## 张祖锦2023年数学专业真题分类70天之第50天
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1128、 6、 设 $\displaystyle A=(a\_\{ij\})\_\{3\times 3\}$ 是一个 $\displaystyle 3$ 阶实对称矩阵, 若 $\displaystyle A$ 的迹
\begin\{aligned\} \mathrm\{tr\} A=a\_\{11\}+a\_\{22\}+a\_\{33\}=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且向量 $\displaystyle \alpha\_1=(1,1,1)^\mathrm\{T\}$ 与 $\displaystyle \alpha\_2=(1,1,-1)^\mathrm\{T\}$ 是 $\displaystyle A$ 的属于特征值 $\displaystyle 1$ 的特征向量. 求矩阵 $\displaystyle A$. (湖南大学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) / 由题设, $\displaystyle A\sim \mathrm\{diag\}(1,1,\lambda)$. 进而 $\displaystyle 0=\mathrm\{tr\} A=2+\lambda\Rightarrow \lambda=-2$. 设 $\displaystyle x$ 为 $\displaystyle A$ 的属于特征值 $\displaystyle -2$ 的特征向量, 则
\begin\{aligned\} \alpha\_i^\mathrm\{T\} x=0, i=1,2\Rightarrow x=k(-1,1,0)^\mathrm\{T\}, k\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故取
\begin\{aligned\} &\alpha\_3=(-1,1,0)^\mathrm\{T\}, P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-1\\\\ 1&1&1\\\\ 1&-1&0\end\{array\}\right)\\\\ \Rightarrow&P^\{-1\}AP=\mathrm\{diag\}(1,1,-2)\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{2\}&\frac\{3\}\{2\}&0\\\\ \frac\{3\}\{2\}&-\frac\{1\}\{2\}&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\}
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1129、 7、 设 $\displaystyle A\in \mathbb\{K\}^\{n\times n\}$ 是数域 $\displaystyle \mathbb\{K\}$ 上的 $\displaystyle n$ 阶矩阵, 按递归方式定义 $\displaystyle A$ 的 $\displaystyle m$ 重伴随矩阵 $\displaystyle A^\{(m\star)\}$ 如下:
\begin\{aligned\} A^\{(0\star)\}=A, A^\{(1\star)\}=A^\star, A^\{(2\star)\}=A^\{\star\star\}, \cdots, A^\{(m\star)\}=\left(A^\{\left((m-1)\star\right)\}\right)^\star, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即 $\displaystyle A^\{(m\star)\}=\underbrace\{\left(\cdots \left(\left(A^\star\right)^\star\right)\cdots\right)^\star\}\_\{m\mbox\{重\}\}$. 证明: (1)、 $\displaystyle A^\{\star\star\}=|A|^\{n-2\}A$; (2)、
\begin\{aligned\} A^\{(m\star)\}=\left\\{\begin\{array\}\{llllllllllll\}|A|^\frac\{(n-1)^m-1\}\{n\}A,&\mbox\{当 $\displaystyle m$ 为偶数\},\\\\ |A|^\frac\{(n-1)^m-n+1\}\{n\}A^\star,&\mbox\{当 $\displaystyle m$ 为奇数\}.\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\}
(湖南大学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)、 当 $\displaystyle n=2$ 时, 直接验证知 $\displaystyle A^\{\star\star\}=A=|A|^\{n-2\}A$. 当 $\displaystyle n\geq 3$ 时, 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,
\begin\{aligned\} |A|=0&\Rightarrow \mathrm\{rank\}(A^\star)\leq 1\left(\leq n-2\right)\\\\ &\Rightarrow (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\}
(2)、 用数学归纳法证明
\begin\{aligned\} A^\{(2k\star)\}=|A|^\frac\{(n-1)^\{2k\}-1\}\{n\}A.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle k=1$ 时, 由第 1 步知结论成立. 设结论对 $\displaystyle k$ 成立, 则
\begin\{aligned\} A^\{(2(k+1)\star)\}=&\left\[A^\{(2k\star)\}\right\]^\{\star\star\} \xlongequal\{\tiny\mbox\{归纳假设\}\} \left(|A|^\frac\{(n-1)^\{2k\}-1\}\{n\}A\right)^\{\star\star\}\\\\ \overset\{\tiny\mbox\{第1步\}\}\{=\}&\left||A|^\frac\{(n-1)^\{2k\}-1\}\{n\}A\right|^\{n-2\}\cdot |A|^\frac\{(n-1)^\{2k\}-1\}\{n\}A\\\\ =&\left(|A|^\{(n-1)^\{2k\}-1\}|A|\right)^\{n-2\}|A|^\frac\{(n-1)^\{2k\}-1\}\{n\}A\\\\ =&|A|^\{(n-1)^\{2k\}(n-2)+\frac\{(n-1)^\{2k\}-1\}\{n\}\}A\\\\ =&|A|^\frac\{(n-1)^\{2k\}[n(n-2)+1]-1\}\{n\}A =|A|^\frac\{(n-1)^\{2(k+1)\}-1\}\{n\}A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
归纳步证毕, 而确有 $\displaystyle (I)$ 成立. 也用数学归纳法证明
\begin\{aligned\} |A|^\{((2k-1)\star)\}=|A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star.\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 $\displaystyle k=1$ 时结论自明. 设结论对 $\displaystyle k$ 成立, 则
