^{T}\\mathrm{diag}\\{E_{r},O\\} B\\) 也可以写为 \\(A = C\\mathrm{diag}\\{E_r,O\\} C^T\\)，因此把中间的分解了即可。笔记 特别的可以取 \\(A = \\alpha^{T}\\alpha\\)，\\(\\alpha\\) 为列向量(列满秩)，此时 \\(A\\) 是半正定的。",
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            "content": "练习7.10半正定矩阵 \\(CC^T\\) 分解 \\((1)\\alpha \\in \\mathbb{R}^n\\) ，证明： \\(\\alpha \\alpha^T\\) 一定是半正定矩阵",
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            "content": "(2) 重点: \\(A\\) 的元素 \\(a_{ij} = (i \\times j)\\), \\(B\\) 的元素 \\(b_{ij} = \\left(\\frac{1}{i \\times j}\\right)\\), 则 \\(A, B\\) 都是半正定矩阵",
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            "content": "证明 (1) 显然 \\(\\alpha \\alpha^T\\) 是实对称矩阵，从而 \\(X^T \\alpha \\alpha^T X = (\\alpha^T X)^T (\\alpha^T X) = (\\alpha^T X)^2 \\geq 0\\)",
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            "content": "(2) \\(A, B\\) 可进行如下分解，根据 (1) 的结论可证是半正定矩阵",
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            "content": "\\[\nA = \\left[ \\begin{array}{c} 1 \\\\ 2 \\\\ \\vdots \\\\ n \\end{array} \\right] \\left[ \\begin{array}{l l l l} 1 & 2 & \\dots & n \\end{array} \\right], \\quad B = \\left[ \\begin{array}{c} 1 \\\\ \\frac {1}{2} \\\\ \\vdots \\\\ \\frac {1}{n} \\end{array} \\right] \\left[ \\begin{array}{l l l l} 1 & \\frac {1}{2} & \\dots & \\frac {1}{n} \\end{array} \\right]\n\\]"
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            "content": "定理 7.19 (半正定矩阵对角性质)"
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            "content": "半正定矩阵 \\(A\\) 有以下性质：",
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            "content": "1. 对角元素非负",
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            "content": "2.  \\( 2|a_{ij}| \\leq a_{ik} + a_{ki} \\)",
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            "content": "3. 若 \\(a_{ii} = 0\\)，则 \\(\\forall k, a_{ik} = a_{ki} = 0\\)",
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            "content": "(2) 取 \\(x = e_i + e_j\\)，得到 \\(a_{ii} + a_{jj} + 2a_{ij} \\geq 0\\)，同理取 \\(x = e_i - e_j\\)，得到 \\(a_{ii} + a_{jj} - 2a_{ij} \\geq 0\\)，因此 \\(2|a_{ij}| \\leq a_{ik} + a_{ki}\\)",
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            "content": "(3) 根据 (2) 可以得到",
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            "content": "7.3.3 \\(C^T C\\) 的 (半) 正定性"
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            "content": "\\( C^{T}C \\)  对应着特殊的二次型  \\( f(X)=\\sum_{i=1}^{n}(a_{i1}x_{1}+\\cdots+a_{in}x_{n})^{2} \\) ，且有着特殊的正定性，是正定二次型中特殊的一类。",
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            "content": "7.3 (半) 正定二次型的研究"
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            "content": "引理7.1 \\((r(A^T A)\\) 与 \\(r(A))\\)"
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            "content": "\\(A\\) 为实数域上的 \\(m \\times n\\) 的矩阵，则",
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            "content": "\\[\nr (A ^ {T} A) = r (A)\n\\]"
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            "content": "证明 在最小二乘法一节中已经给出了证明"
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            "content": "定理 7.20 ( \\( A^{T}A \\) 的正定性)"
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            "content": "考虑 \\(A\\) 为实数域上的 \\(m \\times n\\) 的矩阵，则",
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            "content": "1. \\(A^T A, AA^T\\) 是半正定矩阵",
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            "content": "2. \\(A^T A\\) 是正定矩阵当且仅当 \\(A\\) 列满秩，\\(AA^T\\) 正定当且仅当 \\(A\\) 行满秩",
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            "content": "且  \\( A^{T}A \\)  对应的二次型为  \\( f(X)=\\sum_{i=1}^{n}(a_{i1}x_{1}+a_{i2}x_{2}+\\cdots+a_{in}x_{n})^{2} \\)",
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            "content": "证明 (半) 正定性：显然 \\(A^T A\\) 实对称，因此 \\(X^T A^T A X = (AX)^2 \\geq 0\\)，显然 \\(A^T A\\) 半正定。而 \\(A^T A\\) 正定当且仅当 \\(r(A^T A) = n\\)，根据 (1) 当且仅当 \\(r(A) = n\\)，这里 \\(n\\) 等于 \\(A\\) 的列数，因此等价于 \\(A\\) 列满秩",
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            "content": "二次型表达式：在二次型矩阵一节已经作为练习给出。",
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            "content": "7.3.4 正定矩阵同时对角化"
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            "content": "定理 7.21 (同时合同对角化)"
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            "content": "若 \\(A, B\\) 是 \\(n\\) 阶半正定矩阵，则存在一个 \\(n\\) 阶可逆矩阵 \\(P\\)，使得 \\(P^T AP, P^T BP\\) 同时为对角矩阵",
