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张祖锦2023年数学专业真题分类70天之第57天

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发表于 2023-3-5 13:14:13 | 显示全部楼层 |阅读模式
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## 张祖锦2023年数学专业真题分类70天之第57天 --- 1289、 3、 (10 分) 设 $\displaystyle A,B,C$ 是 $\displaystyle n$ 阶方阵, 满足 $\displaystyle \mathrm\{rank\} A < \mathrm\{rank\} C$, 且 $\displaystyle BC=0$. 证明: 存在非零的 $\displaystyle n$ 维列向量 $\displaystyle X$, 使得 $\displaystyle AX=BX$. 这里 $\displaystyle \mathrm\{rank\} M$ 表示矩阵 $\displaystyle M$ 的秩. (中国矿业大学(徐州)2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先证明一个结论. 设 $\displaystyle A$ 为 $\displaystyle m\times n$ 矩阵, $\displaystyle B$ 为 $\displaystyle n\times s$ 矩阵, 证明: 若 $\displaystyle AB=0$, 则 \begin\{aligned\} \mathrm\{rank\} A+\mathrm\{rank\} B\leq n. \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, 设 $\displaystyle B=(\beta\_1,\cdots,\beta\_l)$, 它的极大无关组为 $\displaystyle \beta\_\{i\_1\},\cdots,\beta\_\{i\_r\}$, 其中 $\displaystyle r=\mathrm\{rank\} B$. 由 $\displaystyle AB=0$ 知 $\displaystyle \beta\_\{i\_j\}\left(1\leq j\leq r\right)$ 是 $\displaystyle Ax=0$ 的解. 于是 \begin\{aligned\} \left\\{\beta\_\{i\_1\},\cdots,\beta\_\{i\_r\}\right\\}\subset \left\\{x; Ax=0\right\\}\equiv V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle \dim V=n-\mathrm\{rank\} A$, 我们知 $\displaystyle r\leq n-\mathrm\{rank\} A\Rightarrow \mathrm\{rank\} A+\mathrm\{rank\} B\leq n$. (2)、 回到题目. 由 $\displaystyle BC=0$ 及第 1 步知 $\displaystyle \mathrm\{rank\} B+\mathrm\{rank\} C\leq n$. 再设 \begin\{aligned\} V\_1=\left\\{X; AX=0\right\\}, V\_2=\left\\{X; BX=0\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\} &\dim(V\_1\cap V\_2)=\dim V\_1+\dim V\_2-\dim(V\_1+V\_2)\\\\ \geq&\left\[n-\mathrm\{rank\} A\right\]+\left\[n-\mathrm\{rank\} B\right\]-n\\\\ =&n-\mathrm\{rank\} A-\mathrm\{rank\} B \geq \mathrm\{rank\} C-\mathrm\{rank\} A > 0. \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\} \exists\ 0\neq X\in V\_1\cap V\_2\Rightarrow AX=0=BX. \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)/ --- 1290、 4、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶方阵, 且 $\displaystyle |A|=0$. 证明: 存在 $\displaystyle n$ 阶非零矩阵 $\displaystyle B$, 使得 $\displaystyle AB=BA=0$. (中国矿业大学(徐州)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 $\displaystyle |A|=0$ 知 $\displaystyle AX=0$ 有非零解 $\displaystyle \alpha$. 同理, $\displaystyle |A^\mathrm\{T\}|=0$ 蕴含 $\displaystyle A^\mathrm\{T\} X=0$ 有非零解 $\displaystyle \beta$. 令 $\displaystyle B=\alpha\beta^\mathrm\{T\}$, 则 $\displaystyle B$ 非零, 且 \begin\{aligned\} AB=A\alpha\beta^\mathrm\{T\}=0\beta^\mathrm\{T\}=0, BA=\alpha\beta^\mathrm\{T\} A=\alpha(A^\mathrm\{T\}\beta)^\mathrm\{T\} =\alpha 0^\mathrm\{T\}=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)/ --- 1291、 5、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶正定矩阵, $\displaystyle B$ 是 $\displaystyle n$ 阶半正定矩阵. (1)、 (10 分) 证明: $\displaystyle A$ 与 $\displaystyle B$ 可同时合同于对角阵; (2)、 (5 分) 证明: $\displaystyle |A+B|\geq |A|+|B|$. (中国矿业大学(徐州)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 $\displaystyle A$ 正定知存在可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\mathrm\{T\} AP=E\_n$. 