## 张祖锦2023年数学专业真题分类70天之第51天
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1151、 9、 设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1&-1\\\\ 1&0&-1&1\\\\ 1&-1&0&1\\\\ -1&1&1&0\end\{array\}\right)$. 求正交矩阵 $\displaystyle U$, 使得 $\displaystyle U^\mathrm\{T\} AU$ 为对角矩阵. (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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,1,-3$. 由
\begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1&1\\\\ 0&0&0&0\\\\ 0&0&0&0\\\\ 0&0&0&0 \end\{array\}\right), -3E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&-1\\\\ 0&1&0&1\\\\ 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 $\displaystyle A$ 的属于特征值 $\displaystyle 1,-3$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\1\\\\0 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\0\\\\0\\\\1 \end\{array\}\right); \xi\_4=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\-1\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle \xi\_1,\xi\_2,\xi\_3,\xi\_4$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3,\eta\_4$. 令
\begin\{aligned\} U=(\eta\_1,\eta\_2,\eta\_3,\eta\_4)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{2\sqrt\{3\}\}&\frac\{1\}\{2\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&-\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\}\\\\ 0&\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\}\\\\ 0&0&\frac\{\sqrt\{3\}\}\{2\}&\frac\{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\}
则 $\displaystyle U$ 正交, 且
\begin\{aligned\} U^\mathrm\{T\} AU=U^\{-1\}AU=\mathrm\{diag\}\left(1,1,1,-3\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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1152、 10、 设 $\displaystyle A$ 是正定矩阵, $\displaystyle B$ 是实对称矩阵. 若 $\displaystyle |(1-t)A+B|=0$ 的根全大于 $\displaystyle 1$. 证明: $\displaystyle B$ 为正定矩阵. (南昌大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 正定知存在实可逆矩阵 $\displaystyle C$, 使得 $\displaystyle C^\mathrm\{T\} AC=E\_n$. 于是
\begin\{aligned\} 0=&|C^\mathrm\{T\}|\cdot |(1-t)A+B|\cdot |C| =|(1-t)E\_n+C^\mathrm\{T\} BC|\\\\ =&(-1)^n |(t-1)E\_n-C^\mathrm\{T\} BC| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的根 $\displaystyle t\_1,\cdots,t\_n$ 都大于 $\displaystyle 1$. 于是 $\displaystyle |\lambda E\_n-C^\mathrm\{T\} BC|=0$ 的根 $\displaystyle t\_1-1, \cdots, t\_n-1$ 都大于 $\displaystyle 0$. 既然实对称矩阵 $\displaystyle C^\mathrm\{T\} BC$ 的特征值都大于 $\displaystyle 0$, 而它正定. 进一步, $\displaystyle B$ 与 $\displaystyle C^\mathrm\{T\} BC$ 合同, 而 $\displaystyle B$ 也正定.跟锦数学微信公众号. [在线资料](
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1153、 2、 $\displaystyle A$ 为 $\displaystyle n$ 阶实矩阵, 且 $\displaystyle A^3=2A+4E$, 求所有满足条件的 $\displaystyle A$. (南方科技大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle x^3-2x-4=(x-2)(x^2+2x+2)$ 知 $\displaystyle A$ 的最小多项式 (1)、 $\displaystyle m(x)=x-2\Rightarrow A=2E\_n$. (2)、 $\displaystyle m(x)=x^2+2x+2$, $\displaystyle A$ 的有理标准型为
\begin\{aligned\} Q=\mathrm\{diag\}(D,\cdots,D), D=\left(\begin\{array\}\{cccccccccccccccccccc\}0&-2\\\\ 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\}
故 $\displaystyle A=PQP^\{-1\}$, 其中 $\displaystyle P$ 为任意 $\displaystyle n$ 阶可逆实矩阵. (3)、 $\displaystyle m(x)=(x-2)(x^2+2x+2)$, $\displaystyle A$ 的有理标准形为
\begin\{aligned\} Q=\mathrm\{diag\}(2E\_r, D,\cdots,D). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A=PQP^\{-1\}$, 其中 $\displaystyle P$ 为任意 $\displaystyle n$ 阶可逆实矩阵.跟锦数学微信公众号. [在线资料](
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1154、 3、 $\displaystyle B$ 是 $\displaystyle n$ 阶可逆实方阵, $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}E&B\\\\ B^\mathrm\{T\}&0\end\{array\}\right)$. 求 $\displaystyle A$ 的正、负和零特征值的个数. (南方科技大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 我们给出一个常用的行列式计算公式. 设 $\displaystyle A,B,C,D$ 为同级方阵且 $\displaystyle AC=CA$, 则
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \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, (1-1)、 若 $\displaystyle A$ 可逆, 则由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\} E&0\\\\ -CA^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} A&B\\\\ C&D\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} E&-A^\{-1\}B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\} A&0\\\\ 0&D-CA^\{-1\}B\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\} A&B\\\\ C&D\end\{array\}\right|&=|A|\cdot |D-CA^\{-1\}B|\\\\ &=|A|\cdot |D-A^\{-1\}CB|\left(AC=CA\Rightarrow CA^\{-1\}=A^\{-1\}C\right)\\\\ &=|A(D-A^\{-1\}CB|\left(|AB|=|A|\cdot |B|\right)\\\\ &=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 若 $\displaystyle A$ 奇异, 则关于 $\displaystyle \lambda$ 的多项式 $\displaystyle |\lambda E+A|=0$ 至多有 $\displaystyle n$ 个复根 $\displaystyle \lambda\_1,\cdots,\lambda\_n$. 而 \begin\{aligned\} &\forall\ \lambda\not\in \left\\{\lambda\_1,\cdots,\lambda\_n\right\\}, |\lambda E+A|\neq 0\\\\ \Rightarrow&\tilde\{A\}=\lambda E+A\mbox\{可逆, 满足\}\tilde\{A\} C=C\tilde\{A\}\\\\ \Rightarrow&\left|\begin\{array\}\{cccccccccc\} \tilde\{A\}&B\\\\ C&D\end\{array\}\right|=|\tilde\{A\} D-CB|\left(\mbox\{由 (1)\}\right)\\\\ \Rightarrow& \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|=|(\lambda E+A)D-CB|. \qquad(210226: eq)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f(\lambda)\equiv \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|-|(\lambda E+A)D-CB|$. 若 $\displaystyle f(\lambda)\not\equiv 0$, 则 $\displaystyle f(\lambda)$ 是次数 $\displaystyle \leq n$ 的多项式, 至多有 $\displaystyle n$ 个复根. 这与 (210226: eq) 矛盾. 故