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            "content": "证明 由于 A, B 半正定，因此  \\( A + B \\)  半正定，存在可逆矩阵 M 使得",
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            "content": "\\[\nM ^ {T} (A + B) M = \\left( \\begin{array}{c c} E _ {r} & O \\\\ O & O \\end{array} \\right) = M ^ {T} A M + M ^ {T} B M\n\\]"
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            "content": "由于 \\(M^T AM, M^T BM\\) 半正定，因此根据半正定矩阵的对角性质，当 \\(r + 1 \\leq i \\leq n\\) ，其第 \\(i\\) 个对角元都非负，因此 \\(r + 1 \\sim n\\) 的对角元都必须为0。再根据对角性质可知 \\(r + 1 \\sim n\\) 行、列元素均为0，因此",
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            "content": "\\[\nM ^ {T} A M = \\left( \\begin{array}{c c} H _ {r} & O \\\\ O & O \\end{array} \\right), M ^ {T} B M = \\left( \\begin{array}{c c} E _ {r} - H _ {r} & O \\\\ O & O \\end{array} \\right)\n\\]"
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            "content": "存在正交矩阵 T 使得  \\( T^{T}H_{r}T = \\Lambda = \\operatorname{diag}\\{\\lambda_{1}, \\cdots, \\lambda_{r}\\} \\) ，令  \\( P = M\\operatorname{diag}\\{T, E_{n-r}\\} \\) ，此时有",
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            "content": "\\[\nP ^ {T} A P = \\left( \\begin{array}{c c} \\Lambda & O \\\\ O & O \\end{array} \\right), P ^ {T} B P = \\left( \\begin{array}{c c} E _ {r} - \\Lambda & O \\\\ O & O \\end{array} \\right)\n\\]"
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            "content": "定理 7.22 (AB = BA 同时相似对角化)"
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            "content": "\\[\nP ^ {- 1} A P, P ^ {- 1} B P\n\\]"
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            "content": "证明 由于 A 可相似对角化，因此存在 M，使得  \\( M^{-1}AM = \\text{diag}\\{\\lambda_{1}E, \\lambda_{2}E, \\cdots, \\lambda_{r}E\\} \\) ，而",
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            "content": "\\[\nM ^ {- 1} A B M = (M ^ {- 1} A M) (M ^ {- 1} B M) = M ^ {- 1} B A M = (M ^ {- 1} B M) (M ^ {- 1} A M)\n\\]"
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            "content": "7.3 (半) 正定二次型的研究"
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            "content": "可以得到",
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            "content": "\\[\nT ^ {- 1} A T = \\operatorname{diag} \\left\\{\\lambda_ {1} E, \\dots , \\lambda_ {r} E \\right\\}, T ^ {- 1} B T = \\operatorname{diag} \\left\\{\\Lambda_ {1}, \\dots , \\Lambda_ {r} \\right\\}\n\\]"
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            "content": "同时为对角矩阵。",
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            "content": "7.3.5 正定矩阵的四则运算性质"
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            "content": "定理 7.23 (正定的四则运算)"
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            "content": "若 \\(A, B\\) 是正定矩阵，则 \\(A^{-1}, A^{k}, A + B, A^{*}\\) 都是正定矩阵，但 \\(AB\\) 不一定正定 (原因在于不一定对称)，若 \\(AB\\) 实对称，则 \\(AB\\) 是正定矩阵",
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            "content": "证明 (1) \\( A^{-1}, A^{k}, A^{*} \\) : 由于 A 正定, 因此 A 的特征值  \\( \\lambda_{i} \\)  均大于 0,  \\( A^{-1} \\)  的特征值  \\( \\lambda_{i}^{-1} > 0 \\) ,  \\( A^{*} \\)  的特征值  \\( \\prod_{j \\neq i} \\lambda_{j} > 0 \\) ,  \\( A^{k} \\)  的特征值  \\( \\lambda_{i}^{k} > 0 \\) , 因此均正定。",
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            "content": "(2) \\(A + B\\) 不能用特征值，只能用合同，显然 \\(A + B\\) 实对称，而任意非零 \\(x \\in \\mathbb{R}^n\\) ，有 \\(X^T (A + B)X = X^T AX + X^T BX > 0\\) ，因此 \\(A + B\\) 实对称。",
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            "content": "(3) \\(AB\\) 正定：由于 \\(AB\\) 实对称，此时 \\((AB)^T = B^TA^T = BA\\) 。由于 \\(A, B\\) 正定，故存在可逆 \\(P, Q\\) 使得 \\(A = P^TP, B = Q^TQ, AB = P^TPQ^TQ\\) ，而",
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            "content": "\\[\nQ A B Q ^ {- 1} = Q P ^ {T} P Q ^ {T} (Q Q ^ {- 1}) = (P Q ^ {T}) ^ {T} (P Q ^ {T})\n\\]"
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            "content": "因此 AB 与  \\( (PQ^{T})^{T}(PQ^{T}) \\)  相似，而后者正定（因为列满秩），因此 AB 为正定矩阵。",
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            "content": "7.3.6 正定矩阵分解进阶"
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            "content": "定理 7.24 (正定矩阵平方分解)"
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            "content": "\\(A\\) 是 \\(n\\) 阶正定矩阵，则存在实对称矩阵 \\(C_1\\) ，使得 \\(A = C_1^2\\)",
