## 张祖锦2023年数学专业真题分类70天之第47天
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1059、 (2)、 设 $\displaystyle A,B$ 为 $\displaystyle 3$ 阶复方阵, $\displaystyle A$ 与 $\displaystyle B$ 相似的充要条件为 $\displaystyle A,B$ 有相同的特征多项式和最小多项式. $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ (安徽大学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) / $\displaystyle \times$. 比如 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&\\\\ &0&&\\\\ &&0&1\\\\ &&&0\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&\\\\ &0&&\\\\ &&0&\\\\ &&&0\end\{array\}\right)$ 的特征多项式都是 $\displaystyle \lambda^4$, 最小多项式都是 $\displaystyle \lambda^2$, 但 $\displaystyle A, B$ 不相似.跟锦数学微信公众号. [在线资料](
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1060、 (3)、 已知复矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&-1&0&0\\\\ 1&0&1&0\\\\ 0&0&0&-1\\\\ 0&0&1&0\end\{array\}\right)$, 求 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$. (安徽大学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 $\displaystyle \mathrm\{ i\},\mathrm\{ i\},-\mathrm\{ i\},-\mathrm\{ i\}$. 由
\begin\{aligned\} A-\mathrm\{ i\} E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-\mathrm\{ i\}&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ 0&0&0&0\end\{array\}\right), A+\mathrm\{ i\} E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&\mathrm\{ i\}&0&0\\\\ 0&0&1&0\\\\ 0&0&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\}
知
\begin\{aligned\} \mathrm\{rank\}(J\pm \mathrm\{ i\} E)=\mathrm\{rank\}(A\pm \mathrm\{ i\} E)=3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}\mathrm\{ i\}&1&&\\\\ &\mathrm\{ i\}&&\\\\ &&-\mathrm\{ i\}&1\\\\ &&&-\mathrm\{ i\}\end\{array\}\right)$ [从秩可以确定出严格上三角部分有没有 $\displaystyle 1$, 因为如果 $\displaystyle \mathrm\{ i\}$ 所在的 Jordan 块没有 $\displaystyle 1$, 则 $\displaystyle \mathrm\{rank\}(J-\mathrm\{ i\} E)=2$, 矛盾].跟锦数学微信公众号. [在线资料](
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1061、 (4)、 设 $\displaystyle A$ 为 $\displaystyle 3$ 阶实对称矩阵, 且特征值为 $\displaystyle 1,1,2$. 已知
\begin\{aligned\} \alpha\_1=(1,-1,0)^\mathrm\{T\}, \alpha\_2=(1,1,-2)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为同属于特征值 $\displaystyle 1$ 的特征向量, 求 $\displaystyle A$. (安徽大学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) / 由实对称矩阵属于不同特征值的特征向量正交知 $\displaystyle A$ 的属于特征值 $\displaystyle 2$ 的特征向量 $\displaystyle \alpha\_3$ 满足
\begin\{aligned\} &\alpha\_i^\mathrm\{T\} \alpha\_3=0, i=1,2\Leftrightarrow A\alpha\_3=0,\\\\ &A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 1&1&-2\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&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\}
故可取 $\displaystyle \alpha\_3=(1,1,1)^\mathrm\{T\}$. 令 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ -1&1&1\\\\ 0&-2&1\end\{array\}\right)$, 则
\begin\{aligned\} AP=P\mathrm\{diag\}(1,1,2)\Rightarrow A=P\mathrm\{diag\}(1,1,2)P^\{-1\}=\frac\{1\}\{3\}\left(\begin\{array\}\{cccccccccccccccccccc\}4&1&1\\\\ 1&4&1\\\\ 1&1&4\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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1062、 4、 证明题 (共 40 分). (1)、 (10 分) 设 $\displaystyle A,B$ 均为 $\displaystyle n$ 阶方阵, 且 $\displaystyle A+B=AB$, 证明: $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\} B$. (安徽大学2023年高等代数考研试题) [矩阵 ]
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\begin\{aligned\} B=AB-A=A(B-E)\Rightarrow \mathrm\{rank\} B\leq \mathrm\{rank\} A,\\\\ A=AB-B=(A-E)B\Rightarrow \mathrm\{rank\} A\leq \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\}
故有结论.跟锦数学微信公众号. [在线资料](
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---
1063、 (2)、 (10 分) 设 $\displaystyle A,B$ 均为 $\displaystyle n$ 阶方阵, $\displaystyle AB=BA$, 且 $\displaystyle A$ 有 $\displaystyle n$ 个不同的特征值. 证明: (2-1)、 $\displaystyle B$ 可对角化; (2-2)、 存在 $\displaystyle f(x), \deg f(x)\leq n-1$, 满足 $\displaystyle B=f(A)$. (安徽大学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 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 为 $\displaystyle A$ 的互不相同的特征值, 其对应的特征向量为 $\displaystyle \alpha\_1,\cdots,\alpha\_n$, 则由
