## 张祖锦2023年数学专业真题分类70天之第53天
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1197、 (4)、 欧氏空间上正交变换在任意一组基下的矩阵为正交矩阵. $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ (上海大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \times$. 要在标注正交基下才是哦.跟锦数学微信公众号. [在线资料](
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1198、 (5)、 设 $\displaystyle A$ 是 $\displaystyle 5$ 阶矩阵, $\displaystyle \lambda I-A$ 的不变因子为
\begin\{aligned\} 1,1,1,\lambda-3,(\lambda-3)^2(\lambda-2)^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}3&1&&&\\\\ &3&&&\\\\ &&2&1&\\\\ &&&2&\\\\ &&&&3\end\{array\}\right)$. $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ (上海大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \surd$. 由题设知 $\displaystyle A$ 的初等因子为 $\displaystyle \lambda-3,(\lambda-3)^2, (\lambda-2)^2$, 而结论成立.跟锦数学微信公众号. [在线资料](
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1199、 (3)、 (15 分) 设 $\displaystyle B$ 和 $\displaystyle C$ 分别为 $\displaystyle 3\times 2$ 和 $\displaystyle 2\times 3$ 矩阵, 且
\begin\{aligned\} BC=\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&0\\\\ 1&1&-1\\\\ 1&-1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
记 $\displaystyle A=CB$, 求 $\displaystyle A^n$. (上海大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F=BC$, 则 $\displaystyle \mathrm\{rank\} B=2\Rightarrow \mathrm\{rank\} B=\mathrm\{rank\} C=2\Rightarrow Bx=0, C^\mathrm\{T\} y=0$ 只有零解, 且
\begin\{aligned\} &F^2=2F\Rightarrow (BC)(BC)=2BC\Rightarrow B(CB-2E)C=0\\\\ \Rightarrow& (CB-2E)C=0 \Rightarrow C^\mathrm\{T\}(CB-2E)^\mathrm\{T\}=0\Rightarrow (CB-2E)^\mathrm\{T\}=0\\\\ \Rightarrow& CB=2E\Rightarrow A=2E\_2\Rightarrow A^n=2^nE\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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1200、 (4)、 (15 分) 设 $\displaystyle \lambda\_1,\lambda\_2$ 为非零实数, $\displaystyle n$ 阶实矩阵
\begin\{aligned\} A=(\lambda\_1\alpha, \lambda\_2\beta)\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1\alpha^\mathrm\{T\}\\\\ \lambda\_2\beta^\mathrm\{T\}\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \alpha,\beta$ 为 $\displaystyle \mathbb\{R\}^n$ 中已知的单位列向量, 且正交. 求 $\displaystyle A+I$ 的特征值 (其中 $\displaystyle I$ 为单位矩阵), 并求出 $\displaystyle A$ 的非零特征值 (若存在) 对应的特征向量. (上海大学2023年高等代数考研试题) [矩阵 ]
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\begin\{aligned\} A\alpha=(\lambda\_1\alpha, \lambda\_2\beta)\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1\\\\0\end\{array\}\right)=\lambda\_1^2\alpha, A\beta=(\lambda\_1\alpha, \lambda\_2\beta)\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\\lambda\_2\end\{array\}\right)=\lambda\_2^2\beta \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle A$ 的形式知 $\displaystyle \mathrm\{rank\} A=2$, 而 $\displaystyle Ax=0$ 有 $\displaystyle n-2$ 个线性无关的特征向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-2\}$, 则
\begin\{aligned\} P=(\alpha,\beta,\eta\_1,\cdots,\eta\_\{n-2\})\Rightarrow AP=P\mathrm\{diag\}(\lambda\_1^2,\lambda\_2^2,0,\cdots,0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 的特征值为 $\displaystyle 0$ ($n-2$ 重), $\displaystyle \lambda\_1^2,\lambda\_2^2$. 从而 $\displaystyle A+I$ 的特征值为 $\displaystyle 1$ ($n-2$ 重), $\displaystyle 1+\lambda\_1^2, 1+\lambda\_2^2$, 且 $\displaystyle A$ 的非零特征值 (若存在) 对应的特征向量为 $\displaystyle \alpha,\beta$.跟锦数学微信公众号. [在线资料](
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1201、 2、 (20 分) 设 $\displaystyle S=\left\\{\alpha\in\mathbb\{R\}^2; \alpha^\mathrm\{T\} \alpha\leq 1\right\\}$, $\displaystyle A$ 为 $\displaystyle 2$ 阶实矩阵, 记集合 $\displaystyle \left\\{A\alpha; \alpha\in S\right\\}$ 为 $\displaystyle AS$. (1)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 1&1\end\{array\}\right)$, 请精确描述 $\displaystyle AS$ 的几何形状; (2)、 设 $\displaystyle a,b$ 为任意实数, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&b\\\\ -b&a\end\{array\}\right)$, 请精确描述 $\displaystyle AS$ 的几何形状; (3)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&b\\\\c&d\end\{array\}\right)$ 是任意实矩阵, 请描述集合 $\displaystyle AS$ 的大致形状并求其面积. (上海交通大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)\Rightarrow A\alpha=(x+y)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\end\{array\}\right)$. 由
\begin\{aligned\} x^2+y^2=1\Rightarrow& (x+y)^2=(1\cdot x+1\cdot y)^2 \stackrel\{\tiny\mbox\{Cauchy\}\}\{\leq\} (1^2+1^2)(x^2+y^2)=2\\\\ \Rightarrow& |x+y|\leq \sqrt\{2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle AS$ 是连接 $\displaystyle [-\sqrt\{2\},-\sqrt\{2\}]$ 到 $\displaystyle [\sqrt\{2\},\sqrt\{2\}]$ 的直线段. (2)、 $\displaystyle \alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)\Rightarrow A\alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}ax+by\\\\ -bx+ay\end\{array\}\right)\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\end\{array\}\right)$. 由
\begin\{aligned\} u^2+v^2=(a^2+b^2)(x^2+y^2)=a^2+b^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle AS$ 是以原点为圆心, $\displaystyle \sqrt\{a^2+b^2\}$ 为半径的圆 (当 $\displaystyle a=b=0$ 时, $\displaystyle AS$ 退化为原点). (3)、 (3-1)、 若 $\displaystyle \mathrm\{rank\} A=0$, 则 $\displaystyle A=0$, $\displaystyle AS$ 就是原点. (3-2)、 若 $\displaystyle \mathrm\{rank\} A=1$, 则不妨设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&b\\\\ ka&kb\end\{array\}\right)$, 而
\begin\{aligned\} \alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\y\end\{array\}\right)\Rightarrow A\alpha=(ax+by)\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\k\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle AS$ 是一条线段. (3-3)、 若 $\displaystyle \mathrm\{rank\} A=2$, 则 $\displaystyle A$ 可逆. 对 $\displaystyle \alpha\in S$, $\displaystyle \alpha^\mathrm\{T\} \alpha\leq 1$. 设 $\displaystyle \beta=A\alpha$, 则
\begin\{aligned\} \alpha=A^\{-1\}\beta\stackrel\{B=A^\{-\mathrm\{T\}\}A^\{-1\}\}\{\Rightarrow\} 1\geq\alpha^\mathrm\{T\} \alpha=\beta^\mathrm\{T\} A^\{-\mathrm\{T\}\}A^\{-1\}\beta=\beta^\mathrm\{T\} B\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle B$ 正定知存在正交阵 $\displaystyle P$ 使得 $\displaystyle P^\mathrm\{T\} BP=\mathrm\{diag\}(\lambda\_1,\lambda\_2), \lambda\_i > 0$. 再设 $\displaystyle \beta=P\gamma, \gamma=\left(\begin\{array\}\{cccccccccccccccccccc\}z\_1\\\\z\_2\end\{array\}\right)$, 则
\begin\{aligned\} 1\geq \lambda\_1z\_1^2+\lambda\_2z\_2^2\Rightarrow AS\mbox\{是一个椭圆\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
面积 $\displaystyle =\frac\{\pi\}\{\sqrt\{\lambda\_1\lambda\_2\}\}=\frac\{\pi\}\{\sqrt\{|B|\}\}=\pi |\det A|=\pi|ad-bc|$.跟锦数学微信公众号. [在线资料](
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1202、 3、 (20 分) 设 $\displaystyle A\in\mathbb\{R\}^\{n\times n\}$, $\displaystyle v\_1,\cdots,v\_n\in\mathbb\{R\}^n$, 证明: $\displaystyle Av\_1,\cdots,Av\_n$ 线性无关的充要条件是 $\displaystyle A$ 可逆且 $\displaystyle v\_1,\cdots,v\_n$ 线性无关. (上海交通大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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\begin\{aligned\} &\mbox\{$Av\_1,\cdots,Av\_n$ 线性无关\} \Leftrightarrow\mathrm\{rank\}(Av\_1,\cdots,Av\_n)=n\\\\ \Leftrightarrow&0\neq \det(Av\_1,\cdots, Av\_n) =\det\left\[A(v\_1,\cdots,v\_n)\right\] =\det A\cdot \det (v\_1,\cdots,v\_n)\\\\ \Leftrightarrow&0\neq \det A, 0\neq \det (v\_1,\cdots,v\_n)\\\\ \Leftrightarrow&0\neq \det A, n=\mathrm\{rank\}(v\_1,\cdots,v\_n) \Leftrightarrow\mbox\{$A$ 可逆且 $\displaystyle v\_1,\cdots,v\_n$ 线性无关\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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1203、 4、 (20 分) 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&x\_1&x\_2\\\\ 0&b&x\_3\\\\ 0&0&a\end\{array\}\right)\in\mathbb\{R\}^\{3\times 3\}$, 试确定 $\displaystyle A$ 的 Jordan (约当) 标准形. (上海交通大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle J$. (1)、 若 $\displaystyle b=a$, 则 $\displaystyle A$ 的特征值为 $\displaystyle a,a,a$. 于是 [通过秩可确定 Jordan 标准形严格上三角部分的 $\displaystyle 1$ 的个数!]
