## 张祖锦2023年数学专业真题分类70天之第49天
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1105、 (5)、 已知 $\displaystyle 2022$ 阶实方阵 $\displaystyle A$, $\displaystyle A$ 第 $\displaystyle (i,j)$ 元素为 $\displaystyle \frac\{(i+j)^\{2022\}\}\{j^\{2022\}\}$, 则 $\displaystyle \det(\mathrm\{e\}^A)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (复旦大学2023年代数(第4,6,7,8题没做)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) / 设 $\displaystyle \mathrm\{e\}$ 为自然常数. 对于 $\displaystyle n$ 阶复方阵 $\displaystyle A$, 我们定义
\begin\{aligned\} \mathrm\{e\}^A=I\_n+A+\frac\{A^2\}\{2!\}+\frac\{A^3\}\{3!\}+\cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 (5-1)、 方阵 $\displaystyle \mathrm\{e\}^A$ 定义合理; (5-2)、 $\displaystyle \det \mathrm\{e\}^A=\mathrm\{e\}^\{\mathrm\{tr\} A\}$. 事实上, (5-1)、 由 Jordan 标准形理论知存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}AP=J=\mathrm\{diag\}\left(J\_\{n\_1\}(\lambda\_1),\cdots,J\_\{n\_s\}(\lambda\_s)\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 $\displaystyle J\_\{n\_i\}(\lambda\_i)$ 是以 $\displaystyle \lambda\_i$ 为对角元的 $\displaystyle n\_i$ 阶 Jordan 块. 令 $\displaystyle N\_\{n\_i\}=J\_\{n\_i\}(\lambda\_i)-\lambda\_iE\_\{n\_i\}$, 则 $\displaystyle N\_\{n\_i\}^\{n\_i\}=0$, 而
\begin\{aligned\} J\_\{n\_i\}(\lambda\_i)^k&=(\lambda\_iE\_\{n\_i\}+N\_\{n\_i\})^k =\sum\_\{0\leq j\leq k\atop j\leq n\_i\}C\_k^j \lambda\_i^\{k-j\}N\_\{n\_i\}^j\\\\ &=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_i^k&k\lambda\_i^\{k-1\}&\cdots&\star \\\\ &\ddots&\ddots&\vdots\\\\ &&\ddots&k\lambda\_i^k\\\\ &&&\lambda\_i^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\}
设 $\displaystyle f\_k(\lambda)=\lambda^k$, 则 $\displaystyle J\_\{n\_i\}(\lambda\_i)$ 中非零元素具有形式
\begin\{aligned\} \frac\{f\_k^\{(p)\}(\lambda\_i)\}\{p!\}\left(p=0,1,2,\cdots,n\_i-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
\begin\{aligned\} \sum\_\{k=0\}^\infty \frac\{1\}\{k!\}\left\[\frac\{f\_k^\{(p)\}(\lambda\_i)\}\{p!\}\right\] =\left.\frac\{1\}\{p!\}\left\[\sum\_\{k=0\}^\infty \frac\{f\_k(\lambda)\}\{k!\}\right\]^\{(p)\}\right|\_\{\lambda=\lambda\_i\} =\frac\{1\}\{p!\}\mathrm\{e\}^\{\lambda\_i\} \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\{e\}^\{J\_\{n\_i\}(\lambda\_i)\}=\left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{e\}^\lambda&\mathrm\{e\}^\{\lambda\}&\frac\{\mathrm\{e\}^\lambda\}\{2\}&\cdots&\frac\{\mathrm\{e\}^\{\lambda\}\}\{(n\_i-1)!\}\\\\ &\ddots&\ddots&\ddots&\vdots\\\\ &&\ddots&\ddots&\frac\{\mathrm\{e\}^\lambda\}\{2\}\\\\ &&&\ddots&\mathrm\{e\}^\lambda\\\\ &&&&\mathrm\{e\}^\lambda\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 $\displaystyle \mathrm\{e\}^A=P^\{-1\}\mathrm\{e\}^J P$ 定义合理. (5-2)、 由 Jordan 标准形理论知存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} &P^\{-1\}AP=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\star \\\\ &\ddots&\vdots\\\\ &&\lambda\_n\end\{array\}\right)\Rightarrow P^\{-1\}A^kP=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1^k&\cdots&\star \\\\ &\ddots&\vdots\\\\ &&\lambda\_n^k\end\{array\}\right)\\\\ \Rightarrow&P^\{-1\}\mathrm\{e\}^A P=\left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{e\}^\{\lambda\_1\}&\cdots&\star \\\\ &\ddots&\vdots\\\\ &&\mathrm\{e\}^\{\lambda\_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\}
故
\begin\{aligned\} |\mathrm\{e\}^A|=\mathrm\{e\}^\{\lambda\_1\}\cdots \mathrm\{e\}^\{\lambda\_n\}=\mathrm\{e\}^\{\lambda\_1+\cdots+\lambda\_n\} =\mathrm\{e\}^\{\mathrm\{tr\} A\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
回到题目.