\begin\{aligned\} AB\alpha\_i=BA\alpha\_i=\lambda\_iB\alpha\_i\quad (1\leq i\leq n) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 ($\dim \left\\{\alpha; A\alpha=\lambda\_i\alpha\right\\}=1$)
\begin\{aligned\} B\alpha\_i=\mu\_i\alpha\_i\quad (\mu\_i\in\mathbb\{R\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是令 $\displaystyle P=(\alpha\_1,\cdots,\alpha\_n)$ 后,
\begin\{aligned\} P^\{-1\}AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n),\quad P^\{-1\}BP=\mathrm\{diag\}(\mu\_1,\cdots,\mu\_n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle B$ 可对角化. 取多项式 $\displaystyle f(x)$, 使得
\begin\{aligned\} f(\lambda\_i)=\mu\_i\quad \left(1\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\}
后则
\begin\{aligned\} P^\{-1\}f(A)P=\mathrm\{diag\}(f(\lambda\_1),\cdots,f(\lambda\_n))=P^\{-1\}BP\Rightarrow f(A)=B. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里, $\displaystyle f$ 可为 Lagrange 插值多项式:
\begin\{aligned\} f(x)=\sum\_\{i=1\}^n \mu\_i\frac\{(x-\lambda\_1)\cdots(x-\lambda\_\{i-1\})(x-\lambda\_\{i+1\})\cdots (x-\lambda\_n)\}\{(\lambda\_i-\lambda\_1)\cdots(\lambda\_i-\lambda\_\{i-1\})(\lambda\_i-\lambda\_\{i+1\})\cdots (\lambda\_i-\lambda\_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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---
1064、 (4)、 (10 分) 设 $\displaystyle A$ 为 $\displaystyle n$ 阶正交矩阵, $\displaystyle \alpha+\mathrm\{ i\}\beta$ 为 $\displaystyle A$ 属于特征值 $\displaystyle \lambda$ 的特征向量, 其中 $\displaystyle \lambda\neq \pm 1, \alpha,\beta$ 为实向量, $\displaystyle \mathrm\{ i\}$ 为纯虚数. 证明: $\displaystyle \alpha$ 与 $\displaystyle \beta$ 的模相等, 且相互正交. (安徽大学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 $\displaystyle \lambda\neq\pm 1$ 知 $\displaystyle \lambda=a+b\mathrm\{ i\}, b\neq 0$. 由 $\displaystyle A$ 正交知
\begin\{aligned\} 1=|\lambda|^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\}
又由题设,
\begin\{aligned\} &A(\alpha+\beta\mathrm\{ i\})=(a+b\mathrm\{ i\})(\alpha+\beta\mathrm\{ i\})\\\\ \Rightarrow& A\alpha=a\alpha-b\beta, A\beta=b\alpha+a\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 (4-1)、 由
\begin\{aligned\} \alpha^\mathrm\{T\}\alpha=&\alpha^\mathrm\{T\} A^\mathrm\{T\} A\alpha=(A\alpha)^\mathrm\{T\} (A\alpha)\\\\ =&(a\alpha^\mathrm\{T\}-b\beta^\mathrm\{T\})(a\alpha-b\beta)\\\\ =&a^2\alpha^\mathrm\{T\} \alpha-2ab\alpha^\mathrm\{T\} \beta+b^2\beta^\mathrm\{T\} \beta \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\} &b^2(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\}\beta)=-2ab\alpha^\mathrm\{T\} \beta\\\\ \Rightarrow&b(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\} \beta)=-2a\alpha^\mathrm\{T\} \beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4-2)、 同理, 由
\begin\{aligned\} \alpha^\mathrm\{T\}\beta=&\alpha^\mathrm\{T\} A^\mathrm\{T\} A\beta=(A\alpha)^\mathrm\{T\} (A\beta)\\\\ =&(a\alpha^\mathrm\{T\}-b\beta^\mathrm\{T\})(b\alpha+a\beta)\\\\ =&ab\alpha^\mathrm\{T\} \alpha+a^2\alpha^\mathrm\{T\} \beta-b^2\alpha^\mathrm\{T\} \beta-ab\beta^\mathrm\{T\} \beta \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\} &2b^2\alpha^\mathrm\{T\}\beta=ab(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\} \beta)\\\\ \Rightarrow&2b\alpha^\mathrm\{T\} \beta=a(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\} \beta). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
联合第 (i), (ii) 步即知
\begin\{aligned\} b(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\} \beta)&=-\frac\{a\}\{b\}\cdot 2\alpha^\mathrm\{T\}\beta =-\frac\{a\}\{b\}\cdot a(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\}\beta). \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\} 0=(a^2+b^2)(\alpha^\mathrm\{T\} \alpha-\beta^\mathrm\{T\} \beta)=\alpha^\mathrm\{T\}\alpha-\beta^\mathrm\{T\}\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即 $\displaystyle \alpha,\beta$ 有相同的模长. 代入第 (ii) 步即知 $\displaystyle \alpha^\mathrm\{T\} \beta=0$.跟锦数学微信公众号. [在线资料](