又 $\displaystyle B$ 半正定 $\displaystyle \Rightarrow P^\mathrm\{T\} BP$ 半正定, 而存在正交阵 $\displaystyle Q$, 使得 \begin\{aligned\} Q^\mathrm\{T\} P^\mathrm\{T\} BPQ=\varLambda =\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\geq 0 \stackrel\{|Q|^2=1\}\{\Rightarrow\}|P|^2\cdot |B|=\lambda\_1\cdots\lambda\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle T=PQ$, 则 $\displaystyle T^\mathrm\{T\} AT=E\_n, T^\mathrm\{T\} BT=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)$, 而 $\displaystyle A$ 与 $\displaystyle B$ 可同时合同于对角阵. 进一步, \begin\{aligned\} &|Q^\mathrm\{T\} P^\mathrm\{T\} (A+B)PQ|=|E\_n+\varLambda|=\prod\_\{k=1\}^n (1+\lambda\_k) \geq 1+\lambda\_1\cdots \lambda\_n\\\\ \stackrel\{|Q|^2=1\}\{\Rightarrow\}&|P|^2 |A+B|\geq 1+\lambda\_1\cdots \lambda\_n \Rightarrow |A+B|\geq |A|+|B|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1292、 7、 (15 分) 记全体实数为 $\displaystyle \mathbb\{R\}$, 设 $\displaystyle \mathbb\{R\}^3$ 上的线性变换 $\displaystyle \sigma$ 在基 \begin\{aligned\} \varepsilon\_1=(1,0,0)^\mathrm\{T\}, \varepsilon\_2=(1,1,0)^\mathrm\{T\}, \varepsilon\_3=(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\} 下的矩阵是 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&2-a\\\\ -3&-2&a-2\\\\ 1&1&1\end\{array\}\right)$. (1)、 (5 分) 若 $\displaystyle \sigma$ 有三个线性无关的特征向量, 求 $\displaystyle a$ 的值. (2)、 (10 分) 若 $\displaystyle \alpha=(2,3,-2)^\mathrm\{T\}$ 是 $\displaystyle \sigma$ 的一个特征向量, 证明 $\displaystyle A$ 不可对角化, 并求 $\displaystyle A$ 的 Jordan 标准形. (中国矿业大学(徐州)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)、 易知 $\displaystyle A$ 的特征值为 $\displaystyle 1,1,-1$. 由 $\displaystyle \sigma$ 有三个线性无关的特征向量知 $\displaystyle A$ 可对角化, 而 \begin\{aligned\} A\sim \mathrm\{diag\}(1,1,-1)\Rightarrow \mathrm\{rank\}(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 $\displaystyle A-E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&0&2-a\\\\ 0&0&0\end\{array\}\right)$ 即知 $\displaystyle a=2$. (2)、 由 \begin\{aligned\} (\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)=(e\_1,e\_2,e\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 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\} 知 \begin\{aligned\} \alpha=&(e\_1,e\_2,e\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\3\\\\-2\end\{array\}\right)=(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 0&1&1\\\\ 0&0&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}2\\\\3\\\\-2\end\{array\}\right)\\\\ =&(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\-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\} 由题设, $\displaystyle \exists\ t\in\mathbb\{R\},\mathrm\{ s.t.