\begin\{aligned\} &f(\lambda)\equiv 0, \forall\ \lambda\in\mathbb\{C\}\\\\ \Rightarrow&f(0)=0\Rightarrow \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 由第 1 步知
\begin\{aligned\} |\lambda E-A|=\left|\begin\{array\}\{cccccccccc\}(\lambda-1)E&-B\\\\ -B^\mathrm\{T\}&\lambda E\end\{array\}\right|=|\lambda(\lambda-1)E-B^\mathrm\{T\} B|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle B$ 可逆知 $\displaystyle Bx=0$ 只有零解, 而
\begin\{aligned\} x\neq 0\Rightarrow y=Bx\neq 0\Rightarrow x^\mathrm\{T\} (B^\mathrm\{T\} B)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 $\displaystyle B^\mathrm\{T\} B$ 正定, 特征值 $\displaystyle \mu\_1,\cdots,\mu\_n$ 都为正实数. 于是 $\displaystyle A$ 的特征值为
\begin\{aligned\} \lambda^2-\lambda=\mu\_i\Leftrightarrow \lambda=\frac\{1\pm\sqrt\{1+4\mu\_i\}\}\{2\}, 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\}
故 $\displaystyle A$ 的正特征值的个数为 $\displaystyle n$, 负特征值的个数为 $\displaystyle n$, 零特征值的个数为 $\displaystyle 0$.跟锦数学微信公众号. [在线资料](
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1155、 4、 若复方阵 $\displaystyle A,B$ 满足 $\displaystyle B^2=A$, 则称 $\displaystyle B$ 是 $\displaystyle A$ 的一个平方根. (1)、 举一个 $\displaystyle 2$ 阶复方阵无平方根的例子; (2)、 证明: 所有二阶可逆复方阵一定有平方根; (3)、 $\displaystyle n$ 阶 ($n\geq 3$) 可逆复方阵是否一定有平方根? 说明理由. (南方科技大学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) / 二阶复方阵 $\displaystyle J\_2(0)$ 没有平方根. 这可用反证法证得. 若存在 $\displaystyle B$ 使得 $\displaystyle B^2=J\_2(0)$, 则 $\displaystyle B$ 的特征值全为 $\displaystyle 0$, $\displaystyle B$ 的 Jordan 标准形为 $\displaystyle 0$ 或 $\displaystyle J\_2(1)$. 从而 $\displaystyle B^2=0$. 这与 $\displaystyle B^2=J\_2(0)$ 矛盾. 故有结论. 我们分三步来证明第 3 问, 而第 2 问自然得证.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. 则 (1)、 $\displaystyle (J\_n(\lambda))^k$ 的若当标准形为 $\displaystyle J\_n(\lambda^k)$. (2)、 $\displaystyle J\_n(\lambda)$ 有 $\displaystyle k$ 次方根, 即存在 $\displaystyle n$ 阶复方阵 $\displaystyle B$ 使得 $\displaystyle B^k=J\_n(\lambda)$; (3)、 任意 $\displaystyle n$ 阶可逆复方阵 $\displaystyle A$ 都有 $\displaystyle k$ 次方根. 事实上, (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\}
(2)、 由第 1 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 $\displaystyle B^k=J\_n(\lambda)$. (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\} B=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 $\displaystyle B^k=A$. 取 $\displaystyle k=2$ 即知第 3 问得证.跟锦数学微信公众号. [在线资料](
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1156、 6、 $\displaystyle A,B$ 是 $\displaystyle n$ 阶复方阵, 满足 $\displaystyle A+B=AB$. 求证: $\displaystyle AB=BA$. (南方科技大学2023年高等代数考研试题) [矩阵 ]
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\begin\{aligned\} &A+B=AB\Rightarrow (E-A)(E-B)=E \Rightarrow E-A\mbox\{可逆\}\\\\ \Rightarrow& (E-B)(E-A)=E\Rightarrow E-A-B+BA=E\\\\ \Rightarrow& AB=A+B=BA . \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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1157、 3、 (15 分) 设 $\displaystyle A\in M\_n(\mathbb\{R\})$. 已知 $\displaystyle A$ 的特征值为 $\displaystyle \lambda\_1,\cdots,\lambda\_n$, 求矩阵
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}A&4I\_n\\\\ I\_n&A\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的特征值. (南京大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 我们给出一个常用的行列式计算公式. 设 $\displaystyle A,B,C,D$ 为同级方阵且 $\displaystyle AC=CA$, 则
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \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, (1-1)、 若 $\displaystyle A$ 可逆, 则由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\} E&0\\\\ -CA^\{-1\}&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} A&B\\\\ C&D\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\} E&-A^\{-1\}B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\} A&0\\\\ 0&D-CA^\{-1\}B\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\} A&B\\\\ C&D\end\{array\}\right|&=|A|\cdot |D-CA^\{-1\}B|\\\\ &=|A|\cdot |D-A^\{-1\}CB|\left(AC=CA\Rightarrow CA^\{-1\}=A^\{-1\}C\right)\\\\ &=|A(D-A^\{-1\}CB|\left(|AB|=|A|\cdot |B|\right)\\\\ &=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 若 $\displaystyle A$ 奇异, 则关于 $\displaystyle \lambda$ 的多项式 $\displaystyle |\lambda E+A|=0$ 至多有 $\displaystyle n$ 个复根 $\displaystyle \lambda\_1,\cdots,\lambda\_n$. 