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1065、 3、 (20 分) 已知矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&c&-1\\\\ 1-c&-a&0\\\\ 5&3&b\end\{array\}\right)$ 有一特征向量 $\displaystyle \alpha=(-1,1,-1)^\mathrm\{T\}$. (1)、 若 $\displaystyle |A|=0$, 求 $\displaystyle a,b,c$ 及 $\displaystyle A$ 的所有特征值. 问 $\displaystyle A$ 是否可以对角化? 为什么? (2)、 若 $\displaystyle |A|=-1$, 求 $\displaystyle a,b,c$ 及 $\displaystyle A$ 的所有特征值, 问 $\displaystyle A$ 是否可以对角化? 此时求 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 及可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP=J$. (北京工业大学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)、 由题设,
\begin\{aligned\} &A\alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}1-a+c\\\\ -1-a+c\\\\ -2-b\end\{array\}\right)\parallel\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\-1\end\{array\}\right)\\\\ \Rightarrow&\exists\ t,\mathrm\{ s.t.\} 1-a+c=-t, -1-a+c=t, -2-b=t\\\\ \Rightarrow&-t-1=c-a=t+1, b=t-2\Rightarrow t=-1, b=-3, c=a\qquad(I)\\\\ \Rightarrow& 0=|A|=a-3\Rightarrow c=a=3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&3&-1\\\\ -2&-3&0\\\\ 5&3&-3\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle 0,-1,-2$, 它们是互异的, 而对应的特征向量是线性无关的. 从而 $\displaystyle A$ 有 $\displaystyle 3$ 个线性无关的特征向量, 是可对角化的. (2)、 由 $\displaystyle (I)$ 知 $\displaystyle -1=|A|=a-3\Rightarrow c=a=2$. 故 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&-1\\\\ -1&-2&0\\\\ 5&3&-3\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle -1,-1,-1$. 由
\begin\{aligned\} A+E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&1\\\\ 0&0&0\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\}
知 $\displaystyle \mathrm\{rank\}(A+E)=2\Rightarrow \mathrm\{rank\}(J+E)=2$. 故 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&-1&1\\\\ 0&0&-1\end\{array\}\right)$. 设可逆矩阵 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)$ 使得
\begin\{aligned\} P^\{-1\}AP=J\Leftrightarrow (A+E)\alpha\_1=0, (A+E)\alpha\_2=\alpha\_1, (A+E)\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 $\displaystyle (I)$ 知可选 $\displaystyle \alpha\_1=(1,-1,1)^\mathrm\{T\}$. 又由
\begin\{aligned\} (A+E,\alpha\_1)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1&-1\\\\ 0&1&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\}
知可选 $\displaystyle \alpha\_2=(-1,2,0)^\mathrm\{T\}$. 再由
\begin\{aligned\} (A+E,\alpha\_2)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1&3\\\\ 0&1&1&-5\\\\ 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\}
知可选 $\displaystyle \alpha\_3=(3,-5,0)^\mathrm\{T\}$. 故
\begin\{aligned\} P=(\alpha\_1,\alpha\_2,\alpha\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&3\\\\ -1&2&-5\\\\ 1&0&0\end\{array\}\right)\Rightarrow P^\{-1\}AP=J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0\\\\ 0&-1&1\\\\ 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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1066、 6、 (20 分) 考虑实对称矩阵 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&1\end\{array\}\right)$, 其中 $\displaystyle \alpha=(a\_1,\cdots,a\_n)^\mathrm\{T\}$. (1)、 若 $\displaystyle A$ 可逆, 证明:
\begin\{aligned\} |B|=|A|(1-\alpha^\mathrm\{T\} A^\{-1\}\alpha) =|A-\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\}
(2)、 证明: 矩阵 $\displaystyle A-\alpha\alpha^\mathrm\{T\}$ 正定当且仅当 $\displaystyle A$ 正定, 且 $\displaystyle \alpha^\mathrm\{T\} A^\{-1\}\alpha < 1$. (北京工业大学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](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -\alpha^\mathrm\{T\} A^\{-1\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A\alpha\\\\ \alpha^\mathrm\{T\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ 0&1-\alpha^\mathrm\{T\} A^\{-1\}\alpha\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E&-\alpha\\\\ 0&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -\alpha^\mathrm\{T\}&1\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}A-\alpha\alpha^\mathrm\{T\}&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\}
知 $\displaystyle |B|=|A|(1-\alpha^\mathrm\{T\} A^\{-1\}\alpha) =|A-\alpha\alpha^\mathrm\{T\}|$. (2)、 矩阵 $\displaystyle A-\alpha\alpha^\mathrm\{T\}$ 正定等价于 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A-\alpha\alpha^\mathrm\{T\}&0\\\\ 0&1\end\{array\}\right)$ 正定, 由第 1 步的结果知这等价于 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&0\\\\ 0&1-\alpha^\mathrm\{T\} A^\{-1\}\alpha\end\{array\}\right)$ 正定, 最终等价于 $\displaystyle A$ 正定, 且 $\displaystyle \alpha^\mathrm\{T\} A^\{-1\}\alpha < 1$.跟锦数学微信公众号. [在线资料](