\begin\{aligned\} \det(\mathrm\{e\}^A)=&\mathrm\{exp\}(\mathrm\{tr\} A) =\mathrm\{exp\}\left\[\sum\_\{i=1\}^\{2022\} \frac\{(2i)^\{2022\}\}\{i^\{2022\}\}\right\] =\mathrm\{e\}^\{2022\cdot 2^\{2022\}\}. \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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1106、 5、 已知 $\displaystyle A$ 是正定实对称矩阵. (1)、 证明: 存在唯一的正定实对称阵 $\displaystyle H$, 使得 $\displaystyle A=H-H^\{-1\}$. (2)、 给定任意正数 $\displaystyle r$, 求所有 $\displaystyle r$ 可能的值, 使存在对称可逆方阵 $\displaystyle R$, 有 $\displaystyle A=R-R^\{-1\}$ 且 $\displaystyle R$ 的符号差为 $\displaystyle r$. (复旦大学2023年代数(第4,6,7,8题没做)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 $\displaystyle A$ 正定知存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} A=P\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)P^\mathrm\{T\}, \lambda\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle a\_i=\frac\{\lambda\_i+\sqrt\{\lambda\_i^2+4\}\}\{2\}$, 则 $\displaystyle a\_i-\frac\{1\}\{a\_i\}=\lambda\_i$. 设
\begin\{aligned\} H=P\mathrm\{diag\}(a\_1,\cdots,a\_n)P^\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 $\displaystyle H$ 正定, 且 $\displaystyle A=H-H^\{-1\}$. 往证唯一性. 设 $\displaystyle H$ 满足题设, 则 $\displaystyle AH=HA$,
\begin\{aligned\} &A=H-H^\{-1\}\Rightarrow AH=H^2-E\_n\\\\ \Rightarrow& H^2-AH=E\_n\Rightarrow \left(H-\frac\{A\}\{2\}\right)^2=E\_n+\frac\{A^2\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle H$ 正定知 $\displaystyle H-\frac\{A\}\{2\}$ 的特征值 $\displaystyle > -\frac\{A\}\{2\}$ 的相应特征值, 而 $\displaystyle -\sqrt\{E\_n+\frac\{A^2\}\{4\}\}$ 的特征值 $\displaystyle < -\frac\{A\}\{2\}$ 的相应特征值. 故
\begin\{aligned\} H-\frac\{A\}\{2\}=\sqrt\{E\_n+\frac\{A^2\}\{4\}\} \Rightarrow H=\frac\{A\}\{2\}+\sqrt\{E\_n+\frac\{A^2\}\{4\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
唯一性得证. (2)、 取
\begin\{aligned\} R=P\mathrm\{diag\}(a\_1,\cdots,a\_n)P^\mathrm\{T\}, a\_i\in\left\\{\frac\{\lambda\_i+\sqrt\{\lambda\_i^2+4\}\}\{2\}, \frac\{\lambda\_i-\sqrt\{\lambda\_i^2+4\}\}\{2\}\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle R-R^\{-1\}=A$, 而正惯性指数 $\displaystyle 0\leq p\leq n$, 负惯性指数 $\displaystyle q=n-p$, 符号差
\begin\{aligned\} r=p-q=2p-n, 0\leq p\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle r$ 的可能取值为 $\displaystyle -n,2-n,4-n,\cdots,n$.跟锦数学微信公众号. [在线资料](
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1107、 (4)、 矩阵 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}3&1&&&\\\\ &3&&\\\\ &&5&\\\\ &&&5\end\{array\}\right)$ 的初等因子组为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (广西大学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) / $\displaystyle (\lambda-3)^2, \lambda-5, \lambda-5$.跟锦数学微信公众号. [在线资料](
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1108、 (8)、 三阶复方阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&1\\\\ 0&1&0\\\\ 1&0&0\end\{array\}\right)$ 的最小多项式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (广西大学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) / $\displaystyle A$ 的特征值为 $\displaystyle 1,1,-1$. 由 $\displaystyle E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)$ 知 [由秩可判定出严格上三角部分 $\displaystyle 1$ 的个数] $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle \mathrm\{diag\}(1,1,-1)$, 最小多项式为 $\displaystyle (\lambda-1)(\lambda+1)$.跟锦数学微信公众号. [在线资料](
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1109、 4、 设 $\displaystyle A=(a\_\{ij\})$ 为实数域上的 $\displaystyle n$ 阶矩阵, 证明: (1)、 若 $\displaystyle |a\_\{ii\}| > \sum\_\{j\neq i\}|a\_\{ij\}|, i=1,\cdots,n$, 则 $\displaystyle |A|\neq 0$; (2)、 若 $\displaystyle a\_\{ii\} > \sum\_\{j\neq i\}|a\_\{ij\}|, i=1,\cdots,n$, 则 $\displaystyle |A| > 0$. (广西大学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)、 用反证法.