而 \begin\{aligned\} &\forall\ \lambda\not\in \left\\{\lambda\_1,\cdots,\lambda\_n\right\\}, |\lambda E+A|\neq 0\\\\ \Rightarrow&\tilde\{A\}=\lambda E+A\mbox\{可逆, 满足\}\tilde\{A\} C=C\tilde\{A\}\\\\ \Rightarrow&\left|\begin\{array\}\{cccccccccc\} \tilde\{A\}&B\\\\ C&D\end\{array\}\right|=|\tilde\{A\} D-CB|\left(\mbox\{由 (1)\}\right)\\\\ \Rightarrow& \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|=|(\lambda E+A)D-CB|. \qquad(210226: eq)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f(\lambda)\equiv \left|\begin\{array\}\{cccccccccc\} \lambda E+A&B\\\\ C&D\end\{array\}\right|-|(\lambda E+A)D-CB|$. 若 $\displaystyle f(\lambda)\not\equiv 0$, 则 $\displaystyle f(\lambda)$ 是次数 $\displaystyle \leq n$ 的多项式, 至多有 $\displaystyle n$ 个复根. 这与 (210226: eq) 矛盾. 故
\begin\{aligned\} &f(\lambda)\equiv 0, \forall\ \lambda\in\mathbb\{C\}\\\\ \Rightarrow&f(0)=0\Rightarrow \left|\begin\{array\}\{cccccccccc\}A&B\\\\ C&D\end\{array\}\right|=|AD-CB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 由 Jordan 标准形理论知存在可逆阵 $\displaystyle P$ 使得
\begin\{aligned\} &P^\{-1\}AP=J=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star&\star\\\\ &\ddots&\star\\\\ &&\lambda\_n\end\{array\}\right)\\\\ \Rightarrow&P^\{-1\}\left\[(\lambda I\_n-A)^2-4I\_n\right\]P=\left(\begin\{array\}\{cccccccccccccccccccc\}(\lambda-\lambda\_1)^2-4&\star&\star\\\\ &\ddots&\star\\\\ &&(\lambda-\lambda\_n)^2-4\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\}
故
\begin\{aligned\} &\left|\lambda I\_\{2n\}-\left(\begin\{array\}\{cccccccccccccccccccc\}A&4I\_n\\\\ I\_n&A\end\{array\}\right)\right|=\left|\begin\{array\}\{cccccccccc\}\lambda I\_n-A&-4I\_n\\\\ -I\_n&\lambda I\_n-A\end\{array\}\right|\\\\ \overset\{\tiny\mbox\{第1步\}\}\{=\}&|(\lambda I\_n-A)^2-4I\_n| =\prod\_\{k=1\}^n [(\lambda-\lambda\_k)^2-4]. \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 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&4I\_n\\\\ I\_n&A\end\{array\}\right)$ 的特征值为 $\displaystyle \lambda\_k\pm 2, 1\leq k\leq n$.跟锦数学微信公众号. [在线资料](
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---
1158、 4、 (15 分) 设
\begin\{aligned\} \alpha=(1,-1,1,-1)^\mathrm\{T\},\quad \beta=(1,3,2,-1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求一个正定矩阵 $\displaystyle A$ 使得 $\displaystyle \beta=A\alpha$. (南京大学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) / 我们给出一般结果. 设 $\displaystyle \alpha,\beta$ 为 $\displaystyle n$ 维欧氏空间 $\displaystyle \mathbb\{R\}^n$ 中的两个非零列向量, 则 $\displaystyle \alpha^\mathrm\{T\} \beta > 0$ 的充分必要条件是存在正定矩阵 $\displaystyle A$ 使得 $\displaystyle \beta=A\alpha$. 事实上, (1)、 $\displaystyle \Leftarrow$: $\displaystyle \alpha^\mathrm\{T\} \beta=\alpha^\mathrm\{T\} A\alpha > 0$. (2)、 $\displaystyle \Rightarrow$: (2-1)、 先设 $\displaystyle \alpha$ 具有形式 $\displaystyle (|\alpha|,0,\cdots,0)^\mathrm\{T\}, \beta=(b\_1,\cdots,b\_n)^\mathrm\{T\}$, 则
\begin\{aligned\} 0 < \alpha^\mathrm\{T\} \beta=|\alpha|b\_1\Rightarrow \frac\{b\_1\}\{|\alpha|\}=|\alpha|b\_1\cdot \frac\{1\}\{|\alpha|^2\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令
\begin\{aligned\} A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{b\_1\}\{|\alpha|\}&\frac\{b\_2\}\{|\alpha|\}&\cdots&\frac\{b\_n\}\{|\alpha|\}\\\\ \frac\{b\_2\}\{|\alpha|\}&a&&\\\\ \vdots&&\ddots&\\\\ \frac\{b\_n\}\{|\alpha|\}&&&a\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a > 0$ 待定. 设 $\displaystyle \gamma=\frac\{(b\_2,\cdots,b\_n)^\mathrm\{T\}\}\{|\alpha|\}$, 则
\begin\{aligned\} A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{b\_1\}\{|\alpha|\}&\gamma^\mathrm\{T\}\\\\ \gamma&aE\_\{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\}
由
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}1&-\frac\{1\}\{a\}\gamma^\mathrm\{T\}\\\\ 0&E\_\{n-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{b\_1\}\{|\alpha|\}&\gamma^\mathrm\{T\}\\\\ \gamma&aE\_\{n-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ -\frac\{1\}\{a\}\gamma&E\_\{n-1\}\end\{array\}\right)\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{b\_1\}\{|\alpha|\}-\frac\{1\}\{a\}\gamma^\mathrm\{T\} \gamma&0\\\\ 0&aE\_\{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\}
知当 $\displaystyle a > \frac\{|\alpha|\}\{b\_1\} \gamma^\mathrm\{T\} \gamma$ 时, $\displaystyle A\_1$ 正定. 此时,
\begin\{aligned\} \beta=A\_1\left(\begin\{array\}\{cccccccccccccccccccc\}|\alpha|\\\\0\\\\\vdots\\\\0\end\{array\}\right) =|\alpha| \left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{b\_1\}\{|\alpha|\}\\\\ \vdots\\\\\frac\{b\_n\}\{|\alpha|\}\end\{array\}\right)=\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 一般的, 将 $\displaystyle \eta\_1=\frac\{\alpha\}\{|\alpha|\}$ 扩充为 $\displaystyle \mathbb\{R\}^n$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_n$. 令