\begin\{aligned\} &|A|^\{((2k+1)\star)\} =\left(|A|^\{((2k-1)\star)\}\right)^\{\star\star\} \xlongequal\{\tiny\mbox\{归纳假设\}\} \left(|A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star\right)^\{\star\star\}\\\\ \overset\{\tiny\mbox\{第1步\}\}\{=\}&\left||A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star\right|^\{n-2\}\cdot |A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star\\\\ =&\left(|A|^\{(n-1)^\{2k-1\}-n+1\}|A^\star|\right)^\{n-2\}\cdot |A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star\\\\ =&\left(|A|^\{(n-1)^\{2k-1\}-n+1\}|A|^\{n-1\}\right)^\{n-2\}\cdot |A|^\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}A^\star. \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\} &\left\[(n-1)^\{2k-1\}-n+1+n-1\right\] (n-2)+\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}\\\\ =&(n-1)^\{2k-1\}(n-2)+\frac\{(n-1)^\{2k-1\}-n+1\}\{n\}\\\\ =&\frac\{(n-1)^\{2k-1\}[n(n-2)+1]-n+1\}\{n\} =\frac\{(n-1)^\{2k1\}-n+1\}\{n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知归纳步证毕, 而 $\displaystyle (II)$ 确实成立.跟锦数学微信公众号. [在线资料](
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1130、 (8)、 若方阵 $\displaystyle A$ 的初等因子组为
\begin\{aligned\} \lambda,\lambda,\lambda^2, \lambda+1, (\lambda+1)^2,(\lambda-1)^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle A$ 的极小多项式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / $\displaystyle A$ 的极小多项式就是 $\displaystyle A$ 的最后一个不变因子 $\displaystyle \lambda^2(\lambda+1)^2(\lambda-1)^2$.跟锦数学微信公众号. [在线资料](
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1131、 (9)、 实对称矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&0\\\\ -2&1&-2\\\\ 0&-2&0\end\{array\}\right)$ 可通过正交相似变换化为对角阵 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华东师范大学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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 4,1,-2$, 而 $\displaystyle A\sim \mathrm\{diag\}(4,1,-2)$.跟锦数学微信公众号. [在线资料](
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1132、 (4)、 证明对任何可逆复矩阵 $\displaystyle A\in GL\_n(\mathbb\{C\})$ 以及任意正整数 $\displaystyle k$, 矩阵方程 $\displaystyle X^k=A$ 一定有解. (华东师范大学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) / 设 $\displaystyle \lambda$ 是非零复数, $\displaystyle k$ 为正整数, $\displaystyle J\_n(\lambda)$ 表示特征值为 $\displaystyle \lambda$ 的 $\displaystyle n$ 阶若当块. (4-1)、 令
\begin\{aligned\} C=\left(\begin\{array\}\{cccccccccccccccccccc\} 0&1&&\\\\ &\ddots&\ddots&\\\\ &&\ddots&1\\\\ &&&0\end\{array\}\right)\_n, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} (J\_n(\lambda))^k&=(\lambda E\_n+C)^k\\\\ &=\lambda^k I\_n+C\_k^1 \lambda^\{k-1\} C +C^2 \lambda^\{k-2\}C^2 +\cdots\\\\ &\quad +C\_k^\{\min\left\\{k,n-1\right\\}\}\lambda^\{k-\min\left\\{k,n-1\right\\}\} C^\{\min\left\\{k,n-1\right\\}\}\\\\ &=\left(\begin\{array\}\{cccccccccccccccccccc\} \lambda^k&k \lambda^\{k-1\}&C\_k^2 \lambda^\{k-2\}&\cdots&\\\\ &\lambda^k&\ddots&\ddots&\vdots\\\\ &&\ddots&\ddots&C\_k^2\lambda^\{k-2\}\\\\ &&&\ddots&k \lambda^\{k-1\}\\\\ &&&&\lambda^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 $\displaystyle (J\_n(\lambda))^k$ 的特征值为 $\displaystyle \lambda^k$, 且
\begin\{aligned\} \mathrm\{rank\}(\lambda^k I\_n-(J\_n(\lambda))^k)=n-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle (J\_n(\lambda))^k$ 的 Jordan 标准形 $\displaystyle J$ 满足
\begin\{aligned\} \mathrm\{rank\}(\lambda^k I\_n-J)=n-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle J-\lambda^kI\_n$ 的对角线上面有 $\displaystyle n-1$ 个 $\displaystyle 1$, 即
\begin\{aligned\} J=\left(\begin\{array\}\{cccccccccccccccccccc\} \lambda^k&1&&\\\\ &\ddots&\ddots&\\\\ &&\ddots&1\\\\ &&&\lambda^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\}
(4-2)、 由第 i 步, $\displaystyle (J\_n(\sqrt[k]\{\lambda\}))^k$ 的 Jordan 标准形为 $\displaystyle J\_n(\lambda)$, 而存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}(J\_n(\sqrt[k]\{\lambda\}))^kP=J\_n(\lambda). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle B=P^\{-1\}J\_n(\sqrt[k]\{\lambda\})P$, 则 $\displaystyle B^k=J\_n(\lambda)$. (4-3)、 由 Jordan 标准形理论, 存在可逆阵 $\displaystyle P$ 使得
\begin\{aligned\} A=P^\{-1\}JP, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle J=\mathrm\{diag\}(J\_1,\cdots,J\_s)$ 是 $\displaystyle A$ 的 Jordan 标准形, 对每一个 Jordan 块 $\displaystyle J\_i$, 由第 2 步, 存在 $\displaystyle B\_i$, 使得 $\displaystyle B\_i^k=J\_i$. 令
\begin\{aligned\} X=P^\{-1\}\mathrm\{diag\}(J\_1,\cdots,J\_s)P, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle X^k=A$.跟锦数学微信公众号. [在线资料](
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1133、 8、 设 $\displaystyle A,B$ 均为正交矩阵, $\displaystyle |A|=-1, |B|=1$. (1)、 证明: $\displaystyle -1$ 为 $\displaystyle A$ 的特征值; (2)、 证明: $\displaystyle |A+B|=0$. (华南理工大学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)、 由