\begin\{aligned\} \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1&x\_2\\\\ 0&x\_3\end\{array\}\right)=0\Leftrightarrow x\_1=x\_2=x\_3=0\Rightarrow& J=aE\_3,\\\\ \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1&x\_2\\\\ 0&x\_3\end\{array\}\right)=1\Leftrightarrow \boxed\{\begin\{array\}\{c\}x\_3=0, x\_1,x\_2\mbox\{不全为\}0\\\\\mbox\{或\} x\_3\neq 0, x\_1=0\end\{array\}\} \Rightarrow& J=\left(\begin\{array\}\{cccccccccccccccccccc\}a&1&\\\\ &a&\\\\ &&a\end\{array\}\right),\\\\ \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}x\_1&x\_2\\\\ 0&x\_3\end\{array\}\right)=2\Leftrightarrow x\_1x\_3\neq 0\Rightarrow& J=\left(\begin\{array\}\{cccccccccccccccccccc\}a&1&\\\\ &a&1\\\\ &&a\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle b\neq a$, 则
\begin\{aligned\} \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}0&x\_1&x\_2\\\\ 0&b-a&x\_3\\\\ 0&0&0\end\{array\}\right)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&x\_2\\\\ 0&1&0\\\\ 0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} x\_2=0\Rightarrow& \mathrm\{rank\}(A-aE)=1\Rightarrow J=\mathrm\{diag\}(b,a,a),\\\\ x\_2\neq 0\Rightarrow& \mathrm\{rank\}(A-aE)=2\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}b&&\\\\ &a&1\\\\ &&a\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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1204、 5、 (20 分) 设 $\displaystyle A,B\in\mathbb\{R\}^\{n\times n\}$ 均为正定矩阵. (1)、 $\displaystyle n=2$ 时, 求 $\displaystyle A,B$ 使得 $\displaystyle AB$ 不是正定矩阵; (2)、 求一个充分必要条件, 使得 $\displaystyle AB$ 是正定矩阵, 并证明. (上海交通大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 取 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2\\\\2&5\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1\\\\-1&2\end\{array\}\right)$, 则 $\displaystyle AB=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&3\\\\ -3&8\end\{array\}\right)$ 不对称, 而不是正定矩阵. (2)、 一言以蔽之, 正定矩阵的乘积正定的充要条件是这两个矩阵可交换. (1)、 $\displaystyle \Rightarrow$:
\begin\{aligned\} AB=(AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\}=BA . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle A$ 正定知存在正交阵 $\displaystyle Q$ 使得
\begin\{aligned\} Q^\mathrm\{T\} AQ=\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \varLambda\_i$ 是以 $\displaystyle \lambda\_i > 0$ 为对角元的对角阵, 且各 $\displaystyle \lambda\_i$ 互异. 于是
\begin\{aligned\} &\quad \ AB=BA\Leftrightarrow Q^\mathrm\{T\} AQ\cdot Q^\mathrm\{T\} BQ=Q^\mathrm\{T\} BQ\cdot Q^\mathrm\{T\} AQ\\\\ &\Leftrightarrow \mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s)\tilde\{B\}=\tilde\{B\}\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s)\left(\tilde\{B\}=Q^\mathrm\{T\} BQ\right)\\\\ &\Leftrightarrow \varLambda\_i\tilde\{B\}\_\{ij\} =\tilde\{B\}\_\{ij\}\varLambda\_j\left(\tilde\{B\}=\left(\begin\{array\}\{cccccccccccccccccccc\}\tilde\{B\}\_\{11\}&\cdots&\tilde\{B\}\_\{1s\}\\\\ \vdots&\ddots&\vdots\\\\ \tilde\{B\}\_\{s1\}&\cdots&\tilde\{B\}\_\{ss\}\end\{array\}\right)\right)\\\\ &\Leftrightarrow \tilde\{B\}\_\{ij\}=0\left(i\neq j\right)\\\\ &\Leftrightarrow \tilde\{B\}=\mathrm\{diag\}(\tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle B$ 正定知 $\displaystyle \tilde\{B\}=Q^\mathrm\{T\} BQ$ 正定, $\displaystyle \tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}$ 也正定. 于是存在正交阵 $\displaystyle R\_i$, 使得
\begin\{aligned\} R\_i^\mathrm\{T\} \tilde\{B\}\_\{ii\}R\_i=D\_i, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle D\_i$ 为对角阵. 取
\begin\{aligned\} P=Q\mathrm\{diag\}(R\_1,\cdots,R\_s), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle P$ 正交, 且
\begin\{aligned\} P^\mathrm\{T\} AP&=\mathrm\{diag\}(R\_1^\mathrm\{T\},\cdots,R\_s^\mathrm\{T\})\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s) \mathrm\{diag\}(R\_1,\cdots,R\_s)\\\\ &=\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s),\\\\ P^\mathrm\{T\} BP&=\mathrm\{diag\}(R\_1^\mathrm\{T\},\cdots,R\_s^\mathrm\{T\})\mathrm\{diag\}(\tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}) \mathrm\{diag\}(R\_1,\cdots,R\_s)\\\\ &=\mathrm\{diag\}(D\_1,\cdots,D\_s). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} P^\mathrm\{T\} ABP=\mathrm\{diag\}(\lambda\_1\mu\_1,\cdots,\lambda\_n\mu\_n), \lambda\_i\mu\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle AB$ 正定.跟锦数学微信公众号. [在线资料](
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1205、 7、 (20 分) 设 $\displaystyle A\in \mathbb\{R\}^\{m\times n\}, \mathrm\{rank\} A=r > 0$. (1)、 证明存在唯一的矩阵 $\displaystyle P\in\mathbb\{R\}^\{m\times m\}$, 使得
\begin\{aligned\} P^\mathrm\{T\} =P^2=P, \mathrm\{rank\} P=\mathrm\{rank\} A, PA=A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 求 $\displaystyle P$ 的特征多项式与特征子空间. (3)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2\\\\ -1&-2\\\\ 0&0\end\{array\}\right)$, 求 $\displaystyle P$ 与矩阵 $\displaystyle B=(P,P)$ 的奇异值分解. (上海交通大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle A$ 的奇异值分解为 $\displaystyle A=U\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)V$, 其中 $\displaystyle U,V$ 为正交阵, $\displaystyle \varLambda$ 为对角元全大于 $\displaystyle 0$ 的 $\displaystyle r$ 阶方阵, 则取 $\displaystyle P=U\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)U^\mathrm\{T\}$, 则它满足题中所有条件. 设 $\displaystyle Q$ 也满足题中条件, 则由 $\displaystyle Q^\mathrm\{T\}=Q$ 知 $\displaystyle Q$ 实对称. 又由 $\displaystyle Q^2=Q, \mathrm\{rank\} Q=r$ 知 $\displaystyle Q$ 有 $\displaystyle r$ 个特征值为 $\displaystyle 1$, $\displaystyle n-r$ 个特征值为 $\displaystyle 0$. 于是存在正交阵 $\displaystyle R$ 使得 $\displaystyle Q=R\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)R^\mathrm\{T\}$. 设 $\displaystyle S=R^\mathrm\{T\} U=\left(\begin\{array\}\{cccccccccccccccccccc\}S\_1&S\_2\\\\ S\_3&S\_4\end\{array\}\right)$, 则