\begin\{aligned\} P=\left(\begin\{array\}\{cccccccccccccccccccc\}\eta\_1^\mathrm\{T\}\\\\\vdots\\\\\eta\_n^\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\}
则 $\displaystyle P$ 正交, 且
\begin\{aligned\} P\alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}\eta\_1^\mathrm\{T\}\\\\\vdots\\\\\eta\_n^\mathrm\{T\}\end\{array\}\right)|\alpha|\eta\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}|\alpha|\\\\0\\\\\vdots\\\\0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 1 步知存在正定矩阵 $\displaystyle A\_1$ 使得
\begin\{aligned\} &A\_1\left(\begin\{array\}\{cccccccccccccccccccc\}|\alpha|\\\\0\\\\\vdots\\\\0\end\{array\}\right)=P\beta \Rightarrow A\_1P\alpha=P\beta\Rightarrow(P^\mathrm\{T\} A\_1P)\alpha=\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle A=P^\mathrm\{T\} A\_1P$, 则 $\displaystyle A$ 正定, 且 $\displaystyle \beta=A\alpha$. 利用 Gram-Schdmit 标准正交过程, 从
\begin\{aligned\} &\eta\_1=\frac\{\alpha\}\{|\alpha|\}=\left(\frac\{1\}\{2\},-\frac\{1\}\{2\},\frac\{1\}\{2\},-\frac\{1\}\{2\}\right)^\mathrm\{T\}, \xi\_2=(0,1,0,0)^\mathrm\{T\},\\\\ &\xi\_3=(0,0,1,0)^\mathrm\{T\}, \xi\_4=(0,0,0,1)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
获得 $\displaystyle \mathbb\{R\}^4$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_4$, 并设
\begin\{aligned\} P=\left(\begin\{array\}\{cccccccccccccccccccc\}\eta\_1^\mathrm\{T\}\\\\\vdots\\\\\eta\_4^\mathrm\{T\}\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2\}&-\frac\{1\}\{2\}&\frac\{1\}\{2\}&-\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\sqrt\{3\}\}&-\frac\{\sqrt\{3\}\}\{2\}&\frac\{1\}\{2\sqrt\{3\}\}&-\frac\{1\}\{2\sqrt\{3\}\}\\\\ -\frac\{1\}\{\sqrt\{6\}\}&0&\frac\{2\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{6\}\}\\\\ \frac\{1\}\{\sqrt\{2\}\}&0&0&\frac\{1\}\{\sqrt\{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\}
则
\begin\{aligned\} P\alpha=\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\0\\\\0\\\\0\end\{array\}\right), \quad P\beta=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{2\}\\\\\frac\{13\}\{2\sqrt\{3\}\}\\\\\frac\{2\}\{\sqrt\{6\}\}\\\\0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由已有结论的构造过程知取
\begin\{aligned\} A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{4\}&\frac\{13\}\{4\sqrt\{3\}\}&\frac\{1\}\{\sqrt\{6\}\}&0\\\\ \frac\{13\}\{4\sqrt\{3\}\}&255&&\\\\ \frac\{1\}\{\sqrt\{6\}\}&&255&\\\\ 0&&&255\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} A=P^\mathrm\{T\} A\_1P=\frac\{1\}\{6\}\left(\begin\{array\}\{cccccccccccccccccccc\} 3067&1029&-1009&1013\\\\ 1029&3035&1025&-1029\\\\ -1009&1025&3075&1009\\\\ 1013&-1029&1009&3067 \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 $\displaystyle A$ 正定, 且 $\displaystyle A\alpha=\beta$. 另解如下. 设 $\displaystyle A=I\_4-\frac\{\alpha\alpha^\mathrm\{T\}\}\{\alpha^\mathrm\{T\} \alpha\}+\frac\{\beta\beta^\mathrm\{T\}\}\{\alpha^\mathrm\{T\}\beta\}$, 则
\begin\{aligned\} A\alpha=\alpha-\frac\{\alpha^\mathrm\{T\} \alpha\}\{\alpha^\mathrm\{T\} \alpha\}\alpha+\frac\{\beta^\mathrm\{T\} \alpha\}\{\alpha^\mathrm\{T\} \beta\}\beta =\alpha-\alpha+\beta=\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle 0\neq x\in\mathbb\{R\}^n$,
\begin\{aligned\} x^\mathrm\{T\} Ax=&x^\mathrm\{T\} x-\frac\{(\alpha^\mathrm\{T\} x)^\mathrm\{T\} (\alpha^\mathrm\{T\} x)\}\{\alpha^\mathrm\{T\} \alpha\} +\frac\{(\beta^\mathrm\{T\} x)^\mathrm\{T\} (\beta^\mathrm\{T\} x)\}\{\alpha^\mathrm\{T\} \beta\}\\\\ =&\frac\{\alpha^\mathrm\{T\} \alpha\cdot x^\mathrm\{T\} x-|\alpha^\mathrm\{T\} x|^2\}\{\alpha^\mathrm\{T\} \alpha\} +\frac\{|\beta^\mathrm\{T\} x|^2\}\{\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\}
(1)、 若 $\displaystyle \alpha\parallel x$, 则 $\displaystyle \beta^\mathrm\{T\} \alpha=1 > 0\Rightarrow \beta^\mathrm\{T\} x\neq 0$, 而
\begin\{aligned\} x^\mathrm\{T\} Ax=\frac\{|\beta^\mathrm\{T\} x|^2\}\{\alpha^\mathrm\{T\} \beta\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \alpha\not\parallel x$, 则由 Schwarz 不等式等号成立的条件知
\begin\{aligned\} x^\mathrm\{T\} Ax\geq \frac\{\alpha^\mathrm\{T\} \alpha\cdot x^\mathrm\{T\} x-|\alpha^\mathrm\{T\} x|^2\}\{\alpha^\mathrm\{T\} \alpha\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle A$ 确实正定.跟锦数学微信公众号. [在线资料](
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---
1159、 5、 (15 分) 设 $\displaystyle A\in M\_\{m\times n\}(\mathbb\{R\}), \mathrm\{rank\} A=m < n$. 证明存在一个 $\displaystyle (n-m)\times n$ 矩阵 $\displaystyle B$ 使得 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 为 $\displaystyle n$ 阶可逆阵, 且 $\displaystyle AB^\mathrm\{T\}=0$. (南京大学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 $\displaystyle \mathrm\{rank\} A=m$ 知 $\displaystyle Ax=0$ 的基础解系有 $\displaystyle n-m$ 个线性无关的向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-m\}, \eta\_i\in\mathbb\{R\}^n$. 设 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}\eta\_1^\mathrm\{T\}\\\\\vdots\\\\\eta\_\{n-m\}\end\{array\}\right)\in M\_\{(n-m)\times n\}(\mathbb\{R\})$, 则设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1\\\\\vdots\\\\\alpha\_m\end\{array\}\right)$ 后,