\begin\{aligned\} &|-E-A|=-[-|-E-A|]=-|A^\mathrm\{T\}|\cdot |-E-A|\\\\ =&-|-A^\mathrm\{T\}-E| \stackrel\{\mbox\{转置\}\}\{=\}-|-E-A| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle |-E-A|=0$, 而 $\displaystyle -1$ 为 $\displaystyle A$ 的特征值. (2)、 由
\begin\{aligned\} -|A+B|=&|A^\mathrm\{T\}|\cdot |A+B| =|E+A^\mathrm\{T\} B|\\\\ =&|(B^\mathrm\{T\}+A^\mathrm\{T\})B|=|B^\mathrm\{T\}+A^\mathrm\{T\}|\cdot |B|=|A+B| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知 $\displaystyle |A+B|=0$.跟锦数学微信公众号. [在线资料](
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1134、 (2)、 $\displaystyle A=(a\_\{ij\})$ 为三阶矩阵, 其特征值为 $\displaystyle 1,-1,-\frac\{1\}\{3\}$, 则 $\displaystyle \det(3A^\star-4A^\{-1\})=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华南师范大学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) / $\displaystyle |A|=\frac\{1\}\{3\}$, 而
\begin\{aligned\} \mbox\{原式\}=&\left|3|A|A^\{-1\}-4A^\{-1\}\right| =|A^\{-1\}-4A^\{-1\}|=|-3A^\{-1\}|\\\\ =&(-3)^3|A|^\{-1\}=-27\cdot 3=-81. \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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1135、 2、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ ($n\geq 2$) 级幂零矩阵, 求下列矩阵的秩:
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}A^\{n-1\}&A^\{n-2\}&\cdots&A&E\\\\ &\ddots&\ddots&&A\\\\ &&\ddots&\ddots&\vdots\\\\ &&&\ddots&A^\{n-2\}\\\\ &&&&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\}
(华中科技大学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) / 第 $\displaystyle n$ 列 $\displaystyle \cdot (-A^\{n-i\})$ 加到第 $\displaystyle i$ 列, $\displaystyle 1\leq i\leq n-1$, 得
\begin\{aligned\} M\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&\cdots&0&E\\\\ 0&\cdots&0&A\\\\ \vdots&&\vdots&\vdots\\\\ 0&\cdots&0&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\}
第 $\displaystyle 1$ 行 $\displaystyle \cdot(-A^\{i-1\})$ 加到第 $\displaystyle i$ 行, $\displaystyle 2\leq i\leq n$ 得
\begin\{aligned\} M\to \left(\begin\{array\}\{cccccccccccccccccccc\}0&\cdots&0&E\\\\ 0&\cdots&0&0\\\\ \vdots&&\vdots&\vdots\\\\ 0&\cdots&0&0\end\{array\}\right)\Rightarrow \mathrm\{rank\} M=n. \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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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1136、 3、 (20 分) 已知矩阵 $\displaystyle A$ 满足 $\displaystyle A^3=A^2+A+2E$, 且 $\displaystyle A$ 在实数域上不可对角化. 矩阵 $\displaystyle B=xA^2+yA+E$, 其中 $\displaystyle x,y$ 为实数. 回答下面问题: (1)、 求 $\displaystyle x,y$ 满足的关系, 使其等价于 $\displaystyle B$ 可逆; (2)、 当 $\displaystyle x=-1, y=1$ 时, 求次数最低的多项式 $\displaystyle f(x)$ 使得 $\displaystyle Bf(A)=E$. (华中科技大学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 $\displaystyle f(x)=x^3-x^2-x-2=(x-2)(x^2+x+1)$. 则 $\displaystyle A$ 的特征值 $\displaystyle \in \left\\{2, \mathrm\{e\}^\frac\{2\pi\mathrm\{ i\}\}\{3\}, \mathrm\{e\}^\{-\frac\{2\pi \mathrm\{ i\}\}\{3\}\}\right\\}$. 由 $\displaystyle A$ 在实数域上不可对角化知 $\displaystyle \mathrm\{e\}^\{2\pi\mathrm\{ i\}\}\{3\}, \mathrm\{e\}^\{-\frac\{2\pi \mathrm\{ i\}\}\{3\}\}$ 是 $\displaystyle A$ 的特征值 (若不然, $\displaystyle A$ 的零化多项式 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{C\}$ 中没有重根, 而可对角化, 只能相似与 $\displaystyle 2E\_n\Rightarrow A=2E\_n$, 矛盾). 这就证明了 $\displaystyle A$ 的特征值为
\begin\{aligned\} \left\\{\lambda\_1,\lambda\_2,\lambda\_3\right\\}=\left\\{2, \mathrm\{e\}^\{2\pi\mathrm\{ i\}\}\{3\}, \mathrm\{e\}^\{-\frac\{2\pi \mathrm\{ i\}\}\{3\}\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} B\mbox\{可逆\}\Leftrightarrow&B\mbox\{的特征值\}x\lambda\_i^2+y\lambda\_i+1\neq 0, \forall\ 1\leq i\leq 3\\\\ \Leftrightarrow&\left\\{\begin\{array\}\{llllllllllll\} 4x+2y+1\neq 0\\\\ x\mathrm\{e\}^\{-\frac\{2\pi\mathrm\{ i\}\}\{3\}\}+y\mathrm\{e\}^\frac\{2\pi \mathrm\{ i\}\}\{3\}+1\neq 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)、 当 $\displaystyle x=-1, y=1$ 时, 设
\begin\{aligned\} g(x)=x^3-x^2-x-2, h(x)=-x^2+x+1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} g(x)=-xh(x)-2\Rightarrow& 0=g(A)=-Ah(A)-2=-AB-2E\\\\ \Rightarrow& B\frac\{A\}\{-2\}=E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle f(x)=-\frac\{x\}\{2\}$, 则 $\displaystyle Bf(A)=E$. 往用反证法证明 $\displaystyle f(x)$ 就是所求. 若不然, 还存在次数为 $\displaystyle 0$ 的多项式 $\displaystyle c$, 使得