\begin\{aligned\} &QA=A\Leftrightarrow R\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)R^\mathrm\{T\} U\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)V= U\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)V\\\\ \Leftrightarrow&S^\mathrm\{T\}\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)S \left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)\\\\ \Leftrightarrow& \left(\begin\{array\}\{cccccccccccccccccccc\}S\_1^\mathrm\{T\} S\_1\varLambda&0\\\\ S\_2^\mathrm\{T\} S\_1\varLambda&0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&0\\\\ 0&0\end\{array\}\right) \Leftrightarrow S\_1^\mathrm\{T\} S\_1=E\_r, S\_2^\mathrm\{T\} S\_1=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &Q=P\Leftrightarrow R\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)R^\mathrm\{T\}=U\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)U^\mathrm\{T\}\\\\ \Leftrightarrow& U^\mathrm\{T\} R\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)R^\mathrm\{T\} U=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)\\\\ \Leftrightarrow& S^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)S=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)\Leftarrow (I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知唯一性得证. (2)、 $\displaystyle P$ 的特征多项式为 $\displaystyle (\lambda-1)^r\lambda^\{m-r\}$. 设 $\displaystyle U=(\eta\_1,\cdots,\eta\_m)$, 则 $\displaystyle P$ 的属于特征值 $\displaystyle 1,0$ 的特征子空间分别为
\begin\{aligned\} L(\eta\_1,\cdots,\eta\_r),\quad L(\eta\_\{r+1\},\cdots,\eta\_m). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 $\displaystyle B=AA^\mathrm\{T\}=\left(\begin\{array\}\{cccccccccccccccccccc\}5&-5&0\\\\ -5&5&0\\\\ 0&0&0\end\{array\}\right)$ 的特征值为 $\displaystyle 10,0,0$. 由
\begin\{aligned\} 10E\_3-B\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right), 0E\_3-B\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle B$ 的属于特征值 $\displaystyle 10,0$ 的单位特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{\sqrt\{2\}\}\\\\\frac\{1\}\{\sqrt\{2\}\}\\\\0\end\{array\}\right);\quad \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{2\}\}\\\\\frac\{1\}\{\sqrt\{2\}\}\\\\0\end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} A=U\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)V\Rightarrow AA^\mathrm\{T\} =U\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda^2&\\\\ &0\end\{array\}\right)U^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle U=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ 0&0&1\end\{array\}\right)$. 从而
\begin\{aligned\} P=U\mathrm\{diag\}(1,0,0)U^\mathrm\{T\}=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2\}&-\frac\{1\}\{2\}&0\\\\ -\frac\{1\}\{2\}&\frac\{1\}\{2\}&0\\\\ 0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
写出 $\displaystyle B$ 后, 设 $\displaystyle B=R\left(\begin\{array\}\{cccccccccccccccccccc\}\varLambda&\\\\ &0\end\{array\}\right)S$, 其中 $\displaystyle R,S$ 为正交阵, $\displaystyle \varLambda$ 为 $\displaystyle \mathrm\{rank\} B=1$ 阶方阵, 则
\begin\{aligned\} BB^\mathrm\{T\}=R\mathrm\{diag\}(\varLambda^2,0)R^\mathrm\{T\}, B^\mathrm\{T\} B=S^\mathrm\{T\} \mathrm\{diag\}(\varLambda^2,0)S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由实对称矩阵的正交标准形容易算出 $\displaystyle \varLambda=\sqrt\{2\}$,
\begin\{aligned\} R=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{2\}\}&0\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ 0&0&1\end\{array\}\right), S=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2\}&\frac\{1\}\{2\}&0&-\frac\{1\}\{2\}&\frac\{1\}\{2\}&0\\\\ \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0&0&0&0\\\\ 0&0&1&0&0&0\\\\ -\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{6\}\}&0&\frac\{2\}\{\sqrt\{6\}\}&0&0\\\\ \frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\sqrt\{3\}\}&0&\frac\{1\}\{2\sqrt\{3\}\}&\frac\{\sqrt\{3\}\}\{2\}&0\\\\ 0&0&0&0&0&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle D=\left(\begin\{array\}\{cccccccccccccccccccc\}\sqrt\{2\}&0&0&0&0&0\\\\ 0&0&0&0&0&0\\\\ 0&0&0&0&0&0\end\{array\}\right)$, 则 $\displaystyle B=RDS$ 就是 $\displaystyle B$ 的奇异值分解.跟锦数学微信公众号. [在线资料](
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---
1206、 6、 设正实数 $\displaystyle p,q$ 满足 $\displaystyle p+q=1$, 考虑 $\displaystyle n$ 阶方阵 ($n\geq 2$)
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&p&0&\cdots&0&0\\\\ q&0&p&\cdots&0&0\\\\ 0&q&0&\cdots&0&0\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&0&0&\cdots&0&p\\\\ 0&0&0&\cdots&q&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: (1)、 若 $\displaystyle \alpha=(a\_1,\cdots,a\_n)^\mathrm\{T\}$ 为 $\displaystyle A$ 的特征向量, 则 $\displaystyle |a\_1|, \cdots, |a\_n|$ 不全相等. (2)、 若 $\displaystyle t$ 是 $\displaystyle A$ 的特征值, 则 $\displaystyle |t| < 1$. (首都师范大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle \alpha$ 对应的特征值为 $\displaystyle t$, 则
\begin\{aligned\} &A\alpha=t \alpha\Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}0&p&0&\cdots&0&0\\\\ q&0&p&\cdots&0&0\\\\ 0&q&0&\cdots&0&0\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&0&0&\cdots&0&p\\\\ 0&0&0&\cdots&q&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}a\_1\\\\a\_2\\\\a\_3\\\\\vdots\\\\a\_n\end\{array\}\right)=t\left(\begin\{array\}\{cccccccccccccccccccc\}a\_1\\\\a\_2\\\\a\_3\\\\\vdots\\\\a\_n\end\{array\}\right)\\\\ \Leftrightarrow&pa\_2=t a\_1, qa\_1+pa\_3=t a\_2, \cdots,\\\\ &qa\_\{n-2\}+pa\_n=t a\_\{n-1\},qa\_\{n-1\}=t a\_n.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
用反证法. 若 $\displaystyle |a\_1|, \cdots, |a\_n|$ 全都相等, 设为 $\displaystyle a$, 则 $\displaystyle \alpha\neq 0\Rightarrow |a\_i|=a > 0$. 由 $\displaystyle (I)$ 的第 1 式和最后一个式子知