\begin\{aligned\} A\eta\_j=0\Rightarrow \alpha\_i\eta\_j=0, \forall\ i,j. \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\} &\sum\_i x\_i\alpha\_i+\sum\_j y\_j\eta\_j^\mathrm\{T\}=0\\\\ \Rightarrow& 0=\left(\sum\_i x\_i\alpha\_i+\sum\_j y\_j\eta\_j^\mathrm\{T\}\right)\alpha\_k^\mathrm\{T\} =\left(\sum\_i x\_i\alpha\_i\right)\alpha\_k^\mathrm\{T\}\\\\ \Rightarrow&\left\Vert \sum\_i x\_i\alpha\_i\right\Vert ^2 =\sum\_i x\_i\alpha\_i \sum\_k x\_k\alpha\_k^\mathrm\{T\}=\sum\_k x\_k \left(\sum\_i x\_i\alpha\_i\right)\alpha\_k^\mathrm\{T\}=0\\\\ \Rightarrow&s\sum\_i x\_i\alpha\_i=0\Rightarrow x\_i=0, \forall\ i\Rightarrow y\_j=0, \forall\ j. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A\\\\B\end\{array\}\right)$ 可逆, 且 $\displaystyle AB^\mathrm\{T\}=A(\eta\_1,\cdots,\eta\_\{n-m\})=0$.跟锦数学微信公众号. [在线资料](
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---
1160、 1、 已知三阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&-2&6\\\\ -1&0&a\\\\ -1&-1&4\end\{array\}\right)$, $\displaystyle f(x)=|xE-A|$ 是 $\displaystyle A$ 的特征多项式, 且 $\displaystyle (x-1)^2$ 是 $\displaystyle A$ 的最小多项式. (1)、 求 $\displaystyle a$ 及 $\displaystyle f(x)$; (2)、 求 $\displaystyle A$ 的初等因子; (3)、 $\displaystyle A$ 是否与对角矩阵相似? 请说明理由. (南京航空航天大学大学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 $\displaystyle A$ 的最小多项式为 $\displaystyle (x-1)^2$ 知 $\displaystyle A$ 的特征值全为 $\displaystyle 1$. 又由 $\displaystyle A$ 是三阶矩阵知 $\displaystyle f(x)=(x-1)^3$. 于是 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&\\\\ &1&1\\\\ &&1\end\{array\}\right)$ [由最小多项式知 $\displaystyle J$ 一定含有 $\displaystyle J\_2(1)$, 从而另一个 Jordan 块为 $\displaystyle J\_1(1)$]. 这也表明 $\displaystyle A$ 不可对角化; $\displaystyle A$ 的初等因子为 $\displaystyle \lambda-1,(\lambda-1)^2$; $\displaystyle \mathrm\{rank\}(A-E)=1$. 由
\begin\{aligned\} A-E=\left(\begin\{array\}\{cccccccccccccccccccc\}-2&-2&6\\\\ -1&-1&a\\\\ -1&-1&3\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-3\\\\ 0&0&a-3\\\\ 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\}
即知 $\displaystyle a=3$.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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---
1161、 2、 已知矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1\\\\ a&2&1\\\\ 1&-1&b\end\{array\}\right)$ 有特征向量 $\displaystyle \beta=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\-1\end\{array\}\right)$. (1)、 求 $\displaystyle a,b$ 的值; (2)、 求可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP$ 为对角阵; (3)、 求 $\displaystyle A^\{2022\}$. (南京航空航天大学大学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\} \exists\ \lambda,\mathrm\{ s.t.\} A\beta=\lambda\beta \Rightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1+a\\\\-b\end\{array\}\right)=\lambda\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\-1\end\{array\}\right)\Rightarrow \lambda=0, a=-1, b=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,0$. 由
\begin\{aligned\} E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1\\\\ 0&0&0\\\\ 0&0&0 \end\{array\}\right), 0E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&1\\\\ 0&0&0 \end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 的属于特征值 $\displaystyle 1,0$ 的特征向量分别为
\begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\0 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\0\\\\1 \end\{array\}\right); \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\-1\\\\1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} P=(\xi\_1,\xi\_2,\xi\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-1\\\\ 1&0&-1\\\\ 0&1&1 \end\{array\}\right)\Rightarrow P^\{-1\}AP=\mathrm\{diag\}\left(1,1,0\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 由 $\displaystyle A\sim \mathrm\{diag\}(1,1,0)$ 知 $\displaystyle A$ 的最小多项式为 $\displaystyle (x-1)x$. 故
\begin\{aligned\} &A^2=A\Rightarrow A^k=A^\{k-2\}A^2=A^\{k-2\}A=A^\{k-1\}\left(\forall\ k\geq 2\right)\\\\ \Rightarrow& A^\{2022\}=A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1\\\\ -1&2&1\\\\ 1&-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\}
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---
1162、 (2)、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 满足 $\displaystyle A^4=E$, 证明: 在复数域 $\displaystyle \mathbb\{C\}$ 上 $\displaystyle A$ 一定可对角化; (南京航空航天大学大学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) / $\displaystyle A$ 的零化多项式 $\displaystyle x^4-1=(x-1)(x-\mathrm\{ i\})(x+1)(x+\mathrm\{ i\})$ 没有重根, 而可对角化.跟锦数学微信公众号. [在线资料](
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---