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            "content": "证明 由于  \\( \\exists P \\)  使得  \\( P^{T}AP = \\Lambda = \\text{diag}\\{\\lambda_{1}, \\cdots, \\lambda_{n}\\} \\) ，根据正定，所有特征值都为正，因此取",
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            "content": "84"
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            "content": "\\[\nC _ {1} = P ^ {T} \\sqrt {\\Lambda} P\n\\]"
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            "content": "即可得到 \\(A = C_1^2\\)",
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            "content": "笔记 上述  \\( A = C_{1}^{2} \\)  也是正定矩阵分解！且是唯一的。",
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            "content": "定理 7.25 (正定矩阵三角分解)"
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            "content": "\\(A\\) 是 \\(n\\) 阶正定矩阵，则存在可逆上三角阵 \\(R\\) ，使得 \\(A = R^{T}R\\)",
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            "content": "证明 首先 \\(A = C^T C\\) ， \\(C\\) 为可逆矩阵，再对 \\(C\\) 做 \\(QR\\) 分解得到 \\(C = QR\\) ，因此 \\(A = R^T Q^T QR = R^TR\\) ，即所求。",
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            "content": "定理 7.26 (极分解定理)"
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            "content": "对任一可逆矩阵 A，一定存在一个正交矩阵 T 和两个正定矩阵  \\( S_{1}, S_{2} \\)  使得",
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            "content": "\\[\nA = T S _ {1} = S _ {2} T\n\\]"
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            "content": "且这种分解是唯一的。",
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            "content": "证明 (1) 存在性: 由于 A 可逆, 因此  \\( A^{T}A \\)  正定, 根据前面结论存在正定矩阵  \\( S_{1} \\) , 满足  \\( A^{T}A = S_{1}^{2} \\) , 设  \\( T = (A^{T})^{-1}S_{1} \\) , 则",
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            "content": "\\[\nA = (A ^ {T}) ^ {- 1} S _ {1} ^ {2} = T S _ {1}\n\\]"
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            "content": "经过验证可发现 \\(TT^{T} = (A^{T})^{-1}S_{1}S_{1}^{T}A^{-1} = (A^{T})^{-1}S_{1}^{2}A^{-1} = (A^{T})^{-1}A^{T}AA^{-1} = E\\) ，因此为正交矩阵。",
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            "content": "定理 7.27 (负定二次型的充要条件)"
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            "content": "\\(A\\) 负定当且仅当 \\(-A\\) 正定，因此充要条件为： \\(A\\) 的所有奇数阶顺序主子式小于0，偶数阶顺序主子式大于0。",
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            "content": "第8章 内积空间"
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            "content": "8.1 Euclid 空间"
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            "content": "8.1.1 Euclid 空间与正交"
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            "content": "定义 8.1 (欧式空间)"
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            "content": "\\(V\\) 是 \\(\\mathbb{R}\\) 上线性空间，定义内积 \\((\\alpha, \\beta)\\) 满足：",
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            "content": "- 线性性：\\((k_{1}\\alpha_{1} + k_{2}\\alpha_{2}, \\beta) = k_{1}(\\alpha_{1}, \\beta) + k_{2}(\\alpha_{2}, \\beta)\\)",
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            "content": "- 对称性： \\((\\alpha ,\\beta) = (\\alpha ,\\beta)\\)",
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            "content": "- 正定性：\\((\\alpha, \\alpha) = 0\\) 当且仅当 \\(\\alpha = 0\\)",
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            "content": "定义 8.2 (度量矩阵)"
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            "content": "给定 Euclid 空间 V 的一组基  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\) ，则定义这组基对应的度量矩阵为",
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            "content": "\\[\nA = \\left[ \\begin{array}{c c c c} (\\epsilon_ {1}, \\epsilon_ {1}) & (\\epsilon_ {1}, \\epsilon_ {2}) & \\dots & (\\epsilon_ {1}, \\epsilon_ {n}) \\\\ (\\epsilon_ {2}, \\epsilon_ {1}) & (\\epsilon_ {2}, \\epsilon_ {2}) & \\dots & (\\epsilon_ {2}, \\epsilon_ {n}) \\\\ \\vdots & \\vdots & & \\vdots \\\\ (\\epsilon_ {n}, \\epsilon_ {1}) & (\\epsilon_ {n}, \\epsilon_ {2}) & \\dots & (\\epsilon_ {n}, \\epsilon_ {n}) \\end{array} \\right]\n\\]"
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            "content": "定理 8.1 (内积的度量矩阵表示)"
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            "content": "若 Euclid 空间中的向量  \\( \\alpha, \\beta \\)  在基  \\( (\\epsilon_{1}, \\cdots, \\epsilon_{n}) \\)  中的坐标分别为 X, Y,  \\( (\\epsilon_{1}, \\cdots, \\epsilon_{n}) \\)  对应的度量矩阵为 A，则",
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            "content": "\\[\n(\\alpha , \\beta) = X ^ {T} A Y\n\\]"