\begin\{aligned\} &p=|t|=q\stackrel\{p+q=1\}\{\Rightarrow\}p=q=\frac\{1\}\{2\}\\\\ \Rightarrow& \frac\{a\_i+a\_\{i+2\}\}\{2\}=t a\_i, 1\leq i\leq n-2, |t|=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-1)、 若存在某个 $\displaystyle i$, 使得 $\displaystyle a\_i,a\_\{i+2\}$ 异号, 则
\begin\{aligned\} &|a\_i|=|a\_\{i+2\}|=a\Rightarrow a\_i=-a\_\{i+2\}\\\\ \Rightarrow& 0=\frac\{a\_i+a\_\{i+2\}\}\{2\}=t a\_i\stackrel\{|t|=\frac\{1\}\{2\}\}\{\Rightarrow\}a\_i=0,\mbox\{矛盾\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 若 $\displaystyle \forall\ i, a\_i,a\_\{i+2\}$ 同号, 则
\begin\{aligned\} a\_1=a\_3=a\_5=\cdots\equiv b, a\_2=a\_4=a\_6=\cdots\equiv c. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} &b=\frac\{a\_1+a\_3\}\{2\}=t a\_2=t c\\\\ \Rightarrow& a=|a\_1|=|b|=|t| \cdot |c|=\frac\{1\}\{2\}|a\_2|=\frac\{1\}\{2\}a \Rightarrow a=0,\mbox\{矛盾\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不论何种情形都得到矛盾. 故有结论. (2)、 设 $\displaystyle \alpha=(a\_1,\cdots,a\_n)^\mathrm\{T\}$ 为对应的特征向量, 则由第 1 步知
\begin\{aligned\} \exists\ 1\leq k\leq n-1,\mathrm\{ s.t.\} |a\_k|\neq |a\_\{k+1\}|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不妨设 $\displaystyle |a\_k| < |a\_\{k+1\}|$ ($ > $ 时类似讨论, 不过是从 $\displaystyle k$ 往小走, 至多到 $\displaystyle 1$). (2-1)、 若 $\displaystyle |a\_\{k+1\}|\geq |a\_\{k+2\}|$, 则
\begin\{aligned\} ta\_\{k+1\}=qa\_k+pa\_\{k+2\} \Rightarrow |t|\leq q\left|\frac\{a\_k\}\{a\_\{k+1\}\}\right|+p\left|\frac\{a\_\{k+2\}\}\{a\_\{k+1\}\}\right| \leq q+p=1.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 若 $\displaystyle |a\_\{k+1\}| < |a\_\{k+2\}|$, 则看 $\displaystyle |a\_\{k+2\}|, |a\_\{k+3\}|$ 的大小. 若 $\displaystyle |a\_\{k+2\}|\geq |a\_\{k+3\}|$, 则类似 $\displaystyle (I)$ 知 $\displaystyle |t|\leq 1$. 若 $\displaystyle |a\_\{k+2\}| < |a\_\{k+3\}|$, 则又看 $\displaystyle |a\_\{k+3\}|, |a\_\{k+4\}|$ 的大小. 如此这般, 要么有限次后类似 $\displaystyle (I)$ 得到 $\displaystyle |t|\leq 1$, 要么
\begin\{aligned\} &|a\_k| < |a\_\{k+1\}| < \cdots < |a\_\{n-1\}| < |a\_n|\\\\ \Rightarrow& q|a\_\{n-1\}|=|t|\cdot|a\_n| > |t|\cdot |a\_\{n-1\}|\Rightarrow 1 > q > |t|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1207、 8、 设 $\displaystyle A,B$ 为实数域上的 $\displaystyle n$ 阶方阵, 满足 $\displaystyle AB=-A-B$. 证明: $\displaystyle AB=BA$. (首都师范大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &E=E+A+B+AB=(E+A)(E+B)\\\\ \Rightarrow&E=(E+B)(E+A)\Rightarrow BA=-A-B\xlongequal\{\tiny\mbox\{题设\}\} AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1208、 9、 求下面矩阵的若尔当标准形:
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&0&0\\\\ 0&0&1&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 1&-5&10&-10&5\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(首都师范大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 按第一列展开知
\begin\{aligned\} |\lambda E-A|=\lambda^5-5\lambda^4+10\lambda^3-10\lambda^2+5\lambda-1=(\lambda-1)^5. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lambda E-A$ 的右上角有一个 $\displaystyle 4$ 阶子式
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}-1&&&\\\\ \lambda&-1&&\\\\ &\lambda&-1&\\\\ &&\lambda&-1\end\{array\}\right|=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的行列式因子为 $\displaystyle 1,1,1,1,(\lambda-1)^5$, 不变因子也是 $\displaystyle 1,1,1,1,(\lambda-1)^5$, 初等因子为 $\displaystyle (\lambda-1)^5$, Jordan 标准形为
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&&\\\\ &1&1&&\\\\ &&1&1&\\\\ &&&1&1\\\\ &&&&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1209、 3、 $\displaystyle \mathrm\{rank\} M$ 表示矩阵 $\displaystyle M$ 的秩. (1)、 设 $\displaystyle M$ 是数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 阶方阵, $\displaystyle n > 1$, 求 $\displaystyle \mathrm\{rank\}\left\[(M^\star)^\star\right\]$, 其中 $\displaystyle M^\star$ 表示 $\displaystyle M$ 的伴随矩阵. (2)、 证明:
\begin\{aligned\} \mathrm\{rank\}(I\_n-A^\mathrm\{T\} A)-\mathrm\{rank\}(I\_m-AA^\mathrm\{T\})=n-m, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle A$ 是 $\displaystyle m\times n$ 型矩阵, $\displaystyle A^\mathrm\{T\}$ 是 $\displaystyle A$ 的转置, $\displaystyle I\_n,I\_m$ 是单位阵. (3)、 设 $\displaystyle h(x)=h\_1(x)h\_2(x)$ 是数域 $\displaystyle \mathbb\{F\}$ 上的多项式, $\displaystyle h\_1(x)$ 与 $\displaystyle h\_2(x)$ 互素, 证明: 对 $\displaystyle n$ 阶方阵 $\displaystyle B$, 有
\begin\{aligned\} \mathrm\{rank\} h\_1(B)+\mathrm\{rank\} h\_2(B)=n+\mathrm\{rank\} h(B). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(四川大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} \mathrm\{rank\} M\leq n-2\Rightarrow& \mathrm\{rank\}(M^\star)=0\Rightarrow (M^\star)^\star=0\Rightarrow \mathrm\{rank\}\left\[(M^\star)^\star\right\]=0,\\\\ \mathrm\{rank\} M=n-1\Rightarrow&\mathrm\{rank\}(M^\star)=1\Rightarrow \left\\{\begin\{array\}\{llllllllllll\}\mathrm\{rank\}\left\[(M^\star)^\star\right\]=1,&n=2,\\\\ \mathrm\{rank\}\left\[(M^\star)^\star\right\]=0,&n\geq 3,\end\{array\}\right.\\\\ \mathrm\{rank\} M=n\Rightarrow&\mathrm\{rank\}(M^\star)=n\Rightarrow \mathrm\{rank\}\left\[(M^\star)^\star\right\]=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由
\begin\{aligned\} \left(\begin\{array\}\{cc\} E\_m&-A\\\\ 0&E\_n \end\{array\}\right) \left(\begin\{array\}\{cc\} E\_m&A\\\\ A^\mathrm\{T\} &E\_n \end\{array\}\right) \left(\begin\{array\}\{cc\} E\_m&0\\\\ -A^\mathrm\{T\} &E\_n \end\{array\}\right)=&\left(\begin\{array\}\{cc\} E\_m-AA^\mathrm\{T\} &0\\\\ 0&E\_n \end\{array\}\right),\\\\ \left(\begin\{array\}\{cc\} E\_m&0\\\\ -A^\mathrm\{T\} &E\_n \end\{array\}\right) \left(\begin\{array\}\{cc\} E\_m&A\\\\ A^\mathrm\{T\} &E\_n \end\{array\}\right) \left(\begin\{array\}\{cc\} E\_m&-A\\\\ 0&E\_n \end\{array\}\right)=&\left(\begin\{array\}\{cc\} E\_m&0\\\\ 0&E\_n-A^\mathrm\{T\} A \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathrm\{rank\}(E\_m-AA^\mathrm\{T\})+n=\mathrm\{rank\}\left(\begin\{array\}\{cc\} E\_m&A\\\\ A^\mathrm\{T\} &E\_n \end\{array\}\right)=m+\mathrm\{rank\}(E\_n-A^\mathrm\{T\} A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 由 $\displaystyle (h\_1,h\_2)=1\Rightarrow \exists\ u\_i,\mathrm\{ s.t.\} u\_1h\_1+u\_2h\_2=1$ 知