1163、 (3)、 设 $\displaystyle A,B$ 是两个 $\displaystyle n$ 阶矩阵, 且满足 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&\\\\ &B\end\{array\}\right)$ 在数域 $\displaystyle \mathbb\{P\}$ 上可对角化, 证明: 在数域 $\displaystyle \mathbb\{P\}$ 上, $\displaystyle A,B$ 均可对角化. (南京航空航天大学大学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) / (3-1)、 先证明一个结论. 设 $\displaystyle \left\\{A\_i\right\\}\_\{i\in I\}, \left\\{B\_i\right\\}\_\{i\in I\}$ 是数域 $\displaystyle \mathbb\{F\}$ 上两个矩阵集合, 称它们在 $\displaystyle \mathbb\{F\}$ 上相似: 如果存在 $\displaystyle \mathbb\{F\}$ 上与 $\displaystyle i\in I$ 无关的可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}A\_iP=B\_i, \forall\ i\in I. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设有数域 $\displaystyle \mathbb\{K\}, \mathbb\{F\}$, 且 $\displaystyle \mathbb\{F\}\subset \mathbb\{K\}$. 对数域 $\displaystyle \mathbb\{F\}$ 上两个矩阵集合 $\displaystyle \left\\{A\_i\right\\}\_\{i\in I\}, \left\\{B\_i\right\\}\_\{i\in I\}$, 如果它们在数域 $\displaystyle \mathbb\{K\}$ 上相似, 则它们在数域 $\displaystyle \mathbb\{F\}$ 上也相似. 事实上, $\displaystyle \forall\ i\in I$, 考虑 $\displaystyle M\_n(\mathbb\{K\})$ 的子空间
\begin\{aligned\} U\_\{i,\mathbb\{K\}\}=\left\\{T\in M\_n(\mathbb\{K\}); A\_iT=TB\_i\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle M\_n(\mathbb\{F\})$ 的子空间
\begin\{aligned\} U\_\{i,\mathbb\{F\}\}=\left\\{T\in M\_n(\mathbb\{F\}); A\_iT=TB\_i\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 $\displaystyle M\_n(\mathbb\{K\})$ 和 $\displaystyle M\_n(\mathbb\{F\})$ 分别表示数域 $\displaystyle \mathbb\{K\}$ 和数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 阶方阵构成的线性空间. 令
\begin\{aligned\} U\_\mathbb\{K\}=\bigcap\_\{i\in I\}U\_\{i,\mathbb\{K\}\}, U\_\mathbb\{F\}=\bigcap\_\{i\in I\}U\_\{i,\mathbb\{F\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由题意知 $\displaystyle U\_\mathbb\{K\}$ 包含了可逆矩阵. 又由所涉及的线性空间的维数都不超过 $\displaystyle n^2$. 因此 $\displaystyle U\_\mathbb\{K\}$ 和 $\displaystyle U\_\mathbb\{F\}$ 实际上可写成有限多个 $\displaystyle U\_\{i,\mathbb\{K\}\}$ 和 $\displaystyle U\_\{i,\mathbb\{F\}\}$ 的交集. 求 $\displaystyle U\_\{i,\mathbb\{K\}\}$ 与 $\displaystyle U\_\{i,\mathbb\{F\}\}$ 的基底实际上就是解线性方程组 $\displaystyle A\_iT=TB\_i$, 并且求它们的基础解系的步骤相同. 因此, 可以取到一组公共基底
\begin\{aligned\} T\_1,\cdots,T\_l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
考虑多项式
\begin\{aligned\} f(x\_1,\cdots,x\_l)=\det \left(\sum\_\{i=1\}^l t\_kT\_k\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由 $\displaystyle U\_\mathbb\{K\}$ 包含了可逆矩阵知
\begin\{aligned\} &\exists\ s\_1,\cdots,s\_l\in\mathbb\{K\},\mathrm\{ s.t.\} f(s\_1,\cdots,s\_n)\neq 0\\\\ \Rightarrow&\mbox\{$f$ 作为 $\displaystyle \mathbb\{F\}$ 上的多项式不是零多项式\}\\\\ \Rightarrow&\exists\ r\_1,\cdots,r\_l\in\mathbb\{F\},\mathrm\{ s.t.\} f(r\_1,\cdots,r\_l)\neq 0\\\\ \Rightarrow&\mbox\{取 $\displaystyle P=\sum\_\{k=1\}^n r\_kT\_k\in U\_\mathbb\{F\}$ 可逆, 而 $\displaystyle \forall\ i\in I, P^\{-1\}A\_iP=B\_i$\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-2)、 设 $\displaystyle C=\mathrm\{diag\}(A,B)$, 则 $\displaystyle C$ 的特征值就是 $\displaystyle A,B$ 的特征值全体. 由 $\displaystyle C$ 在 $\displaystyle \mathbb\{P\}$ 中可对角化知 $\displaystyle C$ (而 $\displaystyle A,B$) 的特征值都在 $\displaystyle \mathbb\{P\}$ 中. 再设 $\displaystyle A,B,C$ 的最小多项式分别为 $\displaystyle m\_A, m\_B, m\_C$. 将 $\displaystyle A,B$ 看成复矩阵, 则
\begin\{aligned\} &0=m\_C(C)=\mathrm\{diag\}\left(m\_C(A),m\_C(B)\right)\\\\ \Rightarrow& m\_C(A)=0, m\_C(B)=0\Rightarrow m\_A\mid m\_C, m\_B\mid m\_C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle C$ 可对角化知 $\displaystyle m\_C$ 在复数域上没有重根. 而由上式知 $\displaystyle m\_A, m\_B$ 在复数域上也没有重根, 而可对角化, 分别相似于
\begin\{aligned\} \mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \mathrm\{diag\}(\mu\_1,\cdots,\mu\_n), \lambda\_i, \mu\_j\in\mathbb\{P\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据第 i 步即知 $\displaystyle A,B$ 在 $\displaystyle \mathbb\{P\}$ 上也可对角化.跟锦数学微信公众号. [在线资料](
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---
1164、 7、 设 $\displaystyle A,B$ 是两个 $\displaystyle n$ 阶矩阵, 且 $\displaystyle AB=A+B$. 证明: (1)、 $\displaystyle AB=BA$; (2)、 $\displaystyle \lambda=1$ 不是 $\displaystyle A$ 的特征值; (3)、 若 $\displaystyle A$ 相似于对角阵, 则存在可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP, P^\{-1\}BP$ 同时为对角阵. (南京航空航天大学大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} &AB-A-B=0\Rightarrow (A-E)(B-E)=E\qquad(I)\\\\ \Rightarrow& (B-E)(A-E)=E\Rightarrow BA=A+B\xlongequal\{\tiny\mbox\{题设\}\} AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle (I)$ 知 $\displaystyle A-E$ 可逆, $\displaystyle |E-A|\neq 0$, 而 $\displaystyle \lambda=1$ 不是 $\displaystyle A$ 的特征值. (3)、
\begin\{aligned\} (I)\Rightarrow (A-E)(B-E)=E\Rightarrow B=E+(A-E)^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由题设, 存在可逆矩阵 $\displaystyle P$, 使得