\begin\{aligned\} B\cdot cE=E\Rightarrow B=\frac\{1\}\{c\}E\Rightarrow -A^2+A+E=\frac\{1\}\{c\}E. \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 $\displaystyle A$ 的互异特征值至多有两个. 与第 1 步的结果矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
1137、 4、 (20 分) 已知 $\displaystyle a$ 为实对称矩阵, $\displaystyle d > 0$ 且方程组 $\displaystyle AX=\alpha$ 无解. 证明: $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&d\end\{array\}\right)$ 为不定矩阵. (华中科技大学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&-\frac\{\alpha\}\{d\}\\\\ 0&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&d\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -\frac\{\alpha^\mathrm\{T\}\}\{d\}&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}&\\\\ &d\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\}
即 $\displaystyle e\_\{n+1\}^\mathrm\{T\} \left(\begin\{array\}\{cccccccccccccccccccc\}A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}&\\\\ &d\end\{array\}\right)e\_\{n+1\}=d > 0$ 知为证 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&d\end\{array\}\right)$ 不定, 仅需证明
\begin\{aligned\} &\exists\ 0\neq x\_0\in\mathbb\{R\}^n,\mathrm\{ s.t.\} x\_0^\mathrm\{T\} \left(A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}\right)x\_0 < 0\qquad(I)\\\\ \Rightarrow&(x\_0^\mathrm\{T\},0)\left(\begin\{array\}\{cccccccccccccccccccc\}A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}&\\\\ &d\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}x\_0\\\\0\end\{array\}\right) < 0\\\\ \Rightarrow&\left(\begin\{array\}\{cccccccccccccccccccc\}A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}&\\\\ &d\end\{array\}\right)\mbox\{不定\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 由 $\displaystyle Ax=\alpha$ 无解知 (都用反证法证明)
\begin\{aligned\} &\mathrm\{rank\} A=r < n, \alpha\neq 0,\\\\ &\mathrm\{rank\} A < \mathrm\{rank\}(A,\alpha)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A^\mathrm\{T\}\\\\\alpha^\mathrm\{T\}\end\{array\}\right)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\\alpha^\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\}
设 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r$ 是 $\displaystyle A$ 的列向量组的一个极大无关组, 则由 $\displaystyle A^\mathrm\{T\} =A$ 知 $\displaystyle \varepsilon\_1^\mathrm\{T\},\cdots,\varepsilon\_r^\mathrm\{T\}$ 是 $\displaystyle A$ 的行向量组的一个极大无关组, 且 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r,\alpha$ 线性无关. 将 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_r,\alpha$ 标准正交化为 $\displaystyle \eta\_1,\cdots,\eta\_r,\eta\_\{r+1\}$, 则
\begin\{aligned\} L(\varepsilon\_1,\cdots,\varepsilon\_i)=&L(\eta\_1,\cdots,\eta\_i), 1\leq i\leq r,\\\\ L(\varepsilon\_1,\cdots,\varepsilon\_r,\alpha)=&L(\eta\_1,\cdots,\eta\_r,\eta\_\{r+1\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle x\_0=\eta\_\{r+1\}\neq 0$, 则
\begin\{aligned\} &x\_0=\eta\_\{r+1\}\perp \eta\_i\left(1\leq i\leq r\right) \Rightarrow x\_0\perp \varepsilon\_i\left(1\leq i\leq r\right)\\\\ \Leftrightarrow&\varepsilon\_i^\mathrm\{T\} x\_0=0\left(1\leq i\leq r\right)\Leftrightarrow Ax\_0=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此外, 还有
\begin\{aligned\} \alpha=\sum\_\{i=1\}^r x\_i\eta\_i+x\_\{r+1\}\eta\_\{r+1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle x\_\{r+1\}\neq 0$ (否则, $\displaystyle \alpha\in L(\eta\_1,\cdots,\eta\_r)=L(\varepsilon\_1,\cdots,\varepsilon\_r)$, 矛盾). 从而
\begin\{aligned\} \alpha^\mathrm\{T\} x\_0=\left(\sum\_\{i=1\}^r x\_i\eta\_i+x\_\{r+1\}\eta\_\{r+1\}\right)^\mathrm\{T\} \eta\_\{r+1\} =x\_\{r+1\}\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们最终得到 $\displaystyle x\_0^\mathrm\{T\} \left(A-\frac\{\alpha\alpha^\mathrm\{T\}\}\{d\}\right)x\_0 =-\frac\{|\alpha^\mathrm\{T\} x\_0|^2\}\{d\} < 0$, $\displaystyle (I)$ 得证.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)/
---
1138、 7、 (20 分) 设矩阵 $\displaystyle A$ 的最小多项式为 $\displaystyle m(\lambda)$, 且 $\displaystyle m(\lambda)\mid h(\lambda)$, 其中 $\displaystyle h(\lambda)=[f(\lambda),g(\lambda)]$. 证明:
\begin\{aligned\} \mathrm\{rank\} f(A)+\mathrm\{rank\} g(A)\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中科技大学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\} &m(\lambda)\mid h(\lambda)\Rightarrow \exists\ k(\lambda),\mathrm\{ s.t.\} h(\lambda)=k(\lambda)m(\lambda),\\\\ &h(\lambda)=\left\[f(\lambda),g(\lambda)\right\]=\frac\{f(\lambda)g(\lambda)\}\{\left(f(\lambda),g(\lambda)\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\} &f(\lambda)g(\lambda)=\left(f(\lambda),g(\lambda)\right)h(\lambda) =\left(f(\lambda),g(\lambda)\right)k(\lambda)m(\lambda)\\\\ \Rightarrow&f(A)g(A)=\left(f(A),g(A)\right)k(A)m(A)=0\\\\ \Rightarrow&\mathrm\{rank\} f(A)+\mathrm\{rank\} g(A)\leq n. \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)/
---
1139、 (2)、 三阶矩阵 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&2&1\\\\ 0&0&3\end\{array\}\right)$ 的逆矩阵为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华中师范大学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) / $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&-\frac\{1\}\{2\}&-\frac\{1\}\{6\}\\\\ 0&\frac\{1\}\{2\}&-\frac\{1\}\{6\}\\\\ 0&0&\frac\{1\}\{3\}\end\{array\}\right)$.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)/
---
1140、 (3)、 设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&-2&-1\\\\ 1&2&1\\\\ 1&2&1\end\{array\}\right)$, 那么 $\displaystyle A^\{2022\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华中师范大学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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,0,0$. 由
\begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1\\\\ 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 $\displaystyle 2,0$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\1 \end\{array\}\right); \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\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)=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&-2&-1\\\\ 1&1&0\\\\ 1&0&1 \end\{array\}\right)\Rightarrow P^\{-1\}AP=\mathrm\{diag\}\left(2,0,0\right)\\\\ \Rightarrow& A^\{2022\}=P\mathrm\{diag\}(2^\{2022\},0,0)P^\{-1\} =\left(\begin\{array\}\{cccccccccccccccccccc\}-2^\{2021\}&-2^\{2022\}&-2^\{2021\}\\\\ 2^\{2021\}&2^\{2022\}&2^\{2021\}\\\\ 2^\{2021\}&2^\{2022\}&2^\{2021\}\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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---
1141、 (4)、 实对称矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&-1\\\\ 1&0&1\\\\ -1&1&0\end\{array\}\right)$ 的正惯性指数是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华中师范大学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) / $\displaystyle A$ 的特征值为 $\displaystyle 1,1,-2$, 而正惯性指数为 $\displaystyle 2$.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1142、 (6)、 二阶 $\displaystyle \lambda$-矩阵 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda^2-3\lambda+2&0\\\\ 0&\lambda^2-4\lambda+3\end\{array\}\right)$ 的等价标准形为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (华中师范大学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) / 由
\begin\{aligned\}\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda^2-3\lambda+2&0\\\\ 0&\lambda^2-4\lambda+3\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}(\lambda-1)(\lambda-2)&\\\\ &(\lambda-1)(\lambda-3)\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\}
知行列式因子为 $\displaystyle \lambda-1,(\lambda-1)^2(\lambda-2)(\lambda-3)$, 而不变因子为 $\displaystyle \lambda-1,(\lambda-1)(\lambda-2)(\lambda-3)$, 等价标准形为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-1&\\\\ &(\lambda-1)(\lambda-2)(\lambda-3)\end\{array\}\right)$.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)/
---
1143、 3、 (20 分) 设 $\displaystyle A,B$ 都是 $\displaystyle n$ 阶实对称矩阵, $\displaystyle n\geq 2$. 证明: $\displaystyle B^3AB^3$ 正定当且仅当仅当 $\displaystyle A$ 正定, $\displaystyle B$ 可逆. (吉林大学2023年高等代数与解析几何考研试题) [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)、 $\displaystyle \Leftarrow$: $\displaystyle B^3AB^3=(B^3)^\mathrm\{T\} AB$ 与正定矩阵 $\displaystyle A$ 合同, 而正定. (2)、 $\displaystyle \Rightarrow$: 先用反证法证明 $\displaystyle B$ 可逆. 若 $\displaystyle B$ 不可逆, 则 $\displaystyle Bx=0$ 有非零解 $\displaystyle \alpha\neq 0$. 从而