\begin\{aligned\} |A|=0\Rightarrow& \mbox\{ $\displaystyle Ax=0$ 有非零解 $\displaystyle x=(x\_1,\cdots,x\_n)^\mathrm\{T\}$\}\\\\ \Rightarrow& \sum\_\{j=1\}^n a\_\{kj\}x\_j=0\left(|x\_k|=\max\_\{1\leq i\leq n\} |x\_i| > 0\right)\\\\ \Rightarrow& |x\_k|=\left|-\sum\_\{j\neq k\} \frac\{a\_\{kj\}\}\{a\_\{kk\}\}x\_j\right|\leq \sum\_\{j\neq k\}\frac\{|a\_\{kj\}|\}\{|a\_\{kk\}|\}|x\_j| \leq |x\_\{k\}| \frac\{\sum\_\{j\neq k\}|a\_\{kj\}|\}\{|a\_\{kk\}|\} < |x\_\{k\}|\\\\ \Rightarrow&\mbox\{矛盾! 故有结论\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle \lambda$ 是 $\displaystyle A$ 的实特征值, 则
\begin\{aligned\} &\exists\ x\neq 0,\mathrm\{ s.t.\} Ax=\lambda x\\\\ \Rightarrow&\sum\_\{j=1\}^n a\_\{kj\}x\_j=\lambda x\_k\left(|x\_k|=\max\_\{1\leq i\leq n\}|x\_i| > 0\right)\\\\ \Rightarrow& (\lambda-a\_\{kk\})x\_k=\sum\_\{j\neq k\} a\_\{kj\}x\_j\\\\ \Rightarrow& |\lambda-a\_\{kk\}|\leq \sum\_\{j\neq k\} |a\_\{kj\}|\frac\{|x\_j|\}\{|x\_k|\} \leq \sum\_\{j\neq k\} |a\_\{kj\}|\\\\ \Rightarrow&\lambda \geq a\_\{kk\}-\sum\_\{j\neq k\} |a\_\{kj\}| > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle A$ 是实矩阵, 其虚特征值成对出现, 而可设它的全部特征值为
\begin\{aligned\} \lambda\_1,\cdots,\lambda\_s, a\_1+b\_1\mathrm\{ i\}, a\_1-b\_1\mathrm\{ i\}, \cdots, a\_\{n-s\}+b\_\{n-s\}\mathrm\{ i\}, a\_\{n-s\}-b\_\{n-s\}\mathrm\{ 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 $\displaystyle \lambda\_i\in\mathbb\{R\}, a\_j,b\_j\in\mathbb\{R\}, b\_j\neq 0$. 如此,
\begin\{aligned\} |A|=\prod\_\{i=1\}^s \lambda\_i\cdot \prod\_\{j=1\}^\{n-s\} (a\_i^2+b\_i^2) > 0. \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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1110、 6、 设 $\displaystyle A,B$ 均为 $\displaystyle n$ 阶正定矩阵, 证明: $\displaystyle AB$ 为正定矩阵的充分必要条件是 $\displaystyle AB=BA$. (广西大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Leftarrow$: 由 $\displaystyle (AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\} =BA$ 知 $\displaystyle AB$ 是实对称矩阵. 设 $\displaystyle \lambda\in\mathbb\{R\}$ 是 $\displaystyle AB$ 的任一特征值, $\displaystyle 0\neq\alpha\in\mathbb\{R\}^n$ 为对应的特征向量, 则
\begin\{aligned\} &AB\alpha=\lambda \alpha\Rightarrow B\alpha=\lambda A^\{-1\}\alpha\\\\ \Rightarrow& 0 < \alpha^\mathrm\{T\} B\alpha=\lambda\alpha^\mathrm\{T\} A^\{-1\}\alpha\Rightarrow \lambda=\frac\{\alpha^\mathrm\{T\} B\alpha\}\{\alpha^\mathrm\{T\} A^\{-1\}\alpha\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
既然实对称矩阵 $\displaystyle AB$ 的特征值都大于 $\displaystyle 0$, 那么 $\displaystyle AB$ 就是正定的. (2)、 $\displaystyle \Rightarrow$: 若 $\displaystyle AB$ 正定, 则 $\displaystyle AB$ 实对称, 而 $\displaystyle AB=(AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\}=BA$.跟锦数学微信公众号. [在线资料](
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1111、 2、 设 $\displaystyle n$ 阶实矩阵 $\displaystyle A$ 满足 $\displaystyle A^2+A+E=0$, $\displaystyle E$ 为单位阵. (1)、 求证: $\displaystyle A$ 没有实特征值; (2)、 求证: 对任意实数 $\displaystyle a$, $\displaystyle A-aE$ 均可逆. (哈尔滨工程大学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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 用反证法. 若 $\displaystyle A$ 有实特征值 $\displaystyle \lambda$, 则由题设知 $\displaystyle 0=\lambda^2+\lambda+1$. 这与方程 $\displaystyle x^2+x+1=0$ 的根为 $\displaystyle \frac\{-1\pm \sqrt\{3\}\mathrm\{ i\}\}\{2\}$ 为虚根矛盾. 故有结论. (2)、 设 $\displaystyle f(x)=x^2+x+1, g(x)=x-a$, 则由 $\displaystyle f(a)=a^2+a+1 > 0$ 知 $\displaystyle (f,g)=1$, 而