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---
1067、 8、 (15 分) 设 $\displaystyle A$ 为 $\displaystyle n$ 阶复方阵, $\displaystyle B$ 为 $\displaystyle m$ 阶复方阵, 且存在秩为 $\displaystyle r$ 的矩阵 $\displaystyle X$ 满足 $\displaystyle AX=XB$, 其中 $\displaystyle 1\leq r\leq \min\left\\{n,m\right\\}$. 证明: $\displaystyle A$ 与 $\displaystyle B$ 至少有 $\displaystyle r$ 个公共的特征值 (重根按重数计算). (北京科技大学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 $\displaystyle \mathrm\{rank\} X=r$ 知存在可逆矩阵 $\displaystyle P,Q$ 使得 $\displaystyle PXQ=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)$. 设
\begin\{aligned\} PAP^\{-1\}=C=\left(\begin\{array\}\{cccccccccccccccccccc\}C\_\{11\}&C\_\{12\}\\\\ C\_\{21\}&C\_\{22\}\end\{array\}\right), Q^\{-1\}BQ=D=\left(\begin\{array\}\{cccccccccccccccccccc\}D\_\{11\}&D\_\{12\}\\\\ D\_\{21\}&D\_\{22\}\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\}
则
\begin\{aligned\} &AX=XB\Leftrightarrow PAP^\{-1\}\cdot PXQ=PXQ\cdot Q^\{-1\}BQ\\\\ \Leftrightarrow& C\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)D \Leftrightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}C\_\{11\}&0\\\\ C\_\{21\}&0\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}D\_\{11\}&D\_\{12\}\\\\ 0&0\end\{array\}\right)\\\\ \Leftrightarrow& C\_\{11\}=D\_\{11\}, D\_\{12\}=0, C\_\{21\}=0. \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\} &PAP^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}C\_\{11\}&C\_\{12\}\\\\ 0&C\_\{22\}\end\{array\}\right), Q^\{-1\}BQ=\left(\begin\{array\}\{cccccccccccccccccccc\}C\_\{11\}&0\\\\ D\_\{21\}&D\_\{22\}\end\{array\}\right)\\\\ \Rightarrow&|\lambda E-A|=|\lambda E\_r-C\_\{11\}|\cdot |\lambda E\_\{n-r\}-C\_\{22\}|\\\\ &|\lambda E-B|=|\lambda E\_r-C\_\{11\}|\cdot |\lambda E\_m-D\_\{22\}|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle C\_\{11\}$ 的特征值为 $\displaystyle \lambda\_1,\cdots,\lambda\_r$, 则它们就是 $\displaystyle A,B$ 的公共特征值.跟锦数学微信公众号. [在线资料](
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---
1068、 9、 (15 分) 已知 $\displaystyle \alpha=(a\_1,\cdots,a\_n)^\mathrm\{T\}, \beta=(b\_1,\cdots,b\_n)^\mathrm\{T\}$ 为两个非零实向量 $\displaystyle (n > 1)$, $\displaystyle A=\alpha\beta^\mathrm\{T\}$. (1)、 求 $\displaystyle A$ 的最小多项式; (2)、 求 $\displaystyle A$ 的若尔当标准形. (北京科技大学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) /
\begin\{aligned\} A\alpha=\alpha\beta^\mathrm\{T\}\alpha=\alpha(\beta^\mathrm\{T\} \alpha)=(\beta^\mathrm\{T\} \alpha)\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由常用公式知
\begin\{aligned\} &\lambda|\lambda E\_n-\alpha\beta^\mathrm\{T\}|=\lambda^n(\lambda-\beta^\mathrm\{T\} \alpha)\\\\ \Rightarrow&|\lambda E\_n-A|=\lambda^\{n-1\} (\lambda-\beta^\mathrm\{T\} \alpha). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle \beta^\mathrm\{T\} \alpha\neq 0$, 则 $\displaystyle A$ 的特征值为 $\displaystyle 0$ ($n-1$ 重), $\displaystyle \beta^\mathrm\{T\} \alpha$ (单重). 由 $\displaystyle \alpha\neq 0, \beta\neq 0$ 知 $\displaystyle \mathrm\{rank\} A=1$. 故 $\displaystyle Ax=0$ 有 $\displaystyle n-1$ 个线性无关的解向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-1\}$. 令 $\displaystyle \eta\_n=\alpha$, 则 $\displaystyle P=(\eta\_1,\cdots,\eta\_n)$ 可逆, 且
\begin\{aligned\} P^\{-1\}AP=\mathrm\{diag\}(0\_\{n-1\},\beta^\mathrm\{T\}\alpha). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle A$ 可对角化, $\displaystyle A$ 的最小多项式为 $\displaystyle \lambda(\lambda-\beta^\mathrm\{T\} \alpha)$, Jordan 标准形为 $\displaystyle \mathrm\{diag\}(0\_\{n-1\},\beta^\mathrm\{T\}\alpha)$. (2)、 若 $\displaystyle \beta^\mathrm\{T\} \alpha=0$, 则 $\displaystyle A$ 的特征值为 $\displaystyle 0$ ($n$ 重). 由 $\displaystyle \mathrm\{rank\} A=1$ 知 $\displaystyle A$ 的 Jordan 标准形也满足 $\displaystyle \mathrm\{rank\} J=1$. 故
\begin\{aligned\} J=\mathrm\{diag\}\left(\left(\begin\{array\}\{cccccccccccccccccccc\}0&1\\\\ &0\end\{array\}\right),0\_\{n-2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle A$ 不可对角化, $\displaystyle A$ 的最小多项式为 $\displaystyle \lambda^2$.跟锦数学微信公众号. [在线资料](