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}h(B)&\\\\ &E\_n\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}h(B)&h\_1(B)\\\\ 0&E\_n\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&h\_1(B)\\\\ -h\_2(B)&E\_n\end\{array\}\right)\\\\ \to& \left(\begin\{array\}\{cccccccccccccccccccc\}0&h\_1(B)\\\\ -h\_2(B)&E\_n-u\_1(B)h\_1(B)\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}0&h\_1(B)\\\\ -h\_2(B)&E\_n-u\_1(B)h\_1(B)-u\_2(B)h\_2(B)\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&h\_1(B)\\\\ h\_2(B)\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\} h\_1(B)+\mathrm\{rank\} h\_2(B)=n+\mathrm\{rank\} h(B)$.跟锦数学微信公众号. [在线资料](
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---
1210、 1、 (20 分) 设 $\displaystyle A$ 是 $\displaystyle n\times n$ 矩阵, 且 $\displaystyle A^2=2023A$. 证明:
\begin\{aligned\} \mathrm\{rank\} A+\mathrm\{rank\}(A-2023E\_n)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&0\\\\ 0&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&\\\\ &A^2-2023A\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&A\\\\ 0&A^2-2023A\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&A\\\\ 2023E\_n-A&0\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2023\}A&A\\\\ 2023E\_n-A&0\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}0&A\\\\ 2023E\_n-A&0\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&A\\\\ A-2023E\_n-A&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \mathrm\{rank\} A+\mathrm\{rank\}(A-2023E\_n)=n$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1211、 3、 (25 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实矩阵, $\displaystyle n\geq 2$, $\displaystyle A^\star$ 是其伴随矩阵. (1)、 证明: $\displaystyle \mathrm\{rank\}(A^\star)=\left\\{\begin\{array\}\{llllllllllll\}n,&\mathrm\{rank\} A=n,\\\\ 1,&\mathrm\{rank\} A=n-1,\\\\ 0,&\mathrm\{rank\} A < n-1.\end\{array\}\right.$ (2)、 若 $\displaystyle A$ 是非零实对称矩阵, 且 $\displaystyle n\geq 3$. 证明: $\displaystyle A$ 与 $\displaystyle A^\star$ 合同的充分必要条件是 $\displaystyle A$ 可逆且负惯性指数为偶数. [题目有问题, 反例及正确表述见参考解答] (苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 (1-1)、 若 $\displaystyle \mathrm\{rank\} A=n$, 则 $\displaystyle |A|\neq 0$, 而由 $\displaystyle AA^\star=|A|E$ 知 $\displaystyle A^\star$ 可逆, $\displaystyle \mathrm\{rank\} A^\star=n$. (1-2)、 若 $\displaystyle \mathrm\{rank\} A=n-1$, 则 $\displaystyle A$ 有一个 $\displaystyle n-1$ 阶子式不等于 $\displaystyle 0$, 而 $\displaystyle A^\star$ (它的元素为 $\displaystyle A$ 的 $\displaystyle n-1$ 阶子式) 非零, $\displaystyle \mathrm\{rank\} A^\star\geq 1$. 再由 $\displaystyle AA^\star=|A|E=0$ 知 $\displaystyle A^\star$ 的列向量组是 $\displaystyle Ax=0$ 的解向量. 故 $\displaystyle \mathrm\{rank\} A^\star\leq n-\mathrm\{rank\} A=1$. 故 $\displaystyle \mathrm\{rank\} A^\star=1$. (1-3)、 若 $\displaystyle \mathrm\{rank\} A < n-1$, 则 $\displaystyle A$ 的所有 $\displaystyle n-1$ 阶子式都为 $\displaystyle 0$, 而 $\displaystyle A^\star=0\Rightarrow \mathrm\{rank\} A^\star=0$. (2)、 取 $\displaystyle A=\mathrm\{diag\}(1,-1)\Rightarrow A^\star=\mathrm\{diag\}(-1,1)$ 即知 $\displaystyle A$ 与 $\displaystyle A^\star$ 合同, 但 $\displaystyle A$ 的负惯性指数为奇数! 故题目不正确. 正确表述如下. $\displaystyle A$ 与 $\displaystyle A^\star$ 合同的充分必要条件是 $\displaystyle A$ 可逆, 负惯性指数为偶数或负惯性指数为奇数且符号差为零. (2-1)、 $\displaystyle \Rightarrow$: 由 $\displaystyle A, A^\star$ 合同知 $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^\star)$. 又由第 1 步知 $\displaystyle \mathrm\{rank\} A=n$, $\displaystyle A$ 可逆. 再由 $\displaystyle A$ 实对称知存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_p,\mu\_1,\cdots,\mu\_q),\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \lambda\_i,\mu\_j\in \mathbb\{R\}, \lambda\_i > 0, \mu\_j < 0$. 于是
\begin\{aligned\} P^\star A^\star (P^\star)^\mathrm\{T\}=\mathrm\{diag\}\left(\frac\{|A|\}\{\lambda\_1\},\cdots,\frac\{|A|\}\{\lambda\_p\}, \frac\{|A|\}\{\mu\_1\},\cdots,\frac\{|A|\}\{\mu\_q\}\right).\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle |A| > 0$ 时, $\displaystyle A$ 的负惯性指数为偶数. 当 $\displaystyle |A| < 0$ 时, $\displaystyle A^\star$ 的正负惯性指数分别为 $\displaystyle q,p$. 由 $\displaystyle A$ 的正负惯性指数分别为 $\displaystyle p,q$ 及 $\displaystyle A,A^\star$ 合同知 $\displaystyle p=q$. (2-2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle A$ 可逆知 $\displaystyle (I), (II)$ 成立. 若 $\displaystyle A$ 的负惯性指数为偶数, 则 $\displaystyle |A| > 0$, 而 $\displaystyle A,A^\star$ 的正负惯性指数都为 $\displaystyle p,q$, 而合同. 若 $\displaystyle A$ 的负惯性指数为奇数, 则 $\displaystyle |A| < 0$, $\displaystyle A^\star$ 的正负惯性指数分别为 $\displaystyle q,p$. 由充分性假设知 $\displaystyle q=p$. 从而 $\displaystyle A, A^\star$ 的正负惯性指数分别为 $\displaystyle p,p$, 是合同的.跟锦数学微信公众号. [在线资料](
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---
1212、 4、 (25 分) 设 $\displaystyle k\in \mathbb\{R\}, \alpha\in\mathbb\{R\}$ 为实单位向量, $\displaystyle A=E\_n-k\alpha\alpha^\mathrm\{T\}$. (1)、 证明: $\displaystyle A$ 是正交阵当且仅当 $\displaystyle k=0$ 或 $\displaystyle 2$. (2)、 证明: $\displaystyle A$ 为正定阵当且仅当 $\displaystyle k < 1$. (苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} &A^\mathrm\{T\} A=(E\_n-k\alpha\alpha^\mathrm\{T\})(E\_n-k\alpha\alpha^\mathrm\{T\})\\\\ =&E\_n-2k\alpha\alpha^\mathrm\{T\}+k^2\alpha(\alpha^\mathrm\{T\} \alpha)\alpha^\mathrm\{T\} =E\_n+(k^2-2k)\alpha\alpha^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 正交