\begin\{aligned\} &\exists\ u,v,\mathrm\{ s.t.\} uf+vg=1 \Rightarrow E=u(A)f(A)+v(A)g(A)=v(A)g(A)\\\\ \Rightarrow& g(A)=A-aE\mbox\{可逆\}. \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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1112、 3、 设 $\displaystyle \alpha$ 是非零的 $\displaystyle n$ 维实列向量, $\displaystyle E$ 为单位阵, $\displaystyle A=E-2\frac\{\alpha\alpha^\mathrm\{T\}\}\{\alpha^\mathrm\{T\} \alpha\}$. (1)、 求证: $\displaystyle A$ 为正交矩阵; (2)、 求证: $\displaystyle A$ 可相似对角化, 并求与 $\displaystyle A$ 相似的对角矩阵 $\displaystyle \varLambda$. (哈尔滨工程大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \eta\_1=\frac\{\alpha\}\{|\alpha|\}$, 则 $\displaystyle A=E-2\eta\_1\eta\_1^\mathrm\{T\}$. 将 $\displaystyle \eta\_1$ 扩充为 $\displaystyle \mathbb\{R\}^n$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_n$, 则
\begin\{aligned\} A\eta\_1=\eta\_1-2\eta\_1=-\eta\_1, A\eta\_i=\eta\_i\left(2\leq i\leq n\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故设 $\displaystyle P=(\eta\_1,\cdots,\eta\_n)$ 后, $\displaystyle P$ 为正交矩阵, 且
\begin\{aligned\} AP=P\left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-1\}\end\{array\}\right)\Rightarrow P^\mathrm\{T\} AP=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-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\}
故 $\displaystyle A$ 可对角化. 进一步, $\displaystyle A=P\left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-1\}\end\{array\}\right)P^\mathrm\{T\}$ 蕴含
\begin\{aligned\} A^\mathrm\{T\} A=P\left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-1\}\end\{array\}\right)P^\mathrm\{T\} \cdot P\left(\begin\{array\}\{cccccccccccccccccccc\}-1&\\\\ &E\_\{n-1\}\end\{array\}\right)P^\mathrm\{T\} =E\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 是正交矩阵.跟锦数学微信公众号. [在线资料](
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---
1113、 10、 设 $\displaystyle 3$ 阶方阵 $\displaystyle A$ 满足 $\displaystyle A^\star=A^\mathrm\{T\}$, 其中 $\displaystyle A^\star$ 为 $\displaystyle A$ 的伴随矩阵, $\displaystyle A^\mathrm\{T\}$ 为 $\displaystyle A$ 的转置矩阵. 求 $\displaystyle |A^2-A|$. (哈尔滨工程大学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\} \mathrm\{rank\} A=\mathrm\{rank\} (A^\mathrm\{T\})=\mathrm\{rank\} (A^\star)=\left\\{\begin\{array\}\{llllllllllll\}3,&\mathrm\{rank\} A=3,\\\\ 1,&\mathrm\{rank\} A=2,\\\\ 0,&\mathrm\{rank\} A\leq 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\}
知 $\displaystyle \mathrm\{rank\} A=3\Rightarrow A$ 可逆, 且
\begin\{aligned\} &|A|E=AA^\star=AA^\mathrm\{T\}\Rightarrow |A|^3=|A|\cdot |A^\mathrm\{T\}|=|A|^2\\\\ \Rightarrow& |A|=1 \Rightarrow AA^\mathrm\{T\}=E, |A^\star|=|A^\mathrm\{T\}|=|A|=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 是正交阵,
\begin\{aligned\} &|A-E|=|A^\mathrm\{T\}|\cdot |A-E|=|E-A^\mathrm\{T\}|=|E-A|=(-1)^3 |A-E|\\\\ \Rightarrow& |A-E|=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} |A^2-A|=|A^\star|\cdot |A^2-A| =||A|A-A^\mathrm\{T\} A| =|A-E|=0. \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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---
1114、 6、 若 $\displaystyle A$ 是数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle m\times n$ 级矩阵, 其中 $\displaystyle m\geq n$. 证明以下命题等价: (1)、 $\displaystyle \mathrm\{rank\} A=n$; (2)、 存在可逆矩阵 $\displaystyle P\_\{m\times m\}$, 使得 $\displaystyle PA=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n\\\\0\end\{array\}\right)\_\{m\times n\}$; (3)、 存在矩阵 $\displaystyle Q\_\{n\times m\}$, 使得 $\displaystyle QA=E\_n$; (4)、 对任意的矩阵 $\displaystyle B\_\{n\times p\}$, $\displaystyle \mathrm\{rank\}(AB)=\mathrm\{rank\} B$; (5)、 对任意的矩阵 $\displaystyle B\_\{n\times p\}, C\_\{n\times p\}$, 若 $\displaystyle AB=AC$, 则 $\displaystyle B=C$. (哈尔滨工业大学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)、 $\displaystyle (1)\Rightarrow (2)$: 由 $\displaystyle \mathrm\{rank\} A=n$ 知 $\displaystyle A$ 的行向量组的秩为 $\displaystyle n$, 通过行调换 (即左乘初等矩阵) 的方法可让 $\displaystyle A$ 的前 $\displaystyle n$ 行线性无关, 其余 $\displaystyle n-m$ 行都是这 $\displaystyle n$ 行的线性组合; 即存在可逆矩阵 $\displaystyle Q$ 使得