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---
1069、 (4)、 若 $\displaystyle A$ 为 $\displaystyle s\times n$ 矩阵, 秩为 $\displaystyle n$; $\displaystyle B$ 为 $\displaystyle l\times m$ 矩阵, 秩为 $\displaystyle m$, 则 $\displaystyle \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&C\\\\ 0&B\end\{array\}\right)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/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\} &\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&C\\\\ 0&B\end\{array\}\right)\left\\{\begin\{array\}\{llllllllllll\}\geq \mathrm\{rank\} A+\mathrm\{rank\} B=n+m\\\\ \leq\mbox\{它的列数\} n+m\end\{array\}\right.\\\\ \Rightarrow& \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&C\\\\ 0&B\end\{array\}\right)=n+m. \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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---
1070、 (6)、 矩阵方程 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}3&4\\\\ 6&8\end\{array\}\right)X^\mathrm\{T\}=\left(\begin\{array\}\{cccccccccccccccccccc\}2&9\\\\ 4&18\end\{array\}\right)$ 的解为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学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) / 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&4\\\\ 6&8\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}2&9\\\\ 4&18\end\{array\}\right)=(\beta\_1,\beta\_2)$, 则由
\begin\{aligned\} (A,B)\to\left(\begin\{array\}\{cccccccccccccccccccc\}3&4&2&9\\\\ 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 $\displaystyle Ax=\beta\_1, Ax=\beta\_2$ 的通解分别为
\begin\{aligned\} k\left(\begin\{array\}\{cccccccccccccccccccc\}-4\\\\3\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{2\}\{3\}\\\\0\end\{array\}\right),\quad \forall\ k; l\left(\begin\{array\}\{cccccccccccccccccccc\}-4\\\\3\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}3\\\\0\end\{array\}\right),\quad \forall\ l. \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\} X^\mathrm\{T\}=\left(\begin\{array\}\{cccccccccccccccccccc\}-4k+\frac\{2\}\{3\}&-4l+3\\\\ 3k&3l\end\{array\}\right)\Rightarrow X=\left(\begin\{array\}\{cccccccccccccccccccc\}-4k+\frac\{2\}\{3\}&3k\\\\ -4l+3&3l\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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---
1071、 (7)、 若 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&0&0&1\\\\ 0&1&0&0\\\\ 0&0&1&0\end\{array\}\right)$, 且 $\displaystyle B=P^\{-1\}AP$, 其中 $\displaystyle P$ 为可逆矩阵, 则 $\displaystyle B^3-A^3=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学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) / 易知 $\displaystyle A^3=E$, 而 $\displaystyle B^3=P^\{-1\}A^3P=E$. 于是 $\displaystyle B^3-A^3=0$.跟锦数学微信公众号. [在线资料](
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---
1072、 (8)、 若 $\displaystyle A$ 为 $\displaystyle 5$ 阶方阵, 且 $\displaystyle A$ 的特征多项式为 $\displaystyle f(\lambda)=(\lambda-5)^3(\lambda-8)^2$, 最小多项式 $\displaystyle m(\lambda)=(\lambda-5)^2(\lambda-8)^2$, 则 $\displaystyle A$ 的 Jordan 阵为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$, 特征值为 $\displaystyle 5$ 的特征子空间的维数为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$, $\displaystyle \mathrm\{rank\}(8I-A)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (北京理工大学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) / 由 $\displaystyle m(\lambda)$ 的形式知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 含有 $\displaystyle J\_2(5), J\_2(8)$. 再由 $\displaystyle f(\lambda)$ 的形式知 $\displaystyle J$ 还含有 $\displaystyle J\_1(5)$. 故 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}5&&&&\\\\ &5&1&&\\\\ &&5&&\\\\ &&&8&1\\\\ &&&&8\end\{array\}\right)$. 进而存在可逆矩阵 $\displaystyle P=(\alpha\_1,\cdots,\alpha\_5)$ 使得 $\displaystyle P^\{-1\}AP=J$. 从而特征值为 $\displaystyle 5$ 的特征子空间 $\displaystyle L(\alpha\_1,\alpha\_2)$ 的维数为 $\displaystyle 2$,
\begin\{aligned\} \mathrm\{rank\}(8I-A)=&\mathrm\{rank\}(8I-J)=\mathrm\{rank\}(J-8I)\\\\ =&\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}-3&&&&\\\\ &-3&1&&\\\\ &&-3&&\\\\ &&&0&1\\\\ &&&&0\end\{array\}\right)=4. \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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---