\begin\{aligned\} \Leftrightarrow A^\mathrm\{T\} A=E\_n\Leftrightarrow k^2-2k=0\Leftrightarrow k=0\mbox\{或\} 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 (2-1)、 我们有如下常用公式: 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&E+BA\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E+AB&0\\\\ 0&E\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知
\begin\{aligned\} |E+BA|=|E+AB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 设 $\displaystyle \alpha=(a\_1,\cdots,a\_n)^\mathrm\{T\}$, $\displaystyle \alpha\_i=(a\_1,\cdots,a\_i)^\mathrm\{T\}$, 则 $\displaystyle A$ 的第 $\displaystyle i$ 个顺序主子式为
\begin\{aligned\} |A\_i|=|E\_i-k\alpha\_i\alpha\_i^\mathrm\{T\}| \stackrel\{\mbox\{第 i 步\}\}\{=\}1-k\alpha\_i^\mathrm\{T\} \alpha\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 正定 $\displaystyle \Leftrightarrow A$ 的所有顺序主子式
\begin\{aligned\} &1-k\alpha\_1^\mathrm\{T\} \alpha\_1 > 0,\cdots, 1-k\alpha\_\{n-1\}^\mathrm\{T\} \alpha\_\{n-1\} > 0, 1-k\alpha^\mathrm\{T\} \alpha > 0\\\\ \Leftrightarrow& 1-k\alpha^\mathrm\{T\} \alpha=1-k > 0\Leftrightarrow k < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1213、 5、 (20 分) 设 $\displaystyle A=(a\_\{ij\})\_\{n\times m\}, B=(b\_\{ij\})\_\{p\times q\}$ 分别为数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n\times m$ 和 $\displaystyle p\times q$ 矩阵, $\displaystyle np\times nq$ 矩阵
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{11\}B&a\_\{12\}B&\cdots&a\_\{1m\}B\\\\ a\_\{21\}B&a\_\{22\}B&\cdots&a\_\{2m\}B\\\\ \vdots&\vdots&&\vdots\\\\ a\_\{n1\}B&a\_\{n2\}B&\cdots&a\_\{nm\}B\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
称为 $\displaystyle A$ 和 $\displaystyle B$ 的 Kronecker 乘积, 记作 $\displaystyle A\otimes B$. (1)、 设 $\displaystyle A,B,C,D$ 分别为 $\displaystyle n\times m, p\times q, m\times l, q\times s$ 矩阵, 证明:
\begin\{aligned\} (A\otimes B)(C\otimes D)=(AC)\otimes(BD). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 证明: $\displaystyle \mathrm\{rank\}(A\otimes B)=\mathrm\{rank\} A\cdot \mathrm\{rank\} B$. (苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle A=(a\_\{ih\}), C=(c\_\{hj\})$, 则
\begin\{aligned\} A\otimes B=(a\_\{ih\}B), C\otimes D=(c\_\{hj\}D). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由分块矩阵的乘法规则知 $\displaystyle (A\otimes B)(C\otimes D)$ 的 $\displaystyle (i,j)$ 块为
\begin\{aligned\} \sum\_h a\_\{ih\}B\cdot c\_\{hj\}D=\sum\_h a\_\{ih\}c\_\{hj\}BD, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
恰好为 $\displaystyle AC$ 的 $\displaystyle (i,j)$ 元乘以 $\displaystyle BD$, 就是 $\displaystyle (AC)\otimes(BD)$ 的 $\displaystyle (i,j)$ 块. 故
\begin\{aligned\} (A\otimes B)(C\otimes D)=(AC)\otimes (BD). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 (2-1)、 先证一个结论. 若 $\displaystyle A,B$ 均可逆, 则
\begin\{aligned\} (A^\{-1\}\otimes B^\{-1\})(A\otimes B)=(A^\{-1\}A)\otimes(B^\{-1\}B)=I\otimes I=I. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即若 $\displaystyle A,B$ 可逆, 则 $\displaystyle A\otimes B$ 也可逆, 且 $\displaystyle (A\otimes B)^\{-1\}=(A^\{-1\}\otimes B^\{-1\})$. (2-2)、 回到题目. 设 $\displaystyle \mathrm\{rank\} A=r, \mathrm\{rank\} B=s$, 则存在可逆矩阵 $\displaystyle P,Q,R,S$ 使得
\begin\{aligned\} PAQ=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right), RBS=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_s&\\\\ &0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &(P\otimes Q)\cdot (A\otimes B)\cdot (R\otimes S) \overset\{\tiny\mbox\{第1步\}\}\{=\} (PAQ)\otimes(QBS)\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)\otimes \left(\begin\{array\}\{cccccccccccccccccccc\}E\_s&\\\\ &0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_\{rs\}&\\\\ &0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle P,Q$ 可逆知 $\displaystyle P\otimes Q$ 可逆. 同理 $\displaystyle R\times S$ 可逆. 故
\begin\{aligned\} \mathrm\{rank\} (A\otimes B)=rs=\mathrm\{rank\} A\cdot \mathrm\{rank\} B . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
1214、 6、 (20 分) $\displaystyle n$ 阶复矩阵 $\displaystyle A$ 称为对合矩阵, 若 $\displaystyle A^2=E\_n$. (1)、 证明: 任意一个对合矩阵都可以相似对角化; (2)、 证明: 两两可交换的互不相同的 $\displaystyle n$ 阶对合矩阵最多有 $\displaystyle 2^n$ 个. (苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设
\begin\{aligned\} V\_1=\left\\{\alpha\in\mathbb\{C\}^n; A\alpha=\alpha\right\\},\quad V\_2=\left\\{\alpha\in\mathbb\{C\}^n; A\alpha=-\alpha\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由
\begin\{aligned\} \alpha\in\mathbb\{C\}^n&\Rightarrow \alpha=\frac\{\alpha+A\alpha\}\{2\}+\frac\{\alpha-A\alpha\}\{2\}\in V\_1+V\_2,\\\\ \alpha\in V\_1\cap V\_2&\Rightarrow A\alpha =\alpha, A\alpha=-\alpha\\\\ &\Rightarrow \alpha=-\alpha\Rightarrow \alpha=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \mathbb\{C\}^n=V\_1\oplus V\_2$, 而
\begin\{aligned\} &n=\dim V\_1+\dim V\_2=\left\[n-\mathrm\{rank\}(E\_n-A)\right\]+\left\[n-\mathrm\{rank\}(E\_n+A)\right\]\\\\ \Rightarrow&\mathrm\{rank\}(E\_n+A)+\mathrm\{rank\}(E\_n-A)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle V\_1$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$, $\displaystyle V\_2$ 的一组基 $\displaystyle \varepsilon\_\{r+1\},\cdots,\varepsilon\_n$, 则 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$ 是 $\displaystyle \mathbb\{P\}^n$ 的一组基. 令 $\displaystyle T=(\varepsilon\_1,\cdots,\varepsilon\_n)$, 则 $\displaystyle T$ 是可逆阵, 且
\begin\{aligned\} T^\{-1\}AT=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle AB=BA$, 则设 $\displaystyle \tilde\{B\}=T^\{-1\}BT=\left(\begin\{array\}\{cccccccccccccccccccc\}\tilde\{B\}\_\{11\}&\tilde\{B\}\_\{12\}\\\\ \tilde\{B\}\_\{21\}&\tilde\{B\}\_\{22\}\end\{array\}\right)$ 后,