\}$ \begin\{aligned\} \sigma(\alpha)=&\sigma(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\-2\end\{array\}\right) =(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)A\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\-2\end\{array\}\right)\\\\ =&\lambda\alpha=\lambda(\varepsilon\_1,\varepsilon\_2,\varepsilon\_3)\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\-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\} \left(\begin\{array\}\{cccccccccccccccccccc\}2a-1\\\\ -2a-3\\\\ 2\end\{array\}\right)=A\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\-2\end\{array\}\right)=\lambda \left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\5\\\\2\end\{array\}\right)\Rightarrow a=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此时, \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&1\\\\ -3&-2&-1\\\\ 1&1&1\end\{array\}\right)\Rightarrow A-E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 0&0&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\} 故 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 ($J$ 的严格上三角部分一定有一个 $\displaystyle 1$, 否则 $\displaystyle \mathrm\{rank\}(J-E)=1$! 矛盾) \begin\{aligned\} \mathrm\{rank\}(J-E)=\mathrm\{rank\}(A-E)=2\Rightarrow J=\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\} 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)/ --- 1293、 8、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实矩阵, 且 $\displaystyle |A|\neq 0$. 证明: $\displaystyle A$ 可以分解为 $\displaystyle A=QT$, 其中 $\displaystyle Q$ 是正交矩阵, $\displaystyle T$ 是对角元全部大于 $\displaystyle 0$ 的上三角矩阵, 并证明分解唯一. (中国矿业大学(徐州)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 QR 分解的存在唯一性. (1)、 存在性. 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_n)$, 则由 $\displaystyle A$ 可逆知 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关. 将它们施行 Gram-Schmidt 标准正交化过程, \begin\{aligned\} \beta\_1&=\frac\{\alpha\_1\}\{|\alpha\_1|\}=\frac\{\alpha\_1\}\{k\_1\},\\\\ \beta\_2&=\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{|\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1|\}=\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{k\_2\},\cdots,\\\\ \beta\_n&=\frac\{\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\}\{|\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}|\}\\\\ &=\frac\{\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\}\{k\_n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此即 \begin\{aligned\} A=(\alpha\_1,\cdots,\alpha\_n)=(\beta\_1,\cdots,\beta\_n) \left(\begin\{array\}\{cccccccccccccccccccc\}k\_1&(\alpha\_2,\beta\_1)&\cdots&(\alpha\_n,\beta\_1)\\\\ &k\_2&\cdots&(\alpha\_n,\beta\_2)\\\\ &&\ddots&\vdots\\\\ &&&k\_n\end\{array\}\right)\equiv QT. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle Q$ 正交, $\displaystyle T$ 上三角, 且对角元大于 $\displaystyle 0$. (2)、 唯一性. 设 $\displaystyle A=QT=Q\_1T\_1$, 其中 $\displaystyle Q\_1$ 正交, $\displaystyle T\_1$ 上三角, 且对角元大于 $\displaystyle 0$, 则 \begin\{aligned\} Q\_1^\{-1\}Q=T\_1T^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 注意到正交 (上三角) 矩阵的逆还是正交 (上三角) 矩阵, 正交 (上三角) 矩阵的乘积还是正交 (上三角) 矩阵, 我们知上式左端是正交阵, 右端是上三角矩阵, 且对角元大于 $\displaystyle 0$. 作为正交阵的 $\displaystyle T\_1T^\{-1\}$ 的第一列是单位向量, 而它只有第一个元素非零且为正数, 而 $\displaystyle =1$. 从而 $\displaystyle T\_1T^\{-1\}$ 的第一行除了第一个元素外全为 $\displaystyle 0$. 继续看 $\displaystyle T\_1T^\{-1\}$ 的第二列. 除了第二个元素非零且为正数. 由于第二列为单位为单位向量, 而它的第二元素也 $\displaystyle =1$. 如此这般即知 $\displaystyle T\_1T^\{-1\}=E\_n\Rightarrow T\_1=T\Rightarrow Q\_1=Q$. 唯一性证得.