1073、 5、 (1)、 设 $\displaystyle A$ 是正定矩阵, 证明: 存在对角元素全大于 $\displaystyle 0$ 的上三角矩阵 $\displaystyle C$, 使得 $\displaystyle A=C^\mathrm\{T\} C$. (2)、 证明 (1) 中的 $\displaystyle C$ 是唯一的. (北京师范大学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)、 对 $\displaystyle n$ 作数学归纳法. 当 $\displaystyle n=1$ 时, 结论自明. 假设结论对 $\displaystyle n-1$ 阶正定矩阵成立, 则对 $\displaystyle n$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}A\_1&\alpha\\\\ \alpha^\mathrm\{T\}&a\end\{array\}\right)$, 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -\alpha^\mathrm\{T\} A\_1^\{-1\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A\_1&\alpha\\\\ \alpha^\mathrm\{T\}&a\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\_1^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A\_1&0\\\\ 0&a-\alpha^\mathrm\{T\} A\_1^\{-1\}\alpha\end\{array\}\right)\ (I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle b=a-\alpha^\mathrm\{T\} A\_1^\{-1\}\alpha > 0$. 由归纳假设, 存在对角元素全大于 $\displaystyle 0$ 的上三角矩阵 $\displaystyle C\_1$, 使得 $\displaystyle A\_1=C\_1^\mathrm\{T\} C\_1$. 设
\begin\{aligned\} C\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}C\_1&\\\\ &\sqrt\{\frac\{a\}\{b\}\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\_1^\{-1\}\alpha\\\\ 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\}
则 $\displaystyle C$ 是对角元素全大于 $\displaystyle 0$ 的上三角矩阵, 且 $\displaystyle (I)$ 蕴含
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ \alpha^\mathrm\{T\} A\_1^\{-1\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A\_1&\\\\ &b\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\_1^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)=C^\mathrm\{T\} C . \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. (2)、 设 $\displaystyle B$ 也是对角元全大于 $\displaystyle 0$ 的上三角阵, 满足 $\displaystyle A=B^\mathrm\{T\} B$, 则 $\displaystyle B^\mathrm\{T\} B=C^\mathrm\{T\} C$. 比较第一行知
\begin\{aligned\} b\_\{11\}\left(b\_\{11\},\cdots,b\_\{1n\}\right)=c\_\{11\}\left(c\_\{11\},\cdots,c\_\{1n\}\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\}
由 $\displaystyle b\_\{11\} > 0, c\_\{11\} > 0$ 及 $\displaystyle (I)$ 的第一个分量知 $\displaystyle b\_\{11\}=c\_\{11\}$. 再由 $\displaystyle (I)$ 的其余分量知 $\displaystyle b\_\{1i\}=c\_\{1i\}, 2\leq i\leq n$. 再比较第 $\displaystyle 2$ 行知
\begin\{aligned\} &\left(b\_\{12\}b\_\{11\},b\_\{12\}^2,\cdots,b\_\{12\}b\_\{1n\}\right) +b\_\{22\}\left(0,b\_\{22\},\cdots,b\_\{2n\}\right)\\\\ =&\left(c\_\{12\}c\_\{11\},c\_\{12\}^2,\cdots,c\_\{12\}c\_\{1n\}\right) +c\_\{22\}\left(0,c\_\{22\},\cdots,c\_\{2n\}\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\} b\_\{22\}\left(0,b\_\{22\},\cdots,b\_\{2n\}\right) =c\_\{22\}\left(0,c\_\{22\},\cdots,c\_\{2n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
联合 $\displaystyle b\_\{22\} > 0, c\_\{22\} > 0$ 即知 $\displaystyle b\_\{22\}=c\_\{22\}$, 进而 $\displaystyle b\_\{2i\}=c\_\{2i\}, 3\leq i\leq n$. 如此这般一行行做下去即知 $\displaystyle B=C$. 唯一性得证.跟锦数学微信公众号. [在线资料](
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---
1074、 6、 设 $\displaystyle A$ 是 $\displaystyle m$ 阶矩阵, $\displaystyle B$ 是 $\displaystyle n$ 阶矩阵, 若 $\displaystyle A,B$ 无公共特征值, 证明: $\displaystyle AX=XB$ 仅有零解. (北京师范大学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 $\displaystyle f,g$ 分别是 $\displaystyle A,B$ 的特征值, 则由 $\displaystyle A,B$ 无公共特征值知 $\displaystyle f,g$ 在 $\displaystyle \mathbb\{C\}$ 上没有公共复根, 而
\begin\{aligned\} &\left(f,g\right)=\left(f,g\right)\_\mathbb\{C\}=1 \Rightarrow\exists\ u,v,\mathrm\{ s.t.\} uf+vg=1\\\\ \Rightarrow&E=u(B)f(B)+v(B)g(B)=u(B)f(B). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f(B)$ 可逆. 据题设,
\begin\{aligned\} AX=XB\Rightarrow A^2X=A\cdot XB=XB^2\Rightarrow \cdots\Rightarrow 0=f(A)X=Xf(B). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f(B)$ 可逆即知 $\displaystyle X=0$.跟锦数学微信公众号. [在线资料](
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---
1075、 2、 用初等变换化矩阵 $\displaystyle A$ 为等价标准形, 其中
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&0&1&2\\\\ 2&0&1&1&0\\\\ 3&1&0&0&4\\\\ 2&2&0&-1&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\}