\begin\{aligned\} &AB=BA\Leftrightarrow T^\{-1\}AT\cdot T^\{-1\}BT=T^\{-1\}BT\cdot T^\{-1\}AT\\\\ \Leftrightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}\tilde\{B\}\_\{11\}&\tilde\{B\}\_\{12\}\\\\ \tilde\{B\}\_\{21\}&\tilde\{B\}\_\{22\}\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\}\tilde\{B\}\_\{11\}&\tilde\{B\}\_\{12\}\\\\ \tilde\{B\}\_\{21\}&\tilde\{B\}\_\{22\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right)\\\\ \Leftrightarrow&\tilde\{B\}\_\{12\}=0, \tilde\{B\}\_\{21\}=0 \Leftrightarrow \tilde\{B\}=\mathrm\{diag\}(\tilde\{B\}\_\{11\},\tilde\{B\}\_\{22\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle B$ 对合知 $\displaystyle \tilde\{B\}\_\{ii\}$ 对合, 而存在可逆矩阵 $\displaystyle P\_i$ 使得 $\displaystyle P\_i^\{-1\}\tilde\{B\}\_\{ii\}P\_i$ 为对角阵, 且对角元为 $\displaystyle \pm 1$. 令 $\displaystyle P=\mathrm\{diag\}(P\_1,P\_2)$, 则 $\displaystyle P$ 可逆, 且 $\displaystyle P^\{-1\}\tilde\{B\}P$ 为对角阵, 且对角元 $\displaystyle =\pm 1$. 再令 $\displaystyle Q=TP$, 则
\begin\{aligned\} &Q^\{-1\}AQ=P^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right)P\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}P\_1^\{-1\}&\\\\ &P\_2^\{-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}P\_1&\\\\ &P\_2\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &-E\_\{n-r\}\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} Q^\{-1\}BQ=&P^\{-1\}\tilde\{B\}P=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\in \left\\{1,-1\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \lambda\_i\in \left\\{-1,1\right\\}$ 知 $\displaystyle \left\\{B; AB=BA\right\\}$ 有 $\displaystyle 2^n$ 个. 故有结论.跟锦数学微信公众号. [在线资料](
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1215、 7、 (20 分) $\displaystyle n$ 阶复方阵 $\displaystyle A$ 称为幂零阵, 若存在正整数 $\displaystyle m$, 使得 $\displaystyle A^m=0$. (1)、 证明: $\displaystyle A$ 是幂零阵当且仅当 $\displaystyle A$ 的特征值全是零; (2)、 证明: 互不相似的 $\displaystyle 5$ 阶幂零阵最多有 $\displaystyle 7$ 个. (苏州大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 (1-1)、 $\displaystyle \Rightarrow$: 由 $\displaystyle A$ 幂零知存在 $\displaystyle m$, 使得 $\displaystyle A^m=0$. 若 $\displaystyle \lambda$ 是 $\displaystyle A$ 的特征值, 对应的特征向量为 $\displaystyle \alpha\neq 0$, 则
\begin\{aligned\} A\alpha=\lambda \alpha\Rightarrow \cdots \Rightarrow 0=A^m\alpha=\lambda^m\alpha\Rightarrow \lambda^m=0\Rightarrow \lambda=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 的特征值全为 $\displaystyle 0$. (1-2)、 $\displaystyle \Leftarrow$: 若 $\displaystyle A$ 的特征值全为 $\displaystyle 0$, 则 $\displaystyle A$ 的 Jordan 标准标形的对角元都是 $\displaystyle 0$, 则由 Jordan 块的性质知 $\displaystyle J^\{\max n\_i\}=0\Rightarrow A^\{\max\_i n\_i\}=0$, 而 $\displaystyle A$ 幂零. 这里 $\displaystyle n\_i$ 是 $\displaystyle J$ 中 Jordan 块的最高阶数. (2)、 $\displaystyle 5$ 阶幂零阵的 Jordan 标准形只有以下 $\displaystyle 7$ 种:
\begin\{aligned\} 0\_7; \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&\\\\ &0&\\\\ &&0\_3\end\{array\}\right); \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&\\\\ &0&1&\\\\ &&0&\\\\ &&&0\_2\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&&\\\\ &0&&&\\\\ &&0&1&\\\\ &&&0&\\\\ &&&&0\end\{array\}\right);\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&&\\\\ &0&1&&\\\\ &&0&1&\\\\ &&&0&\\\\ &&&&0\end\{array\}\right), \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&&\\\\ &0&1&&\\\\ &&0&&\\\\ &&&0&1\\\\ &&&&0\end\{array\}\right); \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&&\\\\ &0&1&&\\\\ &&0&1&\\\\ &&&0&1\\\\ &&&&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而互不相似的 $\displaystyle 5$ 阶幂零阵最多有 $\displaystyle 7$ 个.跟锦数学微信公众号. [在线资料](
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---
1216、 6、 证明: (1)、 若 $\displaystyle A=(a\_\{ij\})$ 的元素 $\displaystyle a\_\{ij\}$ 为整数, $\displaystyle A$ 的各行元素之和为 $\displaystyle a\neq 0$, 则 $\displaystyle A^\star$ 的各行元素之和为 $\displaystyle \frac\{|A|\}\{a\}$, 且 $\displaystyle \frac\{|A|\}\{a\}$ 为整数; (2)、 对 $\displaystyle n$ 阶矩阵 $\displaystyle A$, 有 $\displaystyle (A^\star)^\star=|A|^\{n-2\}A$. (太原理工大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则
\begin\{aligned\} Ae=ae\Rightarrow& |A|e=|A|Ee=A^\star Ae=A^\star(ae)=aA^\star e\\\\ \Rightarrow& A^\star e=\frac\{|A|\}\{a\}e.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle A^\star$ 的各行元素之和为 $\displaystyle \frac\{|A|\}\{a\}$. 注意到 $\displaystyle A^\star$ 的元素作为 $\displaystyle A$ 的代数余子式是整数. 比较 $\displaystyle (I)$ 的第一个分量即知 $\displaystyle \frac\{|A|\}\{a\}\in\mathbb\{Z\}$. (2)、 (2-1)、 由 $\displaystyle |A|=0$ 知 $\displaystyle \mathrm\{rank\}(A)\leq n-1$. 若 $\displaystyle \mathrm\{rank\}(A)\leq n-2$, 则 $\displaystyle A$ 的所有 $\displaystyle n-1$ 阶子式都 $\displaystyle =0$, $\displaystyle A^\star =0\Rightarrow \mathrm\{rank\}(A^\star )=0$. 若 $\displaystyle \mathrm\{rank\}(A)=n-1$, 则 $\displaystyle A$ 由一个 $\displaystyle n-1$ 阶子式 $\displaystyle \neq 0$, $\displaystyle A^\star \neq 0\Rightarrow \mathrm\{rank\}(A^\star )\geq 1$. 又由
\begin\{aligned\} AA^\star =|A|E=0&\Rightarrow \mbox\{ $\displaystyle A^\star $ 的列向量组都是 $\displaystyle AX=0$ 的解\}\\\\ &\Rightarrow \mathrm\{rank\}(A^\star )\leq n-\mathrm\{rank\}(A)=n-(n-1)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\}(A^\star )=1$. (2-2)、 若 $\displaystyle A$ 可逆, 则
\begin\{aligned\} AA^\star =|A|E\Rightarrow |A|\cdot |A^\star |=|A|^n \Rightarrow |A^\star |=|A|^\{n-1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且由公式 $\displaystyle B^\star =|B| B^\{-1\}$ 知
\begin\{aligned\} (A^\star )^\star =&|A^\star |(A^\star )^\{-1\}=|A|^\{n-1\}(|A|A^\{-1\})^\{-1\}\\\\ =&|A|^\{n-1\}|A|^\{-1\}(A^\{-1\})^\{-1\} =|A|^\{n-2\}A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle A$ 不可逆, 则