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            "content": "练习 8.1 度量矩阵 已知 Euclid 空间 V 的一组基为:",
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            "content": "\\[\n\\epsilon_ {1} = e _ {1} + e _ {2}, \\epsilon_ {2} = e _ {1} + e _ {3}, \\epsilon_ {3} = e _ {4} - e _ {1}, \\epsilon_ {4} = e _ {1} - e _ {2} - e _ {3} + e _ {4}\n\\]"
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            "content": "向量  \\( \\alpha,\\beta \\)  在  \\( (\\epsilon_{1},\\cdots,\\epsilon_{4}) \\)  下的坐标分别为  \\( (1,2,3,4)^{T},(2,0,1,0)^{T} \\) ，计算  \\( (\\epsilon_{1},\\cdots,\\epsilon_{4}) \\)  的度量矩阵 A，以及  \\( (\\alpha,\\beta) \\)  解 把  \\( \\epsilon_{1},\\cdots,\\epsilon_{4} \\)  看成  \\( R^{n} \\)  中的坐标：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\epsilon_ {1} = (1, 1, 0, 0) \\\\ \\epsilon_ {2} = (1, 0, 1, 0) \\\\ \\epsilon_ {3} = (- 1, 0, 0, 1) \\\\ \\epsilon_ {4} = (1, - 1, - 1, 1) \\end{array} \\right.\n\\]"
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            "content": "因此根据向量的坐标内积公式可以得到度量矩阵为：",
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            "content": "\\[\nA = \\left[ \\begin{array}{r r r r} 2 & 1 & - 1 & 0 \\\\ 1 & 2 & - 1 & 0 \\\\ - 1 & - 1 & 2 & 0 \\\\ 0 & 0 & 0 & 4 \\end{array} \\right], (\\alpha , \\beta) = 5\n\\]"
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            "content": "定理 8.2 (正交向量线性无关)"
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            "content": "若 Eulid 空间 V 中的向量  \\( \\alpha_{1},\\cdots,\\alpha_{s} \\)  彼此正交，则它们线性无关。",
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            "content": "8.1 Euclid 空间"
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            "content": "证明 不妨设  \\( c_{1}\\alpha_{1}+\\cdots+c_{s}\\alpha_{s}=0 \\) ，则两侧与  \\( \\alpha_{i} \\)  做内积得到：",
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            "content": "\\[\n\\left(c _ {1} \\alpha_ {1} + \\dots + c _ {s} \\alpha_ {s}, \\alpha_ {i}\\right) = (0, \\alpha_ {i}) = 0\n\\]"
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            "content": "根据正交定义可知左侧等于 \\(c_{i}(\\alpha_{i},\\alpha_{i})\\) ，而 \\((\\alpha_{i},\\alpha_{i})\\neq 0\\) ，因此 \\(c_{i} = 0\\) ，根据 \\(i\\) 的任意性可知线性无关。",
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            "content": "8.1.2 标准正交基"
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            "content": "一、标准正交基的概念与存在性"
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            "content": "定义 8.3 (标准正交基)"
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            "angle": 0,
            "content": "将 \\(n\\) 维Eulid空间中 \\(n\\) 个两两正交的单位向量称为标准正交基",
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            "content": "二、标准正交基之间的过渡矩阵"
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            "content": "定理 8.3 (标准正交基的过渡矩阵)"
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            "content": "\\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\) 是 n 维 Euclid 空间 V 的一组基，若 \\( \\eta_{1},\\cdots,\\eta_{n} \\) 下述等式，则 \\( \\eta_{1},\\cdots,\\eta_{n} \\) 是标准正交基当且仅当 T 是正交矩阵。",
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            "content": "\\[\n(\\eta_ {1}, \\dots , \\eta_ {n}) = (\\epsilon_ {1}, \\dots , \\epsilon_ {n}) T\n\\]"
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            "content": "三、标准正交基的计算：Schmidt 正交化"
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            "content": "定理 8.4 (Schmidt 正交化)"
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            "content": "给定一组线性无关的向量  \\( \\alpha_{1},\\cdots,\\alpha_{s} \\) ，要求做出一组相互正交的向量 (单位化单独进行) \\( \\epsilon_{1},\\cdots,\\epsilon_{s} \\) ，可以按照以下步骤：",
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            "content": "\\[\n\\epsilon_ {1} = \\alpha_ {1}\n\\]"
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            "content": "\\[\n\\epsilon_ {2} = \\alpha_ {2} - \\frac {\\left(\\alpha_ {2} , \\epsilon_ {1}\\right)}{\\left(\\epsilon_ {1} , \\epsilon_ {1}\\right)} \\epsilon_ {1}\n\\]"
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            "content": "\\[\n\\epsilon_ {s} = \\alpha_ {s} - \\sum_ {i = 1} ^ {s - 1} \\frac {(\\alpha_ {s} , \\epsilon_ {i})}{(\\epsilon_ {i} , \\epsilon_ {i})} \\epsilon_ {i}\n\\]"
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            "content": "练习 8.2Schmidt 正交化 已知  \\( \\epsilon_{1}=(1,2,-1)^{T},\\epsilon_{2}=(-1,3,1)^{T},\\epsilon_{3}=(4,-1,0)^{T} \\) ，用 Schmidt 正交化将其化为标准正交基",