\begin\{aligned\} &P^\{-1\}AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)\\\\ \Rightarrow&P^\{-1\}BP=P^\{-1\}\left\[E+(A-E)^\{-1\}\right\]P\\\\ &=\mathrm\{diag\} \left(1+\frac\{1\}\{\lambda\_1-1\},\cdots, 1+\frac\{1\}\{\lambda\_n-1\}\right). \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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---
1165、 (3)、 $\displaystyle 4$ 阶矩阵 $\displaystyle A$ 的特征值为 $\displaystyle -2,-1,1,2$, 则 $\displaystyle |A^\{-1\}|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (南京理工大学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) / $\displaystyle |A^\{-1\}|=|A|^\{-1\}=\frac\{1\}\{(-2)\cdot(-1)\cdot 1\cdot 2\}=\frac\{1\}\{4\}$.跟锦数学微信公众号. [在线资料](
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---
1166、 5、 (12 分) 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 满足 $\displaystyle A^2=E$. 证明:
\begin\{aligned\} \mathrm\{rank\}(E-A)+\mathrm\{rank\}(E+A)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南京理工大学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
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&0\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&A-E\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}2E&A-E\\\\ E-A&A-E\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}2E&0\\\\ E-A&(A-E)+\frac\{1\}\{2\}(E-A)(E-A)\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&A^2-E\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} &\mathrm\{rank\}(E-A)+\mathrm\{rank\}(E+A)=\mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)\\\\ =&\mathrm\{rank\}(E\_n)+\mathrm\{rank\}(A^2-E)=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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---
1167、 7、 (12 分) 证明: 实反对称矩阵的特征值为 $\displaystyle 0$ 或纯虚数. (南京理工大学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 $\displaystyle 0\neq \alpha$ 是实反对称矩阵 $\displaystyle B$ 的属于特征值 $\displaystyle \lambda$ 的特征向量, 则由 $\displaystyle B\alpha=\lambda \alpha$ 知
\begin\{aligned\} \alpha^\star B\alpha=\left\\{\begin\{array\}\{llllllllllll\} \alpha^\star \lambda \alpha=\lambda |\alpha|^2,\\\\ \alpha^\mathrm\{T\} B^\mathrm\{T\} \overline\{\alpha\} =-\alpha^\mathrm\{T\} B\overline\{\alpha\} =-\alpha^\mathrm\{T\} \overline\{B\alpha\} =-\alpha^\mathrm\{T\} \overline\{\lambda \alpha\}=\overline\{\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
\begin\{aligned\} (\lambda+\overline\{\lambda\})|\alpha|^2=0\Rightarrow \lambda+\overline\{\lambda\}=0\Rightarrow \lambda=0\mbox\{或纯虚数\}. \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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---
1168、 8、 (12 分) 设 $\displaystyle m\times n$ 阶矩阵 $\displaystyle A$ 的秩为 $\displaystyle r$, 证明: 存在 $\displaystyle m\times r$ 的列满秩矩阵 $\displaystyle F$ 和 $\displaystyle r\times n$ 的行满秩矩阵 $\displaystyle G$, 使得 $\displaystyle A=FG$. (南京理工大学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) / 一言以蔽之, 这就是矩阵的满秩分解. 由 $\displaystyle \mathrm\{rank\} A=r$ 知存在可逆矩阵 $\displaystyle M,N$ 使得
\begin\{aligned\} A=M\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&\\\\ &0\end\{array\}\right)N=M\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r\\\\ 0\end\{array\}\right)\cdot (E\_r,0)N. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle F=M\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r\\\\0\end\{array\}\right), G=(E\_r,0)N$, 则 $\displaystyle F$ 是秩为 $\displaystyle r$ 的列满秩矩阵, $\displaystyle G$ 是秩为 $\displaystyle r$ 的行满秩矩阵, 且 $\displaystyle A=FG$.跟锦数学微信公众号. [在线资料](
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---
1169、 10、 (20 分) 设 $\displaystyle A$ 为三阶实对称矩阵, 齐次线性方程组 $\displaystyle (A-E)X=0$ 有一个非零解 $\displaystyle (1,-1,-1)^\mathrm\{T\}$, 齐次线性方程组 $\displaystyle AX=0$ 有两个线性无关的解. (1)、 求齐次线性方程组 $\displaystyle AX=0$ 的通解. (2)、 求矩阵 $\displaystyle A$. (南京理工大学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) / (1)、 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 $\displaystyle 0$ 的特征向量 $\displaystyle X$ 满足
\begin\{aligned\} \alpha^\mathrm\{T\} X=0, \alpha=(1,-1,-1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
它的基础解系为
\begin\{aligned\} \beta=(1,1,0)^\mathrm\{T\}, \gamma= (1,0,1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就是 $\displaystyle \left\\{X; AX=0\right\\}$ 的一组基, 而是 $\displaystyle AX=0$ 的基础解系. 故 $\displaystyle AX=0$ 的通解为
\begin\{aligned\} k(1,1,0)^\mathrm\{T\}+l(1,0,1)^\mathrm\{T\}, \quad \forall\ k,l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由第 1 步及题设知
\begin\{aligned\} &P=(\alpha,\beta,\gamma)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ -1&1&0\\\\ -1&0&1\end\{array\}\right)\\\\ \Rightarrow& AP=P\mathrm\{diag\}(1,0,0)\Rightarrow A=\frac\{1\}\{3\}\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1\\\\ -1&1&1\\\\ -1&1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