\begin\{aligned\} QA=\left(\begin\{array\}\{cccccccccccccccccccc\}B\_\{n\times n\}\\\\ C\_\{(m-n)\times n\}B\_\{n\times 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\}
其中 $\displaystyle B$ 可逆.令 $\displaystyle P=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&0\\\\ -CB&E\_\{m-n\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}B^\{-1\}&\\\\ &E\_\{m-n\}\end\{array\}\right)Q$, 则
\begin\{aligned\} PA=&\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&0\\\\ -CB&E\_\{m-n\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}B^\{-1\}&\\\\ &E\_\{m-n\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}B\\\\CB\end\{array\}\right)\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n&0\\\\ -CB&E\_\{m-n\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n\\\\CB\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_n\\\\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\}
(2)、 $\displaystyle (2)\Rightarrow (3)$: 设 $\displaystyle P=\left(\begin\{array\}\{cccccccccccccccccccc\}Q\_\{n\times m\}\\\\S\_\{(m-n)\times n\}\end\{array\}\right)$, 则
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E\_n\\\\0\end\{array\}\right)=PA=\left(\begin\{array\}\{cccccccccccccccccccc\}Q\\\\S\end\{array\}\right)A=\left(\begin\{array\}\{cccccccccccccccccccc\}QA\\\\SA\end\{array\}\right)\Rightarrow QA=E\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 $\displaystyle (3)\Rightarrow (4)$: 对 $\displaystyle \forall\ B\_\{n\times p\}$, $\displaystyle \mathrm\{rank\}(AB)\leq \mathrm\{rank\} B$,
\begin\{aligned\} \mathrm\{rank\} B=\mathrm\{rank\}(E\_nB)=\mathrm\{rank\}(QAB)\leq \mathrm\{rank\}(AB). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\}(AB)=\mathrm\{rank\} B$. (4)、 $\displaystyle (4)\to (1)$: 取 $\displaystyle B=E\_n$, 则
\begin\{aligned\} \mathrm\{rank\} A=\mathrm\{rank\}(A E\_n)\stackrel\{(4)\}\{=\}\mathrm\{rank\} E\_n=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(5)、 $\displaystyle (1)\Leftrightarrow (5)$:
\begin\{aligned\} &\mathrm\{rank\} A=n\Leftrightarrow Ax\_\{n\times 1\}=0\mbox\{只有零解\} \Leftrightarrow AX\_\{n\times p\}=0\mbox\{只有零解\}\\\\ \Leftrightarrow&\mbox\{若 $\displaystyle A(B-C)=0$, 则 $\displaystyle B-C=0$\}\Leftrightarrow \mbox\{若 $\displaystyle AB=AC$, 则 $\displaystyle B=C$\}. \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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---
1115、 2、 设 $\displaystyle A^\star$ 是 $\displaystyle n$ 阶 ($n\geq 2$) 复方阵 $\displaystyle A$ 的伴随矩阵, 用 $\displaystyle \mathrm\{rank\} A$ 表示 $\displaystyle A$ 的秩. 证明: (1)、 $\displaystyle \mathrm\{rank\}(A^\star)=\left\\{\begin\{array\}\{llllllllllll\}n,&\mathrm\{rank\} A=n,\\\\ 1,&\mathrm\{rank\} A=n-1,\\\\ 0,&\mathrm\{rank\} A < n-1.\end\{array\}\right.$ (2)、 证明: 若 $\displaystyle \mathrm\{rank\} A=n$, 那么存在复方阵 $\displaystyle B$ 是 $\displaystyle B^\star=A$. (合肥工业大学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)、 若 $\displaystyle \mathrm\{rank\} A=n$, 则 $\displaystyle A$ 可逆,
\begin\{aligned\} AA=|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\}
上述用 $\displaystyle kA^\star$ 代替 $\displaystyle A$ 得
\begin\{aligned\} &(kA^\star)(kA^\star)^\star=|kA^\star|E=k^n|A^\star|E=k^n |A|^\{n-1\}E\\\\ \Rightarrow&A^\star(kA^\star)^\star=(k^\{n-1\}|A|^\{n-2\})\cdot |A|E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle k=\frac\{1\}\{|A|^\frac\{n-2\}\{n-1\}\}$, 则
\begin\{aligned\} A^\star(kA^\star)^\star=|A|E=A^\star A \Rightarrow (kA^\star)^\star=A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle B=\frac\{1\}\{|A|^\frac\{n-2\}\{n-1\}\}A^\star$, 则 $\displaystyle B^\star=A$.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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---