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            "content": "解 最终结果为 \\(\\eta_{1} = \\frac{1}{\\sqrt{6}} (1,2, - 1)^{T},\\eta_{2} = \\frac{1}{\\sqrt{3}} (-1,1,1)^{T},\\eta_{3} = \\frac{1}{\\sqrt{2}} (1,0,1)^{T}\\)",
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            "content": "定理 8.5 (QR 分解)"
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            "content": "证明 假设原本的矩阵 A 的列向量为  \\( \\alpha_{1},\\cdots,\\alpha_{n} \\) ，经过 Schmidt 正交化后得到  \\( \\beta_{1},\\cdots,\\beta_{n} \\) ，则满足：",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\alpha_ {1} = \\beta_ {1} \\\\ \\alpha_ {2} = a _ {1 2} \\beta_ {1} + \\beta_ {2} \\\\ \\quad \\vdots \\\\ \\alpha_ {n} = a _ {1 n} \\beta_ {1} + a _ {2 n} \\beta_ {2} + \\dots + \\beta_ {n} \\end{array} \\right.\n\\]"
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        {
            "type": "header",
            "bbox": [
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            "angle": 0,
            "content": "8.1 Euclid 空间"
        },
        {
            "type": "text",
            "bbox": [
                0.092,
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            "angle": 0,
            "content": "再将 \\(\\beta\\) 进行单位化得到 \\(\\eta_{i} = \\frac{\\beta_{i}}{|\\beta_{i}|}\\)，此时写为",
            "merge_prev": false
        },
        {
            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n(\\alpha_ {1}, \\dots , \\alpha_ {n}) = (\\eta_ {1}, \\dots , \\eta_ {n}) \\left( \\begin{array}{c c c c} b _ {1 1} & b _ {1 2} & \\dots & b _ {1 n} \\\\ & b _ {2 2} & \\dots & b _ {2 n} \\\\ & & \\ddots & \\vdots \\\\ & & & b _ {n n} \\end{array} \\right)\n\\]"
        },
        {
            "type": "text",
            "bbox": [
                0.091,
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            "angle": 0,
            "content": "此时对角  \\( b_{11}=|\\beta_{1}|\\cdots,b_{nn}=|\\beta_{n}| \\) ，而  \\( (\\eta_{1},\\cdots,\\eta_{n}) \\)  是标准正交基。因此是正交矩阵乘上三角阵",
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        },
        {
            "type": "text",
            "bbox": [
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            "content": "练习 8.3QR 分解 将下面的矩阵 A 分解为正交矩阵 T 与对角都是正数的上三角矩阵 B 的乘积",
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            "content": "\\[\nA = \\left[ \\begin{array}{l l l} 1 & 1 & 1 \\\\ 1 & 0 & 1 \\\\ 1 & 2 & 0 \\end{array} \\right]\n\\]"
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            "type": "text",
            "bbox": [
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            "angle": 0,
            "content": "解 先进行  \\( \\alpha_{1}, \\alpha_{2}, \\alpha_{3} \\)  正交化到  \\( \\beta_{1}, \\beta_{2}, \\beta_{3} \\) :",
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        },
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            "type": "page_number",
            "bbox": [
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            "angle": 0,
            "content": "88"
        },
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            "type": "equation",
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            "content": "\\[\n\\left\\{ \\begin{array}{l} \\beta_ {1} = \\alpha_ {1} = (1, 1, 1) ^ {T}, | \\beta_ {1} | = \\sqrt {3} \\\\ \\beta_ {2} = \\alpha_ {2} - \\frac {3}{3} (1, 1, 1) ^ {T} = (0, - 1, 1) ^ {T}, | \\beta_ {2} | = \\sqrt {2} \\\\ \\beta_ {3} = \\alpha_ {3} + \\frac {1}{2} \\beta_ {2} - \\frac {2}{3} \\beta_ {1} = (\\frac {1}{3}, - \\frac {1}{6}, - \\frac {1}{6}), | \\beta_ {3} | = \\sqrt {\\frac {1}{6}} \\end{array} \\right.\n\\]"
        },
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            "angle": 0,
            "content": "进行单位化得到",
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            "type": "equation",
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            "angle": 0,
            "content": "\\[\n\\left\\{ \\begin{array}{l} \\eta_ {1} = (\\frac {\\sqrt {3}}{3}, \\frac {\\sqrt {3}}{3}, \\frac {\\sqrt {3}}{3}) \\\\ \\eta_ {2} = (0, - \\frac {\\sqrt {2}}{2}, \\frac {\\sqrt {2}}{2}) \\\\ \\eta_ {3} = (\\frac {\\sqrt {6}}{3}, - \\frac {\\sqrt {6}}{6}, - \\frac {\\sqrt {6}}{6}) \\end{array} \\right.\n\\]"
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            "angle": 0,
            "content": "反向得到",
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            "type": "equation",
            "bbox": [