(北京邮电大学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) / 取
\begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ -2&1&0&0\\\\ -3&0&1&0\\\\ -2&0&0&1\end\{array\}\right), Q\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-1&-2\\\\ 0&1&0&0&0\\\\ 0&0&1&0&0\\\\ 0&0&0&1&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\}
则 $\displaystyle A\_1=P\_1AQ\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&2&1&-1&-4\\\\ 0&4&0&-3&-2\\\\ 0&4&0&-3&-2\end\{array\}\right)$. 取
\begin\{aligned\} P\_2=E\_4, Q\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&0&1&0&0\\\\ 0&1&0&0&0\\\\ 0&0&0&1&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\}
则 $\displaystyle A\_2=P\_2A\_1Q\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&1&2&-1&-4\\\\ 0&0&4&-3&-2\\\\ 0&0&4&-3&-2\end\{array\}\right)$. 取
\begin\{aligned\} P\_3=E\_4, Q\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&1&-2&1&4\\\\ 0&0&1&0&0\\\\ 0&0&0&1&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\}
则 $\displaystyle A\_3=P\_3A\_2Q\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&1&0&0&0\\\\ 0&0&4&-3&-2\\\\ 0&0&4&-3&-2\end\{array\}\right)$. 取
\begin\{aligned\} P\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&1&0&0\\\\ 0&0&1&0\\\\ 0&0&-1&1\end\{array\}\right), Q\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0&0\\\\ 0&1&0&0&0\\\\ 0&0&1&\frac\{3\}\{4\}&\frac\{1\}\{2\}\\\\ 0&0&0&1&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\}
则 $\displaystyle A\_4=P\_4A\_3Q\_3=\mathrm\{diag\}(1,1,4,0,0)$. 取
\begin\{aligned\} P\_5=E\_4, Q\_5=\mathrm\{diag\}\left(1,1,\frac\{1\}\{4\},1,1\right)\Rightarrow P\_5A\_4Q\_5=\mathrm\{diag\}(1,1,1,0,0). \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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1076、 3、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 满足方程 $\displaystyle A^2+3A+2E=0$, 其中 $\displaystyle E$ 为单位矩阵, $\displaystyle O$ 为零矩阵. 证明: (1)、 $\displaystyle A-E$ 是可逆矩阵, 并求 $\displaystyle (A-E)^\{-1\}$; (2)、 $\displaystyle A+E$ 与 $\displaystyle A+2E$ 不同时可逆. (北京邮电大学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 $\displaystyle f(x)=x^2+3x+2, g(x)=x-1$, 则 $\displaystyle f(x)=(x+4)g(x)+6$, 而
\begin\{aligned\} 0=f(A)=(A+4E)(A-E)+6E\Rightarrow (A-E)\frac\{A+4E\}\{-6\}=E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A-E$ 可逆, 且 $\displaystyle (A-E)^\{-1\}=-\frac\{A+4E\}\{6\}$. (2)、 由题设, $\displaystyle (A+E)(A+2E)=0$, 而 $\displaystyle |A+E|=0$ 或 $\displaystyle |A+2E|=0$. 于是 $\displaystyle A+E$ 与 $\displaystyle A+2E$ 不同时可逆.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)/
---
1077、 6、 已知 $\displaystyle \alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\-1\end\{array\}\right)$ 是矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-1&2\\\\ 5&a&3\\\\ -1&b&-2\end\{array\}\right)$ 的特征向量. (1)、 求 $\displaystyle a,b$ 的值; (2)、 求矩阵 $\displaystyle A$ 的特征值; (3)、 讨论 $\displaystyle A$ 能否与对角矩阵相似. (北京邮电大学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, $\displaystyle \exists\ \lambda,\mathrm\{ s.t.\} A\alpha=\lambda\alpha$, 即
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\ a+2\\\\ b+1\end\{array\}\right)=\lambda \left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\-1\end\{array\}\right)\Rightarrow a=-3, b=0, \lambda=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-1&2\\\\ 5&-3&3\\\\ -1&0&-2\end\{array\}\right)$. 易知 $\displaystyle A$ 的特征值为 $\displaystyle -1,-1,-1$. 由
\begin\{aligned\} -E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&1\\\\ 0&0&0\end\{array\}\right)\Rightarrow \mathrm\{rank\}(-E-A)=2\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 不可对角化. 这可用反证法证得. 若 $\displaystyle A$ 可对角化, 则
\begin\{aligned\} A\sim \mathrm\{diag\}(-1,-1,-1)\Rightarrow A=-E\Rightarrow \mathrm\{rank\}(-E-A)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与 $\displaystyle (I)$ 矛盾, 故有结论.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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---