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)/ --- 1294、 (5)、 设 $\displaystyle A,B$ 是 $\displaystyle n$ 阶可逆方阵, $\displaystyle A^\star$ 是 $\displaystyle A$ 的伴随矩阵, 则 $\displaystyle (AB)^\star=B^\star A^\star$. (中国人民大学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) / $\displaystyle \surd$. \begin\{aligned\} (AB)^\star=&|AB| (AB)^\{-1\} =|A| |B| B^\{-1\}A^\{-1\}\\\\ =&|B|B^\{-1\}\cdot |A|A^\{-1\}=B^\star A^\star. \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)/ --- 1295、 (6)、 对任意一个实矩阵 $\displaystyle A\_\{m\times n\}$, 都有 $\displaystyle E\_n+A^\mathrm\{T\} A$ 正定. (中国人民大学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) / $\displaystyle \surd$. 对 $\displaystyle \forall\ 0\neq x\in \mathbb\{R\}^n$, 令 $\displaystyle y=Ax$, 则 \begin\{aligned\} x^\mathrm\{T\} (E\_n+A^\mathrm\{T\} A)x=x^\mathrm\{T\} x+y^\mathrm\{T\} y\geq x^\mathrm\{T\} x > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle E\_n+A^\mathrm\{T\} A$ 正定.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1296、 (5)、 若 $\displaystyle A,B$ 是 $\displaystyle 3\times 2$ 和 $\displaystyle 2\times 3$ 矩阵, 且 $\displaystyle AB=\left(\begin\{array\}\{cccccccccccccccccccc\}3&0&3\\\\ 0&6&0\\\\ 3&0&3\end\{array\}\right)$, 则 $\displaystyle A$ 的秩 $\displaystyle \mathrm\{rank\} A=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国人民大学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\} 2=\mathrm\{rank\}(AB)\leq \mathrm\{rank\} A\leq 2\Rightarrow \mathrm\{rank\} A=2. \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)/ --- 1297、 7、 设 $\displaystyle A$ 是 $\displaystyle n\times m$ 矩阵, $\displaystyle B$ 是 $\displaystyle m\times n$ 矩阵. 如果 $\displaystyle E\_n-AB$ 是可逆矩阵, 试讨论 $\displaystyle E\_m-BA$ 的可逆性. (中国人民大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E\_m&0\\\\ -A&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E\_m&B\\\\ A&E\_n\end\{array\}\right)&=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_m&B\\\\ 0&E\_n-AB\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E\_m&-B\\\\ 0&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E\_m&B\\\\ A&E\_m\end\{array\}\right)&=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_m-BA&0\\\\ A&E\_n\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} &|E\_m-BA|=\left|\begin\{array\}\{cccccccccc\}E\_m-BA&0\\\\ A&E\_n\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\}E\_m&B\\\\ A&E\_m\end\{array\}\right|\\\\ =&\left|\begin\{array\}\{cccccccccc\}E\_m&B\\\\ 0&E\_n-AB\end\{array\}\right|=|E\_n-AB|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle E\_n-AB$ 可逆 $\displaystyle \Leftrightarrow |E\_n-AB|\neq 0\Leftrightarrow |E\_m-BA|\neq 0\Leftrightarrow E\_m-BA$ 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)/ --- 1298、 8、 解答如下问题. (1)、 证明: 实对称矩阵正定的充分必要条件是所有顺序主子式都大于 $\displaystyle 0$. (中国人民大学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, 实对称矩阵正定的充要条件是所有顺序主子式都大于零. (1-1)、 $\displaystyle \Rightarrow$: 设 $\displaystyle A$ 的 $\displaystyle k$ 阶顺序主子式为 $\displaystyle A\_k$, 则对 $\displaystyle \forall\ 0\neq x=(x\_1,\cdots,x\_k)^\mathrm\{T\}\in\mathbb\{R\}^k$, \begin\{aligned\} x^\mathrm\{T\} A\_kx=\sum\_\{i,j=1\}^k a\_\{ij\}x\_ix\_j =(x,0\_\{1,n-k\})^\mathrm\{T\} A\left(\begin\{array\}\{cccccccccccccccccccc\}x\\\\0\_\{n-k,1\}\end\{array\}\right) > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A\_k$ 正定, 行列式大于 $\displaystyle 0$. (1-2)、 $\displaystyle \Leftarrow$: 当 $\displaystyle n=1$ 时, $\displaystyle a\_\{11\} > 0\Rightarrow A$ 正定, 结论自明. 