1116、 5、 $\displaystyle A$ 为 $\displaystyle n$ 阶方阵. 证明: 存在可逆矩阵 $\displaystyle B$ 与幂等矩阵 $\displaystyle C$, 使得 $\displaystyle A=BC$. ($C$ 知满足 $\displaystyle C^2=C$ 的 $\displaystyle n$ 阶方阵.) (合肥工业大学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) / 一言以蔽之, 任意矩阵可分解为可逆矩阵与幂等矩阵的乘积. 设 $\displaystyle \mathrm\{rank\} A=r$, 则存在可逆矩阵 $\displaystyle P,Q$ 使得
\begin\{aligned\} A&=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)Q=PQ\cdot Q^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)Q. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令
\begin\{aligned\} B=PQ, C=Q^\{-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)Q, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle B$ 可逆, $\displaystyle C$ 幂等 ($C^2=C$), 且 $\displaystyle A=BC$.跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1117、 8、 求
\begin\{aligned\} A(\lambda)=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda&&&&a\_0\\\\ -1&\lambda&&&a\_1\\\\ &\ddots&\ddots&&\vdots\\\\ &&\ddots&\lambda&a\_\{n-2\}\\\\ &&&-1&\lambda+a\_\{n-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\}
的行列式因子与不变因子. (合肥工业大学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) / 按第一行展开知
\begin\{aligned\} |A(\lambda)|=&\lambda \left|\begin\{array\}\{cccccccccc\}\lambda&&&&a\_1\\\\ -1&\lambda&&&a\_2\\\\ &\ddots&\ddots&&\vdots\\\\ &&\ddots&\lambda&a\_\{n-2\}\\\\ &&&-1&\lambda+a\_\{n-1\}\end\{array\}\right|+(-1)^\{1+n\}a\_0\cdot(-1)^\{n-1\}\\\\ =&\cdots=\lambda^n+a\_\{n-1\}\lambda^\{n-1\}+\cdots+a\_1\lambda+a\_0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle A(\lambda)$ 的左下角有一个 $\displaystyle n-1$ 阶子式 $\displaystyle \left|\begin\{array\}\{cccccccccc\}-1&\lambda&&&\\\\ &\ddots&\ddots&\\\\ &&\ddots&\lambda\\\\ &&&-1\end\{array\}\right|=(-1)^\{n-1\}$ 知 $\displaystyle A(\lambda)$ 的行列式因子为
\begin\{aligned\} 1,\cdots,1,\lambda^n+a\_\{n-1\}\lambda^\{n-1\}+\cdots+a\_1\lambda+a\_0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而不变因子也是
\begin\{aligned\} 1,\cdots,1,\lambda^n+a\_\{n-1\}\lambda^\{n-1\}+\cdots+a\_1\lambda+a\_0. \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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---
1118、 4、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&1&0\\\\ 1&a&-1\\\\ 0&1&a\end\{array\}\right)$, 且 $\displaystyle A^3=0$. 求 $\displaystyle a$. 若矩阵 $\displaystyle X$ 满足
\begin\{aligned\} X-AX-XA^2+AXA^2=E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle X$. (河北工业大学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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle a,a,a$. 由 $\displaystyle A^3=0$ 即知
\begin\{aligned\} a^3=0\Rightarrow a=0\Rightarrow A=\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\}
进而
\begin\{aligned\} &\quad \ E=(E-A)X-(E-A)XA^2 \stackrel\{\mbox\{左乘\}(E-A)^\{-1\}\}\{\Rightarrow\}(E-A)^\{-1\}=X-XA^2\\\\ &\stackrel\{\mbox\{右乘\}(E-A)\}\{\Rightarrow\}E=X(E-A)-XA^2(E-A) \xlongequal\{\tiny\mbox\{题设\}\} X(E-A-A^2)\\\\ &\Rightarrow X=(E-A-A^2)^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}3&1&-2\\\\ 1&1&-1\\\\ 2&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\}
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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---
1119、 5、 设 $\displaystyle A,B$ 分别为 $\displaystyle m\times n, n\times m$ 阶矩阵, 且 $\displaystyle E\_m-AB$ 可逆. 求证: $\displaystyle E\_n-BA$ 可逆, 并求 $\displaystyle (E\_n-BA)^\{-1\}$. (河北工业大学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&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\_n-BA|=|E\_m-AB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle E\_m-AB$ 可逆蕴含 $\displaystyle E\_n-BA$ 也可逆. 进一步, 由
\begin\{aligned\} &(E\_n-BA)\left\[E\_n+B(E\_m-AB)^\{-1\}A\right\]\\\\ =&E\_n-BA+B(E\_m-AB)^\{-1\}A-BAB(E\_m-AB)^\{-1\}A\\\\ =&E\_n-BA+B\left(E\_m-AB\right)(E\_m-AB)^\{-1\}A\\\\ =&E\_n-BA+BA=E\_n \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\} (E\_n-BA)^\{-1\}=E\_n+B(E\_m-AB)^\{-1\}A. \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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---