\begin\{aligned\} \alpha^\mathrm\{T\} B^3AB^3\alpha=(B^3\alpha)^\mathrm\{T\} A(B^3\alpha)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle B^3AB^3$ 正定矛盾. 故有结论. 进步, 实对称矩阵 $\displaystyle A$ 与正定矩阵 $\displaystyle B^3AB^3=(B^3)^\mathrm\{T\} AB^3$ 合同, 而正定.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)/
---
1144、 5、 (20 分) 设 $\displaystyle A$ 是 $\displaystyle 3n$ 阶复矩阵, $\displaystyle n\geq 2$, $\displaystyle \mu(x)$ 是 $\displaystyle A$ 的极小多项式, $\displaystyle f(x)$ 是 $\displaystyle A$ 的特征多项式, 且满足 $\displaystyle f(x)=\mu^3(x)$. 设 $\displaystyle A$ 的全体互异特征值的个数为 $\displaystyle k$, $\displaystyle A$ 的线性无关的特征向量的最大个数为 $\displaystyle s$. (1)、 证明: $\displaystyle A$ 可对角化当且仅当 $\displaystyle k=n$; (2)、 证明: $\displaystyle 3k\leq s\leq 2n+k$. (吉林大学2023年高等代数与解析几何考研试题) [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 $\displaystyle \mu(x)=\prod\_\{i=1\}^k (\lambda-\lambda\_i)^\{n\_i\}, n\_i\geq 1$, 则 $\displaystyle f(x)=\prod\_\{i=1\}^k(\lambda-\lambda\_i)^\{3n\_i\}$. 从而
\begin\{aligned\} \sum\_\{i=1\}^k 3n\_i=3n\Rightarrow \sum\_\{i=1\}^k n\_i=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 $\displaystyle A$ 可对角化等价于 $\displaystyle A$ 的最小多项式 $\displaystyle \mu(x)$ 没有重根 $\displaystyle \Leftrightarrow n\_i=1\Leftrightarrow k=n$. 最后一个 $\displaystyle \Leftrightarrow$ 的理由如下. (1-1)、 $\displaystyle \Leftarrow$: 设 $\displaystyle k=n$, 则
\begin\{aligned\} \sum\_\{i=1\}^n n\_i=n\Rightarrow n=\sum\_\{i=1\}^n n\_i\geq \sum\_\{i=1\}^n 1=n \Rightarrow n\_i=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 $\displaystyle \Rightarrow$: 设 $\displaystyle n\_i=1$, 则
\begin\{aligned\} n=\sum\_\{i=1\}^k n\_i=\sum\_\{i=1\}^k 1=k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-3)、 由 $\displaystyle \mu(x)$ 的形式知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 中含有 Jordan 块 $\displaystyle J\_\{n\_i\}(\lambda\_i), 1\leq i\leq k$, 且所有以 $\displaystyle \lambda\_i$ 为对角元的 Jordan 块的阶数 $\displaystyle \leq n\_i$. 再由 $\displaystyle f(x)$ 的形式知以 $\displaystyle \lambda\_i$ 为对角元的 Jordan 块的阶数和为 $\displaystyle 3n\_i$. 于是可设
\begin\{aligned\} J=&\mathrm\{diag\}\left(J\_\{n\_1\}(\lambda\_1), \cdots, J\_\{n\_k\}(\lambda\_k), J\_\{m\_\{11\}\}(\lambda\_1), \cdots, J\_\{m\_\{1t\_1\}\}(\lambda\_1),\right.\\\\ &\left.\cdots, J\_\{m\_\{k1\}\}(\lambda\_k), \cdots, J\_\{m\_\{kt\_k\}\}(\lambda\_k)\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
\begin\{aligned\} 1\leq m\_\{ij\}\leq n\_i, t\_i\geq 0, n\_i+\sum\_\{j=1\}^\{t\_i\}m\_\{ij\}=3n\_i.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到每个 $\displaystyle J\_\{n\_i\}(\lambda\_i), J\_\{m\_\{ij\}\}(\lambda\_i)$ 对应于 $\displaystyle A$ 的一个特征值, 我们知
\begin\{aligned\} s=k+\sum\_\{i=1\}^k t\_i \leq k+\sum\_\{i=1\}^k \sum\_\{j=1\}^\{t\_i\}m\_\{ij\}=k+2n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (I)$ 知以 $\displaystyle \lambda\_i$ 为对角元的 Jordan 块有
\begin\{aligned\} J\_\{n\_1\}(\lambda\_i), J\_\{m\_\{i1\}\}(\lambda\_i), \cdots, J\_\{m\_\{it\_i\}\}(\lambda\_i). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为使 $\displaystyle s$ 最小, 应让 Jordan 块的个数尽可能少, 由 $\displaystyle \sum\_\{j=1\}^\{t\_i\}m\_\{ij\}=2n\_i$ 知应让 Jordan 块的阶数尽可能大. 注意到以 $\displaystyle \lambda\_i$ 为对角元的 Jordan 块的阶数最大为 $\displaystyle n\_i$, 而 $\displaystyle s$ 最小时, 各 $\displaystyle m\_\{ij\}=n\_i$. 此时各 $\displaystyle t\_i=2$, $\displaystyle s=k+\sum\_\{i=1\}^k t\_i=3k$.跟锦数学微信公众号. [在线资料](
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---
1145、 3、 已知矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&-3\\\\ -1&4&-3\\\\ 1&k&5\end\{array\}\right)$ 有一个二重特征值, 求 $\displaystyle k$ 的值并回答 $\displaystyle A$ 能否对角化? (暨南大学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\} &f(\lambda)=\lambda^3-10\lambda^2-3k\lambda+34\lambda-48 \Rightarrow f'(\lambda)=3x^2-20x+3k+34\\\\ \Rightarrow&f(\lambda)=\left(\frac\{x\}\{3\}-\frac\{10\}\{9\}\right)f'(x) +\frac\{16\}\{9\}-\frac\{8k\}\{3\}+\left(\frac\{4\}\{9\}+2k\right)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 的二重特征值 $\displaystyle \lambda\_0$ 作为 $\displaystyle f(\lambda)$ 的重根, 满足
\begin\{aligned\} &\frac\{16\}\{9\}-\frac\{8k\}\{3\}+\left(\frac\{4\}\{9\}+2k\right)\lambda\_0=0 \Rightarrow \lambda\_0=\frac\{4(3k-2)\}\{9k+2\}\\\\ \Rightarrow&0=f'(\lambda\_0)=\frac\{81(k+2)^2(3k+2)\}\{(9k+2)^2\} \Rightarrow k=-2\mbox\{或\} -\frac\{2\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 当 $\displaystyle k=-2$ 时, $\displaystyle A$ 的特征值为 $\displaystyle 6,2,2$. 由