\begin\{aligned\} |A|=0&\Rightarrow \mathrm\{rank\}(A^\star )\leq 1\left(\leq n-2\right)\left(\mbox\{第 i 步\}\right)\\\\ &\Rightarrow \left\\{\begin\{array\}\{llllllllllll\} |A^\star |=0\\\\ (A^\star )^\star =0 \end\{array\}\right.\left(\mbox\{第 i 步的证明\}\right)\\\\ &\Rightarrow \mbox\{ $\displaystyle |A^\star |=0=|A|^\{n-1\}$, 且 $\displaystyle (A^\star )^\star =0=|A|^\{n-2\}A$.\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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1217、 7、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ a&1&0&0\\\\ 1&1&2&0\\\\ -1&-1&b&2\end\{array\}\right)$. (1)、 讨论 $\displaystyle a,b$ 为何值时, $\displaystyle A$ 可对角化, 并求可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP$ 为对角阵; (2)、 当 $\displaystyle a=1, b=0$ 时, 求 $\displaystyle A$ 的若尔当标准形. (太原理工大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,2,2$. 若 $\displaystyle A$ 可对角化, 则 $\displaystyle A\sim\mathrm\{diag\}(1,1,2,2)$. 故 $\displaystyle \mathrm\{rank\}(A-E)=\mathrm\{rank\}(A-2E)=2$. 由
\begin\{aligned\} A-E=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&0&0\\\\ a&0&0&0\\\\ 1&1&1&0\\\\ -1&-1&b&1\end\{array\}\right), A-2E=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0&0&0\\\\ a&-1&0&0\\\\ 1&1&0&0\\\\ -1&-1&b&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a=0, b=0$. 此时, 由
\begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-1\\\\ 0&0&1&1\\\\ 0&0&0&0\\\\ 0&0&0&0 \end\{array\}\right), 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&0&0\\\\ 0&1&0&0\\\\ 0&0&0&0\\\\ 0&0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,2$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\0\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\-1\\\\1 \end\{array\}\right);\quad \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\1\\\\0 \end\{array\}\right), \xi\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\0\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} P=(\xi\_1,\xi\_2,\xi\_3,\xi\_4)=\left(\begin\{array\}\{cccccccccccccccccccc\} -1&1&0&0\\\\ 1&0&0&0\\\\ 0&-1&1&0\\\\ 0&1&0&1\end\{array\}\right)\Rightarrow P^\{-1\}AP=\mathrm\{diag\}\left(1,1,2,2\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 当 $\displaystyle a=1, b=0$ 时,
\begin\{aligned\} A-E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&1&0&-1\\\\ 0&0&1&1\\\\ 0&0&0&0\end\{array\}\right), A-2E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&1&0&0\\\\ 0&0&0&0\\\\ 0&0&0&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 也满足 $\displaystyle \mathrm\{rank\}(J-E)=3, \mathrm\{rank\}(J-2E)=2$. 分析对角线上方 $\displaystyle 1$ 的个数即知 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &1&&\\\\ &&2&\\\\ &&&2\end\{array\}\right)$.跟锦数学微信公众号. [在线资料](
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---
1218、 2、 (1)、 计算行列式
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}1+x\_1y\_1&x\_1y\_2&\cdots&x\_1y\_n\\\\ x\_2y\_1&1+x\_2y\_2&\cdots&x\_2y\_n\\\\ \vdots&\vdots&\ddots&\vdots\\\\ x\_ny\_1&x\_ny\_2&\cdots&1+x\_ny\_n\end\{array\}\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \alpha\beta^\mathrm\{T\}\neq -1$, 证明: $\displaystyle I+\beta\alpha^\mathrm\{T\}$ 可逆. (3)、 若 $\displaystyle \alpha\beta^\mathrm\{T\}=-1$, 证明: $\displaystyle \mathrm\{rank\}(I+\beta\alpha^\mathrm\{T\})=n-1$. (天津大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} \mbox\{原式\}&=\left|\begin\{array\}\{cccc\} 1&y\_1&\cdots&y\_n\\\\ 0&1+x\_1y\_1&\cdots&x\_1y\_n\\\\ \vdots&\vdots&\ddots&\vdots\\\\ 0&x\_ny\_1&\cdots&1+x\_ny\_n \end\{array\}\right|\\\\ &=\left|\begin\{array\}\{cccc\} 1&y\_1&\cdots&y\_n\\\\ -x\_1&1&\cdots&0\\\\ \vdots&\vdots&\ddots&\vdots\\\\ -x\_n&0&\cdots&1 \end\{array\}\right| =\left|\begin\{array\}\{cccc\} \{1+\sum\_\{i=1\}^nx\_iy\_i\}&y\_1&\cdots&y\_n\\\\ 0&1&\cdots&0\\\\ \vdots&\vdots&\ddots&\vdots\\\\ 0&0&\cdots&1 \end\{array\}\right|\\\\ &=1+\sum\_\{i=1\}^nx\_iy\_i. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设
\begin\{aligned\} \alpha=(x\_1,\cdots,x\_n), \beta=(y\_1,\cdots,y\_n)^\mathrm\{T\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由第 1 步知
\begin\{aligned\} |I+\beta\alpha^\mathrm\{T\}|=1+\alpha\beta^\mathrm\{T\}\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle I+\beta\alpha^\mathrm\{T\}$ 可逆. (3)、 若 $\displaystyle \alpha\beta^\mathrm\{T\}=-1$, 则由第 2 步知 $\displaystyle I+\beta\alpha^\mathrm\{T\}$ 奇异, $\displaystyle \mathrm\{rank\}(I+\beta\alpha^\mathrm\{T\})\leq n-1$. 又由 $\displaystyle \alpha\beta^\mathrm\{T\}=-1$ 知 $\displaystyle x\_iy\_i$ 不能同时为 $\displaystyle 0$. 不妨设 $\displaystyle x\_ny\_n\neq 0$. 则 $\displaystyle I+\beta\alpha^\mathrm\{T\}$ 的第 $\displaystyle n-1$ 个顺序主子式按照第 1 问的求解方法知其
\begin\{aligned\} =1+\sum\_\{i=1\}^\{n-1\}x\_iy\_i=-x\_ny\_n\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\}(I+\beta\alpha^\mathrm\{T\})\geq n-1$. 联合已有结果即知 $\displaystyle \mathrm\{rank\}(I+\beta\alpha^\mathrm\{T\})=n-1$.跟锦数学微信公众号. [在线资料](
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---
1219、 3、 已知 $\displaystyle f(x)=x^2+ax+4, a\in\mathbb\{R\}$, 且 $\displaystyle f(A)=0$. (1)、 若对 $\displaystyle \forall\ \lambda\in\mathbb\{R\}$, 有 $\displaystyle A+\lambda I$ 可逆, 求 $\displaystyle a$ 的取值范围. (2)、 若 $\displaystyle A+\lambda I$ 不可逆, 且 $\displaystyle B=A^3+2023I$. 证明: $\displaystyle B$ 可逆, 且求 $\displaystyle g(x)\in \mathbb\{R\}[x]$ 使得 $\displaystyle B^\{-1\}=g(A)$. [本小问有问题, 跟锦数学微信公众号没法做哦.] (天津大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} &\exists\ \lambda\_0\in\mathbb\{R\},\mathrm\{ s.t.\} |A+\lambda\_0I|=0\Leftrightarrow A+\lambda\_0I\mbox\{有特征值\}0\\\\ \Leftrightarrow&A\mbox\{有特征值\}-\lambda\_0 \Rightarrow 0=f(-\lambda\_0)=\lambda\_0^2-a\lambda\_0+4=0\\\\ \Rightarrow&\lambda^2-a\lambda+4=0\mbox\{有实根\} \Leftrightarrow \Delta=a^2-16\geq 0\Leftrightarrow |a|\geq 4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
反之亦成立. 故 $\displaystyle a$ 的取值范围是 $\displaystyle |a| < 4$. (2)、 本小问有问题, 跟锦数学微信公众号没法做哦.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 13:12:28