1078、 7、 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_r)$ 是 $\displaystyle n\times r$ 矩阵, $\displaystyle B=(\beta\_1,\cdots,\beta\_s)$ 是 $\displaystyle n\times s$ 矩阵, 矩阵 $\displaystyle A$ 的秩为 $\displaystyle r$, 矩阵 $\displaystyle B$ 的秩为 $\displaystyle s$. 证明: 如果 $\displaystyle r+s > n$, 则存在非零向量 $\displaystyle \xi$, 使得 $\displaystyle \xi$ 既可以由 $\displaystyle \alpha\_1,\cdots,\alpha\_r$ 线性表出, 又可以由 $\displaystyle \beta\_1,\cdots,\beta\_s$ 线性表出. (北京邮电大学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
\begin\{aligned\} \mathrm\{im\} A=&L(Ae\_1,\cdots, Ae\_s)=L(\alpha\_1,\cdots, \alpha\_r),\\\\ \mathrm\{im\} B=&L(Be\_1,\cdots, Be\_s)=L(\beta\_1,\cdots, \beta\_s), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \dim\mathrm\{im\} A=\mathrm\{rank\} A=r, \dim \mathrm\{im\} B=\mathrm\{rank\} B=s$. 于是
\begin\{aligned\} \dim(\mathrm\{im\} A\cap \mathrm\{im\} B)=&\dim \mathrm\{im\} A+\dim \mathrm\{im\} B-\dim(\mathrm\{im\} A+\mathrm\{im\} B)\\\\ \geq& r+s-n > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \exists\ 0\neq \xi\in \mathrm\{im\} A\cap \mathrm\{im\} B$. 此 $\displaystyle \xi$ 既可以由 $\displaystyle \alpha\_1,\cdots,\alpha\_r$ 线性表出, 又可以由 $\displaystyle \beta\_1,\cdots,\beta\_s$ 线性表出.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)/
---
1079、 (3)、 设 $\displaystyle A$ 是一个 $\displaystyle n$ 阶实对称矩阵, $\displaystyle \lambda\_n$ 是 $\displaystyle A$ 最大的特征值. 证明:
\begin\{aligned\} \lambda\_n=\max\_\{0\neq X\in\mathbb\{R\}^n\}\frac\{X^\mathrm\{T\} AX\}\{X^\mathrm\{T\} X\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mathbb\{R\}^n$ 为实 $\displaystyle n$ 维列向量的集合. (大连理工大学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) / 由 $\displaystyle A$ 实对称知存在正交矩阵 $\displaystyle P=(\alpha\_1,\cdots,\alpha\_n)$, 使得
\begin\{aligned\} P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \quad \lambda\_i\in\mathbb\{R\}, \lambda\_1\leq \cdots\leq \lambda\_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 $\displaystyle \forall\ 0\neq X\in\mathbb\{R\}^n$, 设 $\displaystyle X=PY$ 后,
\begin\{aligned\} X^\mathrm\{T\} AX=&Y^\mathrm\{T\} \mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)Y =\sum\_i \lambda\_iy\_i^2\leq \lambda\_n\sum\_i y\_i^2\\\\ =&\lambda\_nY^\mathrm\{T\} Y=\lambda\_nX^\mathrm\{T\} X. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由
\begin\{aligned\} A\alpha\_n=\lambda\_n\alpha\_n\Rightarrow \alpha\_n^\mathrm\{T\} A\alpha\_n=\lambda\_n\alpha\_n^\mathrm\{T\} \alpha\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \lambda\_n=\max\_\{0\neq X\in\mathbb\{R\}^n\}\frac\{X^\mathrm\{T\} AX\}\{X^\mathrm\{T\} X\}$.跟锦数学微信公众号. [在线资料](
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---
1080、 (8)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&12&2022\\\\ 0&0&25\\\\ 0&0&0\end\{array\}\right)$, 证明: 矩阵方程 $\displaystyle X^2=A$ 无解, 其中 $\displaystyle X\in \mathbb\{C\}^\{3\times 3\}$. (大连理工大学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) / 用反证法. 若存在 $\displaystyle X$ 使得 $\displaystyle X^2=A$, 则 $\displaystyle X$ 的特征值 $\displaystyle \lambda$ 满足 $\displaystyle \lambda^2=0\Rightarrow \lambda=0$. 故 $\displaystyle X$ 的 Jordan 标准形为 $\displaystyle J\_3(0)$ 或 $\displaystyle \mathrm\{diag\}\left(J\_2(0),0\right)$ 或 $\displaystyle 0$. 从而
\begin\{aligned\} A=X^2\sim \left(\begin\{array\}\{cccccccccccccccccccc\}0&0&1\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right)\mbox\{或\} 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 $\displaystyle \mathrm\{rank\} A=2$ 矛盾. 故有结论.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)/
---
1081、 (2)、 设 $\displaystyle 2023$ 阶矩阵 $\displaystyle A$ 满足 $\displaystyle A^8=0$, 则秩 $\displaystyle \mathrm\{rank\} A$ 的最大值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学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) / 设 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle J$, 则 $\displaystyle J$ 中 Jordan 块的阶数小于等于 $\displaystyle 8$. 为了让 $\displaystyle \mathrm\{rank\} A$ 最大, 我们应让更多的 $\displaystyle 1$ 布满 $\displaystyle J$ 的上三角部分. 由 $\displaystyle 2023=252\times 8+7$, $\displaystyle \mathrm\{rank\} J\_8(0)=7$ 知
\begin\{aligned\} \max\mathrm\{rank\} A=252\times 7+6=1770. \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)/
发表于 2023-3-5 09:18:25