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            "angle": 0,
            "content": "\\[\n\\left\\{ \\begin{array}{l} \\alpha_ {1} = \\beta_ {1} \\\\ \\alpha_ {2} = \\beta_ {1} + \\beta_ {2} \\\\ \\alpha_ {3} = \\frac {2}{3} \\beta_ {1} - \\frac {1}{2} \\beta_ {2} + \\beta_ {3} \\end{array} \\right.\n\\]"
        },
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            "angle": 0,
            "content": "因此",
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            "content": "\\[\n(\\alpha_ {1}, \\alpha_ {2}, \\alpha_ {3}) = (\\beta_ {1}, \\beta_ {2}, \\beta_ {3}) \\left( \\begin{array}{c c c} 1 & 1 & \\frac {2}{3} \\\\ 0 & 1 & - \\frac {1}{2} \\\\ 0 & 0 & 1 \\end{array} \\right) = (\\eta_ {1}, \\eta_ {2}, \\eta_ {3}) \\left( \\begin{array}{c c c} | \\beta_ {1} | & & \\\\ & | \\beta_ {2} | & \\\\ & & | \\beta_ {3} | \\end{array} \\right) \\left( \\begin{array}{c c c} 1 & 1 & \\frac {2}{3} \\\\ 0 & 1 & - \\frac {1}{2} \\\\ 0 & 0 & 1 \\end{array} \\right)\n\\]"
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            "content": "8.1.3 正交和与正交补"
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            "content": "定义 8.4 (正交和、正交补)"
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            "angle": 0,
            "content": "\\( V_{1},\\cdots,V_{s} \\)  为一列子空间，若  \\( \\forall i,j \\)  有  \\( V_{i}\\perp V_{j} \\) ，则称  \\( V_{1}+\\cdots+V_{s} \\)  为正交和。若  \\( V=V_{1}+V_{2} \\)  且为正交和，则  \\( V_{1},V_{2} \\)  互为正交补。",
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            "angle": 0,
            "content": "定理 8.6 (正交和必为直和)"
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            "angle": 0,
            "content": "若  \\( V_{1} + \\cdots + V_{s} \\)  为正交和，则其必为直和。",
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            "angle": 0,
            "content": "证明 设  \\( 0 = \\alpha_{1} + \\cdots + \\alpha_{s} \\) ，两侧与  \\( \\alpha_{i} \\)  做内积可知",
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            "content": "\\[\n0 = (\\alpha_ {i}, \\alpha_ {i})\n\\]"
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            "content": "8.1 Euclid 空间"
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            "angle": 0,
            "content": "因此  \\( \\alpha_{i}=0 \\) ，故零元素分解唯一，为直和。",
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            "content": "定理 8.7 (正交补的唯一性)"
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            "angle": 0,
            "content": "Euclid 空间 V 的任一子空间  \\( V_{1} \\)  有唯一的正交补  \\( W^{\\perp} \\)",
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            "bbox": [
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            "content": "证明  \\( V_{1}=\\{0\\} \\)  时，即  \\( V_{1}^{\\perp}=V \\) 。否则设  \\( V_{1}=L(\\epsilon_{1},\\cdots,\\epsilon_{s}) \\)  为标准正交基，将其扩充为 V 的标准正交基：",
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            "content": "\\[\nV = L \\left(\\epsilon_ {1}, \\dots , \\epsilon_ {n}\\right)\n\\]"
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            "angle": 0,
            "content": "则  \\( V_{1}^{\\perp}=L(\\epsilon_{s+1},\\cdots,\\epsilon_{n}) \\)  为正交补",
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            "bbox": [
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            "angle": 0,
            "content": "89"
        },
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            "bbox": [
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            "content": "唯一性: 若  \\( V_{1} \\)  有正交补  \\( V_{2}, V_{3} \\) ，则  \\( \\forall \\alpha \\in V_{2} \\subset V_{1} \\oplus V_{3}, \\exists \\alpha_{1} \\in V_{1}, \\alpha_{3} \\in V_{3} \\)  使得  \\( \\alpha = \\alpha_{1} + \\alpha_{3} \\) ，而  \\( \\alpha \\perp \\alpha_{1}, \\alpha_{1} \\perp \\alpha_{3} \\) ，因此",
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            "bbox": [
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            "content": "\\[\n0 = (\\alpha , \\alpha_ {1}) = (\\alpha_ {1}, \\alpha_ {1}) + (\\alpha_ {3}, \\alpha_ {1}) = (\\alpha_ {1}, \\alpha_ {1})\n\\]"
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            "content": "即 \\(\\alpha_{1} = 0, \\alpha = \\alpha_{3} \\in V_{3}\\)，得到 \\(V_{2} \\subseteq V_{3}, V_{3} \\subseteq V_{2}\\)，因此 \\(V_{2} = V_{3}\\)",
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            "content": "定理 8.8 (正交补的具体表达式)"