\begin\{aligned\} A-2E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&3\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)\Rightarrow \mathrm\{rank\}(A-2E)=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A\sim \mathrm\{diag\}(6,2,2)$ [Jordan 标准形严格上三角部分没有 $\displaystyle 1$, 否则与上述秩等式矛盾]. (2)、 当 $\displaystyle k=-\frac\{2\}\{3\}$ 时, $\displaystyle A$ 的特征值为 $\displaystyle 4,4,2$. 由
\begin\{aligned\} A-4E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&3\\\\ 0&1&3\\\\ 0&0&0\end\{array\}\right)\Rightarrow \mathrm\{rank\}(A-4E)=2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A\sim \left(\begin\{array\}\{cccccccccccccccccccc\}4&1&\\\\ &4&\\\\ &&2\end\{array\}\right)$, 而不可对角化.跟锦数学微信公众号. [在线资料](
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---
1146、 6、 若 $\displaystyle A\_\{m\times n\}$ 的秩为 $\displaystyle n$, 则 $\displaystyle A^\mathrm\{T\} A$ 是否一定可逆? 请证明或举反例. (暨南大学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 $\displaystyle \mathrm\{rank\} A=n$ 在 $\displaystyle Ax=0$ 只有零解. 对 $\displaystyle \forall\ 0\neq x\in\mathbb\{R\}^n$, $\displaystyle y=Ax\neq 0$, 而
\begin\{aligned\} x^\mathrm\{T\}(A^\mathrm\{T\} A)x=y^\mathrm\{T\} y > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A^\mathrm\{T\} A$ 正定, 而可逆.跟锦数学微信公众号. [在线资料](
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---
1147、 8、 在什么条件下, 有分块矩阵的行列式满足 $\displaystyle \left|\begin\{array\}\{cccccccccc\}A&C\\\\ &B\end\{array\}\right|=|A|\cdot |B|$, 并证明你的结论. (暨南大学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) / 只要 $\displaystyle A,B$ 是方阵, 结论就成立. 设 $\displaystyle A$ 是 $\displaystyle m$ 阶方阵, $\displaystyle B$ 是 $\displaystyle n$ 阶方阵, 题中左端矩阵为 $\displaystyle D$, 则按列展开, 并注意到只要 $\displaystyle i\geq m+1, i\leq m$, $\displaystyle d\_\{ij\}=0$, 我们知
\begin\{aligned\} \mbox\{左端\}=&\sum\_\{\left\\{i\_1,\cdots,i\_m\right\\}=\left\\{1,\cdots,m\right\\}\atop \left\\{i\_\{m+1\},\cdots,i\_\{m+n\}\right\\}=\left\\{m+1,\cdots,m+n\right\\}\}\boxed\{\begin\{array\}\{c\}(-1)^\{\tau(i\_1\cdots i\_m,i\_\{m+1\},\cdots,i\_\{m+n\})\}\\\\ d\_\{i\_11\}\cdots d\_\{i\_mm\}d\_\{i\_\{m+1\},m+1\}\cdots d\_\{i\_\{m+n-1\}\}d\_\{m+n\}\end\{array\}\}\\\\ =&\sum\_\{\left\\{i\_1,\cdots,i\_m\right\\}=\left\\{1,\cdots,m\right\\}\}(-1)^\{\tau(i\_1\cdots i\_m)\}d\_\{i\_11\}\cdots d\_\{i\_mm\}\\\\ &\sum\_\{\left\\{i\_\{m+1\},\cdots,i\_\{m+n\}\right\\}=\left\\{m+1,\cdots,m+n\right\\}\} (-1)^\{\tau(i\_\{m+1\},\cdots,i\_\{m+n\})\}d\_\{i\_\{m+1\},m+1\}\cdots d\_\{i\_\{m+n\},m+n\}\\\\ =&\sum\_\{\left\\{i\_1,\cdots,i\_m\right\\}=\left\\{1,\cdots,m\right\\}\}(-1)^\{\tau(i\_1\cdots i\_m)\}a\_\{i\_11\}\cdots a\_\{i\_mm\}\\\\ &\sum\_\{\left\\{i\_\{m+1\},\cdots,i\_\{m+n\}\right\\}=\left\\{m+1,\cdots,m+n\right\\}\} (-1)^\{\tau(i\_\{m+1\},\cdots,i\_\{m+n\})\}b\_\{i\_\{m+1\},m+1\}\cdots b\_\{i\_\{m+n\},m+n\}\\\\ =&|A|\cdot |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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---
1148、 4、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&0&\cdots&0&1\\\\ 1&0&0&\cdots&0&0\\\\ 0&1&0&\cdots&0&0\\\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\\\ 0&0&0&\cdots&0&0\\\\ 0&0&0&\cdots&1&0\end\{array\}\right)$. 求 $\displaystyle A$ 的不变因子, 初等因子以及 $\displaystyle A$ 的若尔当标准形. (南昌大学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 $\displaystyle 1$ 行展开知
\begin\{aligned\} |\lambda E\_n-A|=\lambda\cdot \lambda^\{n-1\}+(-1)^\{1+n\}(-1)\cdot(-1)^\{n-1\} =\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 $\displaystyle A$ 的特征值为全体 $\displaystyle n$ 次单位根 $\displaystyle \mathrm\{e\}^\{\mathrm\{ i\}\frac\{2k\pi\}\{n\}\}, 0\leq k\leq n-1$. 由它们互异知 $\displaystyle A$ 可对角化, 即 $\displaystyle A$ 的 Jordan 标准形为
\begin\{aligned\} \mathrm\{diag\}(1,\mathrm\{e\}^\{\mathrm\{ i\} \frac\{2\pi\}\{n\}\},\cdots, \mathrm\{e\}^\{\mathrm\{ i\} \frac\{2(n-1)\pi\}\{n\}\}), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle A$ 的初等因子为
\begin\{aligned\} \lambda-1, \lambda- \mathrm\{e\}^\{\mathrm\{ i\} \frac\{2\pi\}\{n\}\},\cdots, \lambda-\mathrm\{e\}^\{\mathrm\{ i\} \frac\{2(n-1)\pi\}\{n\}\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不变因子为