假设结论对 $\displaystyle n-1$ 阶实对称矩阵成立, 则对 $\displaystyle n$ 阶矩阵 $\displaystyle A$, 设 \begin\{aligned\} B=\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{ij\}\end\{array\}\right)\_\{1\leq i,j\leq n-1\}, \alpha=(a\_\{1n\},\cdots,a\_\{n-1,n\})^\mathrm\{T\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}B&\alpha\\\\ \alpha^\mathrm\{T\}&a\_\{nn\}\end\{array\}\right)$. 既然 $\displaystyle A$ 的顺序主子式大于 $\displaystyle 0$, 而 $\displaystyle B$ 的顺序主子式也大于 $\displaystyle 0$. 由归纳假设, $\displaystyle B$ 正定, 而存在可逆阵 $\displaystyle P$ 使得 $\displaystyle P^\mathrm\{T\} BP=E\_\{n-1\}$. 令 $\displaystyle Q\_1=\mathrm\{diag\}(P,1)$, 则 \begin\{aligned\} Q\_1^\mathrm\{T\} AQ\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_\{n-1\}&P^\mathrm\{T\} \alpha\\\\ \alpha^\mathrm\{T\} P&a\_\{nn\}\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\} 令 $\displaystyle Q\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_\{n-1\}&-P^\mathrm\{T\} \alpha\\\\ 0&1\end\{array\}\right)$, 则 \begin\{aligned\} Q\_2^\mathrm\{T\} Q\_1^\mathrm\{T\} AQ\_1Q\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}E\_\{n-1\}&\\\\ &a\_\{nn\}-\alpha^\mathrm\{T\} PP^\mathrm\{T\} \alpha\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\} 令 $\displaystyle Q=Q\_1Q\_2, a=a\_\{nn\}-\alpha^\mathrm\{T\} PP^\mathrm\{T\} \alpha$, 则 \begin\{aligned\} Q^\mathrm\{T\} AQ=\mathrm\{diag\}(E\_\{n-1\},a)\stackrel\{|A| > 0\}\{\Rightarrow\}a > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle Q\_3=\mathrm\{diag\}\left(E\_\{n-1\},\frac\{1\}\{\sqrt\{a\}\}\right)$, 则 $\displaystyle Q\_3^\mathrm\{T\} QAQQ\_3=E\_n$. 这表明 $\displaystyle A$ 合同于单位矩阵, 是正定的.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)/ --- 1299、 (2)、 判断: 矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{1\}\{i+j-1\}\end\{array\}\right)\_\{n\times n\}$ 的正定性. (中国人民大学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) / 对 $\displaystyle \forall\ x=(x\_1,\cdots,x\_n)^\mathrm\{T\}\neq 0$, \begin\{aligned\} x^\mathrm\{T\} Ax=&\sum\_\{i,j=1\}^n \frac\{1\}\{i+j-1\}x\_ix\_j =\sum\_\{i,j=1\}^n \int\_0^1 t^\{i+j-1\}\mathrm\{ d\} t\cdot x\_ix\_j\\\\ =&\int\_0^1 \left(\sum\_\{i=1\}^n t^\{i-\frac\{1\}\{2\}\}x\_i\right)^2\mathrm\{ d\} t > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 最后一步是因为 $\displaystyle t^\frac\{1\}\{2\},\cdots,t^\{n-\frac\{1\}\{2\}\}$ 线性无关, 而 \begin\{aligned\} x\neq 0\Rightarrow \sum\_\{i=1\}^n x\_i t^\{i-\frac\{1\}\{2\}\}\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这就证明了 $\displaystyle A$ 正定.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1300、 5、 (16 分) 设 $\displaystyle A\_1,A\_2,B\_1,B\_2$ 均是 $\displaystyle n$ 阶实矩阵, 且 $\displaystyle A\_2,B\_2$ 可逆. 证明: 存在可逆 矩阵 $\displaystyle P,Q$ 使得 $\displaystyle B\_1=PA\_1Q, B\_2=PA\_2Q$ 的充分必要条件是 $\displaystyle A\_1A\_2^\{-1\}$ 与 $\displaystyle B\_1B\_2^\{-1\}$ 相似. (中南大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: \begin\{aligned\} &\left\\{\begin\{array\}\{llllllllllll\}B\_1=PA\_1Q\\\\ B\_2=PA\_2Q\Rightarrow B\_2^\{-1\}=Q^\{-1\}A\_2^\{-1\}P^\{-1\}\end\{array\}\right.