1120、 1、 填空题. (1)、 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 的每行元素之和为 $\displaystyle 1$, 则 $\displaystyle A^\{-1\}$ 的每行元素之和为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (黑龙江大学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) / 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则
\begin\{aligned\} Ae=e\Rightarrow e=A^\{-1\}e\Rightarrow A^\{-1\} \mbox\{的每行元素之和为\}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](
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---
1121、 (3)、 设三阶矩阵 $\displaystyle A\sim \left(\begin\{array\}\{cccccccccccccccccccc\}2&0&0\\\\ 0&2&1\\\\ 0&0&2\end\{array\}\right)$, 则 $\displaystyle A$ 的最小多项式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (黑龙江大学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) / $\displaystyle (\lambda-2)^2$.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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---
1122、 (4)、 设 $\displaystyle A$ 是三阶矩阵, $\displaystyle B$ 是 $\displaystyle 2$ 阶矩阵, $\displaystyle |A|=2, |B|=1$, 则 $\displaystyle \left|\begin\{array\}\{cccccccccc\}0&2A\\\\ B&0\end\{array\}\right|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (黑龙江大学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) / 将 $\displaystyle A$ 的第 $\displaystyle i$ 列向左移动 $\displaystyle 2$ 列, $\displaystyle i=1,2,3$, 得
\begin\{aligned\} \mbox\{原式\}=(-1)^\{2\times 3\}\left|\begin\{array\}\{cccccccccc\}2A&\\\\ &B\end\{array\}\right|=2^3\left|\begin\{array\}\{cccccccccc\}A&\\\\ &B\end\{array\}\right|=2^3\cdot 2\cdot 1=16. \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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1123、 2、 计算题. (1)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&3\\\\ 0&2&1\\\\ 0&0&2\end\{array\}\right)$, 计算 $\displaystyle A^\{100\}$. (黑龙江大学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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2$. 由 $\displaystyle \mathrm\{rank\}(A-2E)=2$ 在 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 $\displaystyle \mathrm\{rank\}(J-2E)=2$. 这表明 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&\\\\ &2&1\\\\ &&2\end\{array\}\right)$, 且存在可逆矩阵 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)$, 使得 $\displaystyle P^\{-1\}AP=J$, 即
\begin\{aligned\} (A-2E)\alpha\_1=0, (A-2E)\alpha\_2=\alpha-1, (A-2E)\alpha\_3=\alpha\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle A-2E\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ 0&0&1\\\\ 0&0&0\end\{array\}\right)$ 知可取 $\displaystyle \alpha\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\0\end\{array\}\right)$. 又由
\begin\{aligned\} (A-2E,\alpha\_1)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&1\\\\ 0&0&1&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\}
知可取 $\displaystyle \alpha\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\\\\0\end\{array\}\right)$. 再由
\begin\{aligned\} (A-2E,\alpha\_2)\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&-3\\\\ 0&0&1&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\}
知可取 $\displaystyle \alpha\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\-3\\\\1\end\{array\}\right)$. 故
\begin\{aligned\} P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&-3\\\\ 0&0&1\end\{array\}\right)\Rightarrow A=PJP^\{-1\} \Rightarrow A^n=PJ^n P^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle N=J-2E$, 则
\begin\{aligned\} J^n&=(2E+N)^n=2^nE+n 2^\{n-1\}N+C\_n^2 2^\{n-2\}N^2\\\\ &=\left(\begin\{array\}\{cccccccccccccccccccc\}2^n&n2^\{n-1\}&n(n-1)2^\{n-3\}\\\\ 0&2^n&n2^\{n-1\}\\\\ 0&0&2^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\}