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            "content": "\\(V_{1}\\) 是线性空间 \\(V\\) 的子空间，则 \\(V_{1}^{\\perp}\\) 恰由所有与 \\(V_{1}\\) 正交的向量组成，即",
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            "content": "\\[\nV _ {1} ^ {\\perp} = \\{\\alpha \\in V, \\alpha \\perp V _ {1} \\}\n\\]"
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            "content": "证明 令 \\(W = \\{\\alpha \\in V : \\alpha \\perp V_1\\}\\)，下面证明 \\(V_1^\\perp = W\\)。首先 \\(V_1 \\perp V_1^\\perp\\)，因此 \\(V_1^\\perp \\subseteq W\\)。",
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            "content": "反之，\\(\\forall \\alpha \\in W\\)，根据 \\(V = V_{1} \\oplus V_{1}^{\\perp}\\)，因此可分解 \\(\\alpha = \\alpha_{1} + \\alpha_{2}\\)，得到",
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            "content": "\\[\n\\left(\\alpha_ {1}, \\alpha_ {1}\\right) = \\left(\\alpha - \\alpha_ {2}, \\alpha_ {1}\\right) = \\left(\\alpha , \\alpha_ {1}\\right) - \\left(\\alpha_ {2}, \\alpha_ {1}\\right) = 0\n\\]"
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            "angle": 0,
            "content": "因此 \\(\\alpha_{1} = 0\\) ，即 \\(\\alpha \\in V_1^\\perp\\)",
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            "content": "定理 8.9 (正交补的性质)"
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            "content": "\\[\n(1) (U ^ {\\perp}) ^ {\\perp} = U (2) (U _ {1} + U _ {2}) ^ {\\perp} = U _ {1} ^ {\\perp} \\cap U _ {2} ^ {\\perp} (3) (U _ {1} \\cap U _ {2}) ^ {\\perp} = U _ {1} ^ {\\perp} + U _ {2} ^ {\\perp}\n\\]"
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            "content": "证明 (1) 根据正交补的具体表达式一下得出",
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            "content": "(2) 显然 \\((U_{1} + U_{2})^{\\perp}\\subseteq U_{1}^{\\perp}\\) ， \\((U_{1} + U_{2})^{\\perp}\\subseteq U_{2}^{\\perp}\\) ，因此 \\((U_{1} + U_{2})^{\\perp}\\subseteq U_{1}^{\\perp}\\cap U_{2}^{\\perp}\\) 。",
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            "content": "另一方面，\\(\\forall \\alpha \\in U_1^\\perp \\cap U_2^\\perp, \\beta \\in U_1 + U_2\\)，有 \\(\\beta = \\beta_1 + \\beta_2\\)，因此",
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            "content": "\\[\n(\\alpha , \\beta) = (\\alpha , \\beta_ {1} + \\beta_ {2}) = (\\alpha , \\beta_ {1}) + (\\alpha , \\beta_ {2}) = 0\n\\]"
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            "angle": 0,
            "content": "因此 \\(U_{1}^{\\perp}\\cap U_{2}^{\\perp}\\subseteq (U_{1} + U_{2})^{\\perp}\\)",
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            "content": "(3) 可转换为 (2)",
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            "content": "8.1.4 正交变换"
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            "content": "定义 8.5 (正交变换)"
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            "content": "\\(V\\) 是 \\(n\\) 维Eulid空间， \\(\\mathcal{A}\\) 是 \\(V\\) 上线性变换，若 \\(\\forall \\alpha ,\\beta \\in V\\) 有 \\((\\mathcal{A}\\alpha ,\\mathcal{A}\\beta) = (\\alpha ,\\beta)\\) ，则称 \\(\\mathcal{A}\\) 为正交变换",
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            "content": "定理 8.10 (正交变换充要条件)"
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            "content": "\\(\\mathcal{A}\\) 是正交变换与下面几个命题等价：",
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            "content": "- \\(\\mathcal{A}\\) 在标准正交基下矩阵为正交矩阵",
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            "content": "- 标准正交基在 \\(\\mathcal{A}\\) 的作用下仍为标准正交基",
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            "content": "- \\(\\mathcal{A}\\) 保持向量长度不变",
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            "content": "若 \\(\\epsilon_1, \\dots, \\epsilon_n\\) 是 \\(V\\) 的标准正交基，则 \\(\\mathcal{A}(\\epsilon_1), \\dots, \\mathcal{A}(\\epsilon_n)\\) 也是 \\(V\\) 的标准正交基",
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            "content": "若对 \\(\\forall \\alpha \\in V\\) ， \\(\\mathcal{A}:V\\to V\\) 满足",
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            "content": "\\[\n\\mathcal {A} \\alpha = \\alpha - 2 (\\eta , \\alpha) \\eta\n\\]"
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            "content": "其中 \\(\\eta\\) 是单位向量",
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            "angle": 0,
            "content": "镜面反射是正交变换",
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            "content": "证明",
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            "content": "定理 8.11 (镜面反射的特征)",