\\\\ \Rightarrow&B\_1B\_2^\{-1\}=PA\_1A\_2^\{-1\}P^\{-1\}\Rightarrow A\_1A\_2^\{-1\}\sim B\_1B\_2^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle \Leftarrow$: 由 $\displaystyle A\_1A\_2^\{-1\}\sim B\_1B\_2^\{-1\}$ 知存在可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} B\_1B\_2^\{-1\}=P(A\_1A\_2^\{-1\})P^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle Q=A\_2^\{-1\}P^\{-1\}B\_2$, 则 $\displaystyle Q$ 可逆, 且 \begin\{aligned\} &B\_1=PA\_1(A\_2^\{-1\}P^\{-1\}B\_2)=PA\_1Q,\\\\ &PA\_2Q=PA\_2\cdot A\_2^\{-1\}P^\{-1\}B\_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\} 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)/ --- 1301、 6、 (20 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实对称正定矩阵, $\displaystyle B$ 是 $\displaystyle n$ 阶反对称矩阵. 证明: $\displaystyle \det(A+B) > 0$. (中南大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 一言以蔽之, 实正定矩阵与实反对称矩阵的和的行列式大于零. (1)、 设 $\displaystyle \lambda$ 是反对称实矩阵 $\displaystyle B$ 的特征值, 则存在 $\displaystyle \alpha\neq 0$ 使 \begin\{aligned\} &B \alpha =\lambda\alpha \\\\ \Rightarrow& \lambda \overline\{\alpha\} ^\mathrm\{T\}\alpha =\overline\{\alpha\} ^\mathrm\{T\} B \alpha =-\overline\{\alpha\} ^\mathrm\{T\} B ^\mathrm\{T\}\alpha =-\left(B \overline\{\alpha\} \right)^\mathrm\{T\}\alpha =-\left(\overline\{B \alpha \}\right)^\mathrm\{T\}\alpha =-\overline\{\lambda\} \overline\{\alpha\}^\mathrm\{T\}\alpha \\\\ \Rightarrow& \lambda=-\overline\{\lambda\} \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\} (2)、 由 $\displaystyle A$ 正定知存在可逆矩阵 $\displaystyle P$ 使得 $\displaystyle P^\mathrm\{T\} AP=E\_n$. 由第 1 步知可设实反对称矩阵 $\displaystyle P^\mathrm\{T\} BP$ 的特征值为 \begin\{aligned\} \underbrace\{0,\cdots,0\}\_r, \pm b\_1\mathrm\{ i\}, \cdots, \pm b\_s \mathrm\{ i\}\left(0\neq b\_i\in\mathbb\{R\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle E+P^\mathrm\{T\} BP$ 的特征值为 \begin\{aligned\} \underbrace\{1,\cdots,1\}\_r, 1\pm b\_1\mathrm\{ i\}, \cdots, 1\pm b\_s \mathrm\{ i\}\left(0\neq b\_i\in\mathbb\{R\}\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\} |P^\mathrm\{T\}(A+B)P|=|E+P^\mathrm\{T\} BP|=\prod\_\{i=1\}^s (1+b\_i^2) > 0 \Rightarrow |A+B| > 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)/ --- 1302、 2、 (15 分) 已知矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}4&1&0&-1\\\\ 0&3&0&1\\\\ 0&0&4&0\\\\ 1&0&0&5\end\{array\}\right)$. (1)、 求 $\displaystyle A$ 的特征多项式 $\displaystyle f\_A(x)$; (2)、 求 $\displaystyle A$ 的最小多项式 $\displaystyle m\_A(x)$; (3)、 求 $\displaystyle A$ 的若尔当标准形. (中山大学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\} &\lambda E-A=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-4&-1&0&1\\\\ 0&\lambda-3&0&-1\\\\ 0&0&\lambda-4&0\\\\ -1&0&0&\lambda-5\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-4&&&\\\\ &\lambda-4&-1&1\\\\ &0&\lambda-3&-1\\\\ &-1&0&\lambda-5\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-4&&&\\\\ &0&-1&\lambda^2-9\lambda+21\\\\ &0&\lambda-3&-1\\\\ &-1&0&\lambda-5\end\{array\}\right) \to\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-4&&&\\\\ &0&-1&\lambda^2-9\lambda+21\\\\ &0&0&(\lambda-4)^3\\\\ &1&0&0\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-4&&&\\\\ &0&1&0\\\\ &0&0&(\lambda-4)^3\\\\ &1&0&0\end\{array\}\right)\to \mathrm\{diag\}\left(1,1,\lambda-4,(\lambda-4)^3\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的不变因子为 $\displaystyle 1,1,\lambda-4,(\lambda-4)^3$, 而初等因子为 $\displaystyle \lambda-4,(\lambda-4)^3$, Jordan 标准形为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}4&1&&\\\\ &4&1&\\\\ &&4&\\\\ &&&4\end\{array\}\right)$, 最小多项式为 $\displaystyle (\lambda-4)^3$, 特征多项式为 $\displaystyle (\lambda-4)^4$.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)/ --- 1303、 6、 (15 分) 设 $\displaystyle A,B$ 是 $\displaystyle n$ 阶复方阵, 证明: $\displaystyle A,B$ 有公共特征值当且仅当存在 $\displaystyle n$ 阶非零复方阵 $\displaystyle X$, 使得 $\displaystyle AX=XB$. (中山大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle \Rightarrow$: 设 $\displaystyle \lambda$ 是 $\displaystyle A,B$ 的公共特征值, 则由 $\displaystyle |\lambda E-B|=|\lambda E-B^\mathrm\{T\}|$ 知 $\displaystyle \lambda$ 也是 $\displaystyle B^\mathrm\{T\}$ 的特征值. 设 $\displaystyle 0\neq \alpha\in\mathbb\{C\}^n, 0\neq \beta\in\mathbb\{C\}^n$ 分别是 $\displaystyle A,B^\mathrm\{T\}$ 的属于特征值 $\displaystyle \lambda$ 的特征向量, 令 $\displaystyle X=\alpha\beta^\mathrm\{T\}\neq 0$, 则 \begin\{aligned\} AX=A\alpha\beta^\mathrm\{T\}=\lambda \alpha\beta^\mathrm\{T\}=\alpha(\lambda \beta)^\mathrm\{T\} =\alpha(B^\mathrm\{T\} \beta)^\mathrm\{T\}=XB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 设 $\displaystyle 0\neq X$ 使得 $\displaystyle AX=XB$, 则由 \begin\{aligned\} A^kX=XB^k\Rightarrow A^\{k+1\}X=A\cdot A^kX=AXB^k=XB^\{k+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及数学归纳法知 $\displaystyle A^kX=XB^k, \forall k\geq 1$. 令 $\displaystyle f(\lambda)=|\lambda E-A|=\prod\_\{i=1\}^n (\lambda-\lambda\_i)$ 是 $\displaystyle A$ 的特征多项式, 则 \begin\{aligned\} 0=0X=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 $\displaystyle X\neq 0$ 知 $\displaystyle f(B)$ 不可逆, \begin\{aligned\} 0=|f(B)|=\prod\_\{i=1\}^n |B-\lambda\_iE|\Rightarrow \exists\ k,\mathrm\{ s.t.\} |B-\lambda\_kE|=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此 $\displaystyle \lambda\_k$ 就是 $\displaystyle A,B$ 的公共特征值.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)/ --- 1304、 7、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, $\displaystyle A^2=A$. 证明: $\displaystyle \mathrm\{rank\} A+\mathrm\{rank\}(A-I)=n$. (中山大学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&\\\\ &A-E\end\{array\}\right)\to&\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ &A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ -A&-E\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}A-A^2&\\\\ -A&-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A-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\}(A)+\mathrm\{rank\}(A-E)=\mathrm\{rank\}(A-A^2)+n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} \mathrm\{rank\}(A)+\mathrm\{rank\}(A-E)=n&\Leftrightarrow \mathrm\{rank\}(A-A^2)=0\\\\ &\Leftrightarrow A-A^2=0\Leftrightarrow A=A^2. \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)/ --- 1305、 8、 (15 分) 设 $\displaystyle A$ 为 $\displaystyle n$ 阶复方阵, $\displaystyle \lambda$ 为 $\displaystyle A$ 的特征值. 