1170、 5、 设 $\displaystyle A,B$ 均是正定矩阵. 证明: (1)、 方程 $\displaystyle |xA-B|=0$ 的根均大于 $\displaystyle 0$; (2)、 方程 $\displaystyle |xA-B|$ 的所有根都等于 $\displaystyle 1$ 当且仅当 $\displaystyle A=B$. (南京师范大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle A$ 正定知存在实可逆矩阵 $\displaystyle C$ 使得 $\displaystyle C^\mathrm\{T\} AC=E\_n$. 另外, 实对称矩阵 $\displaystyle C^\mathrm\{T\} BC$ 与正定矩阵 $\displaystyle B$ 合同, 而正定, 它的特征值全大于 $\displaystyle 0$, 即
\begin\{aligned\} &0=|xE\_n-C^\mathrm\{T\} BC|=|xC^\mathrm\{T\} AC-C^\mathrm\{T\} BC|=|C^\mathrm\{T\}|\cdot |xA-B|\cdot |C|\\\\ \Leftrightarrow& 0=|xA-B| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的根全大于 $\displaystyle 0$. (2)、 (2-1)、 $\displaystyle \Leftarrow$: 若 $\displaystyle A=B$, 则
\begin\{aligned\} &0=|xA-B|=|xB-B|=|(x-1)B|=(x-1)^n|B|\\\\ \Leftrightarrow& (x-1)^n=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
它的根全为 $\displaystyle 1$. (2-2)、 $\displaystyle \Rightarrow$: 若
\begin\{aligned\} |xA-B|=0\Leftrightarrow 0=|C^\mathrm\{T\}|\cdot |xA-B|\cdot |C| =|xE\_n-C^\mathrm\{T\} BC| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的根全为 $\displaystyle 1$, 则正定矩阵 $\displaystyle C^\mathrm\{T\} BC$ 的特征值全为 $\displaystyle 1$. 从而存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\mathrm\{T\} (C^\mathrm\{T\} BC)P=E\_n\Rightarrow C^\mathrm\{T\} BC=PP^\mathrm\{T\}=E\_n=C^\mathrm\{T\} AC\Rightarrow B=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](
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1171、 8、 证明: 不存在 $\displaystyle n$ 阶正交矩阵 $\displaystyle A,B$ 使得 $\displaystyle A^2=AB+B^2$. (南京师范大学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) / 用反证法. 若存在满足 $\displaystyle A^2=AB+B^2$ 的正交矩阵 $\displaystyle A,B$, 则右乘 $\displaystyle B^\mathrm\{T\}$ 得
\begin\{aligned\} A^2B^\mathrm\{T\}=A+B; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
左乘 $\displaystyle A^\mathrm\{T\}$ 得
\begin\{aligned\} A=B+A^\mathrm\{T\} B^2\Rightarrow A-B=A^\mathrm\{T\} B^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由于正交矩阵的转置还是正交矩阵, 正交矩阵的乘积还是正交矩阵, 我们可由上述两式知道 $\displaystyle A\pm B$ 都是正交矩阵:
\begin\{aligned\} E&=(A+ B)^\mathrm\{T\}(A- B) =(A^\mathrm\{T\} + B^\mathrm\{T\})(A+ B)\\\\ &=2E+ A^\mathrm\{T\} B+ B^\mathrm\{T\} A,\\\\ E&=(A- B)^\mathrm\{T\}(A- B) =(A^\mathrm\{T\} - B^\mathrm\{T\})(A- B)\\\\ &=2E- A^\mathrm\{T\} B- B^\mathrm\{T\} A . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上述两个式子加起来得到 $\displaystyle 2E=4E\Rightarrow 2E=0$. 这是一个矛盾. 故有结论. 另证如下. 左乘 $\displaystyle A^\mathrm\{T\}$, 右乘 $\displaystyle B^\mathrm\{T\}$ 得
\begin\{aligned\} AB^\mathrm\{T\} =E+A^\mathrm\{T\} B.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由迹的性质知
\begin\{aligned\} \mathrm\{tr\}(A^\mathrm\{T\} B)=\mathrm\{tr\}(BA^\mathrm\{T\})=\mathrm\{tr\}\left\[(BA^\mathrm\{T\})^\mathrm\{T\}\right\]=\mathrm\{tr\}(AB^\mathrm\{T\}), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle (I)$ 取迹后知 $\displaystyle 0=n$. 这是一个矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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1172、 4、 (20 分) 如果矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}4&-3&0\\\\ 2&-1&0\\\\ a&-1&1\end\{array\}\right)$ 与矩阵 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}4&-5&4\\\\ 2&-3&4\\\\ 1&-2&3\end\{array\}\right)$ 相似, 求实数 $\displaystyle a$ 的值. (南开大学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 $\displaystyle A,B$ 的特征值都为 $\displaystyle 2,1,1$. 由
\begin\{aligned\} E-A=&\left(\begin\{array\}\{cccccccccccccccccccc\}-3&3&0\\\\ -2&2&0\\\\ -a&1&0\end\{array\}\right),\\\\ E-B\to&\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-2\\\\ 0&1&-2\\\\ 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\} A\sim B\Rightarrow \mathrm\{rank\}(E-A)=\mathrm\{rank\}(E-B)=2\Leftrightarrow a\neq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
反之, 若 $\displaystyle a\neq 1$, 则 $\displaystyle A,B$ 的 Jordan 标准形都是 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}2&&\\\\ &1&1\\\\ &&1\end\{array\}\right)$, 而在 $\displaystyle \mathbb\{C\}$ 中相似, 进而在 $\displaystyle \mathbb\{R\}$ 中相似. 综上, $\displaystyle a\neq 1$.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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1173、 7、 (10 分) 是否存在矩阵 $\displaystyle A\in \mathbb\{Q\}^\{2\times 2\}$ 使得 $\displaystyle A^6=E$ (单位 矩阵), $\displaystyle A^3\neq E, A^2\neq E$, 且 $\displaystyle A$ 中的所有元素之和为 $\displaystyle 2023$? 如果存在, 请给出一个具体的例子; 如果不存在, 请说明理由. (南开大学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) / 存在. 比如
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1-\frac\{1\}\{2023\}&2023-1+\frac\{1\}\{2023\}\\\\ -\frac\{1\}\{2023\}&\frac\{1\}\{2023\}\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\}
其实可以构造所有元素和为任意非零数的满足题设的 $\displaystyle A$ 如下:
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1-\frac\{1\}\{a\}&a-1+\frac\{1\}\{a\}\\\\ -\frac\{1\}\{a\}&\frac\{1\}\{a\}\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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 13:11:35