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            "content": "8.2 Unitary 空间"
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            "content": "8.2.1 Unitary 空间概念"
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            "content": "定义 8.7 (Unitary 空间)"
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            "angle": 0,
            "content": "\\(V\\) 是 \\(\\mathbb{C}\\) 上内积空间，其上定义内积 \\((\\alpha, \\beta)\\) 满足 \\(\\forall k_1, k_2 \\in \\mathbb{C}\\)",
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            "content": "\\[\n\\bullet \\left(k _ {1} \\alpha_ {1} + k _ {2} \\alpha_ {2}, \\beta\\right) = k _ {1} (\\alpha_ {1}, \\beta) + k _ {2} (\\alpha_ {2}, \\beta)\n\\]"
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            "content": "\\[\n\\bullet (\\alpha , \\beta) = \\overline {{(\\beta , \\alpha)}}\n\\]"
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            "content": "\\((\\alpha, \\alpha) \\geq 0\\) 且 \\((\\alpha, \\alpha) = 0\\) 当且仅当 \\(\\alpha = 0\\)",
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            "content": "定义 8.8 (酉矩阵)"
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            "content": "\\(\\mathbb{C}\\) 上矩阵 \\(U^{-1} = \\overline{U}^T\\) ，则称 \\(U\\) 为酉矩阵。",
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            "content": "定义 8.9 (酉变换)"
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            "content": "\\(\\mathcal{U}\\) 满足 \\((\\mathcal{U}\\alpha, \\mathcal{U}\\beta) = (\\alpha, \\beta)\\)，则称 \\(\\mathcal{U}\\) 为酉变换",
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            "content": "定义 8.10 (共轭变换)"
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            "content": "第9章 对偶空间"
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            "content": "9.1 对偶空间与对偶基"
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            "content": "\\(V\\) 是 \\(K\\) 上的线性空间，将定义于 \\(V\\) 上的所有有界线性泛函组成的空间称为对偶空间 \\(V^{*}\\) 。",
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            "content": "给定 V 的一组基  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\) ，定义线性泛函  \\( f_{i} \\)  满足：",
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            "content": "\\[\nf _ {i} (\\epsilon_ {j}) = \\left\\{ \\begin{array}{l l} 1, & i = j \\\\ 0, & i \\neq j \\end{array} \\right.\n\\]"
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            "content": "则  \\( \\{f_{1},\\cdots,f_{n}\\} \\)  是  \\( V^{*} \\)  的一组基，称其为  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  的对偶基。",
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            "content": "笔记 计算对偶基时，将  \\( f_{i} \\)  视为一个向量，即  \\( f_{i}(\\alpha)=f_{i1}x_{1}+\\cdots+f_{in}x_{n}=f_{i}^{T}\\alpha \\) 。计算  \\( \\epsilon_{1},\\cdots,\\epsilon_{n} \\)  的对偶基只需要满足：",
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            "content": "\\[\n\\left[ \\begin{array}{c} f _ {1} ^ {T} \\\\ f _ {2} ^ {T} \\\\ \\vdots \\\\ f _ {n} ^ {T} \\end{array} \\right] \\left[ \\begin{array}{c c c c} \\epsilon_ {1} & \\epsilon_ {2} & \\dots & \\epsilon_ {n} \\end{array} \\right] = E\n\\]"
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            "content": "练习9.1给定 \\(\\epsilon_{1} = [1,1,0]^{T},\\epsilon_{2} = [0,1,1]^{T},\\epsilon_{3} = [0,1, - 1]^{T}\\) ，求其对偶基",
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            "content": "解显然 \\(f_{1}, f_{2}, f_{3}\\) 都是三个系数组成的线性函数，不妨记 \\(A = [f_{1}, f_{2}, f_{3}]^{T}, P = [\\epsilon_{1}, \\epsilon_{2}, \\epsilon_{3}]\\)，则",
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            "content": "\\[\nA P = \\left[ \\begin{array}{l} f _ {1} ^ {T} \\\\ f _ {2} ^ {T} \\\\ f _ {3} ^ {T} \\end{array} \\right] [ \\epsilon_ {1}, \\epsilon_ {2}, \\epsilon_ {3} ] = E\n\\]"
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            "content": "因此 \\(A = P^{-1}\\) ，只需求出 \\(P^{-1}\\) 即可。最终求出：",
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            "content": "\\[\nP ^ {- 1} = \\left[ \\begin{array}{c c c} 1 & 0 & 0 \\\\ - \\frac {1}{2} & \\frac {1}{2} & \\frac {1}{2} \\\\ - \\frac {1}{2} & \\frac {1}{2} & - \\frac {1}{2} \\end{array} \\right] \\Rightarrow \\left\\{ \\begin{array}{l} f _ {1} ^ {T} = (1, 0, 0) \\\\ f _ {2} ^ {T} = (- \\frac {1}{2}, \\frac {1}{2}, \\frac {1}{2}) \\\\ f _ {3} ^ {T} = (- \\frac {1}{2}, \\frac {1}{2}, - \\frac {1}{2}) \\end{array} \\right.\n\\]"
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