于是
\begin\{aligned\} &A^n=PJ^nP^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}2^n&n2^\{n-1\}&n(n+11)2^\{n-3\}\\\\ 0&2^n&n2^\{n-1\}\\\\ 0&0&2^n\end\{array\}\right)\\\\ \Rightarrow& A^\{100\}=\left(\begin\{array\}\{cccccccccccccccccccc\}2^\{100\}&100\cdot 2^\{99\}&11100\cdot 2^\{97\}\\\\ 0&2^\{100\}&100\cdot 2^\{99\}\\\\ 0&0&2^\{100\}\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\}
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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1124、 3、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&0\\\\ 2&0&2\\\\ 0&2&1\end\{array\}\right)$. 求正交矩阵 $\displaystyle Q$, 使得 $\displaystyle Q^\mathrm\{T\} AQ$ 为对角矩阵. [$A$ 的特征值带根号. 回忆版有点问题. 考研试题一般不会出现这种情况. 如果你有原版, 请告知我, 我修改就是.] (黑龙江大学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) / $\displaystyle A$ 的特征值为 $\displaystyle \frac\{1\pm \sqrt\{33\}\}\{2\},1$. 感觉回忆版有点问题. 考研试题一般不会出现这种情况. 如果你有原版, 请告知我, 我修改就是.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1125、 5、 证明题. (1)、 证明: $\displaystyle n$ 阶矩阵 $\displaystyle A$ 既可以表示成幂等矩阵与可逆矩阵之积, 也可表示成对称矩阵与可逆矩阵之积. (黑龙江大学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) / 设 $\displaystyle \mathrm\{rank\} A=r$, 则存在可逆矩阵 $\displaystyle P,Q$ 使得
\begin\{aligned\} A=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&0\\\\ 0&0\end\{array\}\right)Q. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令
\begin\{aligned\} B=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&0\\\\ 0&0\end\{array\}\right)P^\{-1\}, C=PQ, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle B$ 幂等, $\displaystyle C$ 可逆, 且 $\displaystyle A=BC$. 再令
\begin\{aligned\} B=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&0\\\\ 0&0\end\{array\}\right)P^\mathrm\{T\}, C=(P^\mathrm\{T\})^\{-1\}Q, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle B$ 实对称, $\displaystyle C$ 可逆, 且 $\displaystyle A=BC$.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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---
1126、 (3)、 证明: 实对称矩阵的特征值为实数. (黑龙江大学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) / 设 $\displaystyle \lambda\in\mathbb\{C\}$ 是实对称矩阵 $\displaystyle A$ 的特征值, $\displaystyle 0\neq\alpha\in\mathbb\{C\}^n$ 是对应的特征向量, 则 $\displaystyle A\alpha=\lambda \alpha$, 而
\begin\{aligned\} \bar\{\alpha\}^\mathrm\{T\} A\alpha=\left\\{\begin\{array\}\{llllllllllll\}\bar\{\alpha\}^\mathrm\{T\} (\lambda \alpha)=\lambda |\alpha|^2,\\\\ =(\bar\{\alpha\}^\mathrm\{T\} A\alpha)^\mathrm\{T\} =\alpha^\mathrm\{T\} A^\mathrm\{T\} \bar\{\alpha\} =\alpha^\mathrm\{T\} A\bar\{\alpha\} =\alpha^\mathrm\{T\} \overline\{A\alpha\} =\bar\{\lambda\}|\alpha|^2.\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 $\displaystyle (\lambda-\bar\{\lambda\})|\alpha|^2=0\Rightarrow \lambda-\bar\{\lambda\}=0\Rightarrow\lambda\in\mathbb\{R\}$.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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---
1127、 9、 $\displaystyle A,B$ 是实对称矩阵, $\displaystyle A$ 的特征值全大于 $\displaystyle 1$, $\displaystyle B$ 的特征值全大于 $\displaystyle 2$. 证明: $\displaystyle A+B$ 的特征值全大于 $\displaystyle 3$. (黑龙江大学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 $\displaystyle A$ 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\_n), \quad \lambda\_i > 1\\\\ \Rightarrow&P^\mathrm\{T\} (A-E)P=\mathrm\{diag\}(\lambda\_1-1,\cdots,\lambda\_n-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A-E$ 正定. 同理, $\displaystyle B-2E$ 正定. 于是 $\displaystyle (A+B)-3E=(A-E)+(B-2E)$ 正定, 它的特征值全大于 $\displaystyle 0$. 这表明 $\displaystyle A+B$ 的特征值全大于 $\displaystyle 3$.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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发表于 2023-3-5 13:09:54