\begin\{aligned\} 1,\cdots,1, \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\}
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---
1149、 5、 设 $\displaystyle A$ 为实对称矩阵, $\displaystyle B$ 为半正定矩阵, $\displaystyle |A+\mathrm\{ i\} B|=0$. 证明: (0-30)、 方程组 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}Ax=0,\\\\ Bx=0\end\{array\}\right.$ 有非零实向量解; (0-31)、 $\displaystyle |A|=|B|=0$. (南昌大学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) / (0-32)、 先证一个结论. 对半正定矩阵 $\displaystyle B$ 而言, $\displaystyle x^\mathrm\{T\} Bx=0\Leftrightarrow Bx=0$. 事实上, ‘$\Leftarrow$‘显然成立. 往证’$\Rightarrow$‘. 由 $\displaystyle B$ 半正定知存在实矩阵 $\displaystyle C$ 使得 $\displaystyle B=C^\mathrm\{T\} C$. 而
\begin\{aligned\} &x^\mathrm\{T\} Bx=0\Rightarrow 0=x^\mathrm\{T\} C^\mathrm\{T\} Cx=(Cx)^\mathrm\{T\} (Cx)\\\\ \Rightarrow& Cx=0\Rightarrow Bx=C^\mathrm\{T\} Cx=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(0-33)、 回到题目. 由 $\displaystyle |A+\mathrm\{ i\} B|=0$ 知 $\displaystyle (A+\mathrm\{ i\} B)x=0$ 有非零解 $\displaystyle x=\alpha+\mathrm\{ i\} \beta, \alpha,\beta\in\mathbb\{R\}^n, \alpha,\beta$ 不全为 $\displaystyle 0$. 于是
\begin\{aligned\} &0=(A+\mathrm\{ i\} B)(\alpha+\mathrm\{ i\} \beta) =(A\alpha-B\beta)+\mathrm\{ i\}(A\beta+B\alpha)\\\\ \Rightarrow&A\alpha=B\beta, A\beta=-B\alpha\\\\ \Rightarrow& 0=-\alpha^\mathrm\{T\} B\alpha=\alpha^\mathrm\{T\} A\beta=\beta^\mathrm\{T\} A\alpha=\beta^\mathrm\{T\} B\beta\geq 0\\\\ \Rightarrow&\alpha^\mathrm\{T\} B\alpha=\beta^\mathrm\{T\} B\beta=0\stackrel\{\mbox\{第 1 步\}\}\{\Rightarrow\}B\alpha=B\beta=0\\\\ \Rightarrow&\left\\{\begin\{array\}\{llllllllllll\}A\alpha=0,\\\\ B\alpha=0,\end\{array\}\right.\mbox\{且\} \left\\{\begin\{array\}\{llllllllllll\}A\beta=0,\\\\ B\beta=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 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}Ax=0,\\\\ Bx=0\end\{array\}\right.$ 有非零解 $\displaystyle \alpha$ 或 $\displaystyle \beta$. (0-34)、 由第 2 步知 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}Ax=0,\\\\ Bx=0\end\{array\}\right.$ 有非零解 $\displaystyle x$, 则 (0-1)、 $\displaystyle Ax=0$ 有非零解 $\displaystyle x$, 而 $\displaystyle |A|=0$. (0-2)、 $\displaystyle Bx=0$ 有非零解 $\displaystyle x$, 而 $\displaystyle |B|=0$.跟锦数学微信公众号. [在线资料](
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---
1150、 6、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 的特征多项式为 $\displaystyle (\lambda-1)^n$. 证明: 对任意正整数 $\displaystyle k: 2\leq k\leq n$, $\displaystyle A$ 与 $\displaystyle A^k$ 相似. (南昌大学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)、 先证明对任意正整数 $\displaystyle k$, $\displaystyle J\_m^k(1)$ 的 Jordan 标准形为 $\displaystyle J\_m(1)$. 事实上, 设 $\displaystyle N=J\_m(0)$, 则
\begin\{aligned\} J\_m^k(1)=&\left(E\_m+N\right)^k =E\_m+C\_k^1 N+C\_k^2 N^2+\cdots\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}1&k&\star&\star\\\\ &\ddots&\ddots&\star\\\\ &&\ddots&k\\\\ &&&1\end\{array\}\right)\_m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mathrm\{rank\}\left(J\_m^k(1)-E\_m\right)=n-1$. 这表明 $\displaystyle J\_m^k(1)$ 的 Jordan 标准形 $\displaystyle J$ 也满足 $\displaystyle \mathrm\{rank\}(J-E\_m)=n-1$, 从而 $\displaystyle J=J\_m(1)$. (2)、 由 $\displaystyle A$ 的特征多项式为 $\displaystyle (\lambda-1)^n$ 知存在可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} &P^\{-1\}AP=\mathrm\{diag\}\left(J\_\{n\_1\}(1),\cdots, J\_\{n\_s\}(1)\right)\\\\ \Rightarrow&P^\{-1\}A^kP=\mathrm\{diag\}\left(J\_\{n\_1\}^k(1),\cdots,J\_\{n\_s\}^k(1)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 1 步知存在可逆矩阵 $\displaystyle Q\_i$ 使得
\begin\{aligned\} Q\_i^\{-1\}J\_\{n\_i\}^k(1)Q\_i=J\_\{n\_i\}(1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle Q=\mathrm\{diag\}(Q\_1,\cdots,Q\_s)$, 则 $\displaystyle Q$ 可逆, 且
\begin\{aligned\} Q^\{-1\}P^\{-1\}A^kPQ=&Q^\{-1\}\mathrm\{diag\}\left(J\_\{n\_1\}^k(1),\cdots,J\_\{n\_s\}^k(1)\right)Q\\\\ =&\mathrm\{diag\}\left(J\_\{n\_1\}(1),\cdots,J\_\{n\_s\}(1)\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle A,A^k$ 都相似于 $\displaystyle \mathrm\{diag\}\left(J\_\{n\_1\}(1),\cdots,J\_\{n\_s\}(1)\right)$, 而 $\displaystyle A\sim A^k$.跟锦数学微信公众号. [在线资料](
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发表于 2023-3-5 13:10:49