假设存在正数 $\displaystyle M$, 使得对任意正整数 $\displaystyle p$, $\displaystyle A^p$ 的所有元素模长都小于 $\displaystyle M$, 证明: $\displaystyle |\lambda|\leq 1$, 且当 $\displaystyle \lambda=1$ 时, $\displaystyle \lambda$ 的几何重数等于代数重数. (中山大学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 Jordan 标准形理论, 存在可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} &P^\{-1\}AP=J=\mathrm\{diag\}\left(J\_\{n\_1\}(\lambda\_1),\cdots,J\_\{n\_s\}(\lambda\_s)\right)\\\\ \Rightarrow&P^\{-1\}A^pP=\mathrm\{diag\}\left(J\_\{n\_1\}^p(\lambda\_1),\cdots,J\_\{n\_s\}^p(\lambda\_s)\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中各 $\displaystyle \lambda\_i\neq 0$, $\displaystyle J\_\{n\_i\}(\lambda\_i)$ 是以 $\displaystyle \lambda\_i$ 为对角元的 $\displaystyle n\_i$ 阶 Jordan 块. 设 $\displaystyle P, P^\{-1\}$ 各元素的模长的上界为 $\displaystyle c$, 则由矩阵乘法知 $\displaystyle P^\{-1\}A^pP$ 的元素 \begin\{aligned\} |\lambda\_i^p|=|\left(P^\{-1\}A^pP\right)\_\{ii\}| =\left|\sum\_\{k,l\}(P^\{-1\})\_\{ik\}(A^p)\_\{kl\}(P)\_\{li\}\right| \leq n^2 c^2 M, \forall\ p\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle |\lambda\_i|\leq 1$. 若 $\displaystyle \lambda=1$, 往用反证法证明 $\displaystyle J$ 中以 $\displaystyle \lambda$ 为对角元的 Jordan 块的阶数都为 $\displaystyle 1$, 而 $\displaystyle \lambda$ 的几何重数等于代数重数. 若不然, $\displaystyle J$ 中含有 \begin\{aligned\} J\_m(1)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&&\\\\ &\ddots&\ddots&\\\\ &&\ddots&1\\\\ &&&1\end\{array\}\right), m\geq 2\Rightarrow J\_m^p(1)=\left(\begin\{array\}\{cccccccccccccccccccc\}1&p&\star&\star\\\\ &\ddots&\ddots&\star\\\\ &&\ddots&p\\\\ &&&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\} 这与 $\displaystyle P^\{-1\}A^pP=J^p$ 的元素的模长一致有界矛盾. 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)/ --- 1306、 4、 (12 分) 设 $\displaystyle A,B$ 是同解方阵且有公共特征值. 证明: 矩阵方程 $\displaystyle AX=XB$ 有非零解. (重庆大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \lambda$ 是 $\displaystyle A,B$ 的公共特征值, 则由 $\displaystyle |\lambda E-B|=|\lambda E-B^\mathrm\{T\}|$ 知 $\displaystyle \lambda$ 也是 $\displaystyle B^\mathrm\{T\}$ 的特征值. 设 $\displaystyle \alpha\neq 0, \beta\neq 0$ 分别是 $\displaystyle A,B^\mathrm\{T\}$ 的属于特征值 $\displaystyle \lambda$ 的特征向量. 令 $\displaystyle X=\alpha\beta^\mathrm\{T\}\neq 0$, 则 \begin\{aligned\} AX=A\alpha\beta^\mathrm\{T\}=\lambda \alpha\beta^\mathrm\{T\} =\alpha(\lambda \beta)^\mathrm\{T\}=\alpha (B^\mathrm\{T\} \beta)^\mathrm\{T\} =\alpha\beta^\mathrm\{T\} B=XB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle AX=XB$ 确有非零解.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)/ --- 1307、 5、 (15 分) 设 $\displaystyle A\in M\_n(\mathbb\{R\})$ 且 $\displaystyle \det A\neq 0$. 证明: 存在唯一的 $\displaystyle Q$ 和 $\displaystyle T$ 使得 $\displaystyle A=QT$, 其中 $\displaystyle Q$ 为正交矩阵, $\displaystyle T$ 为上三角矩阵且对角线元素大于 $\displaystyle 0$. (重庆大学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 QR 分解的存在唯一性. (1)、 存在性. 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_n)$, 则由 $\displaystyle A$ 可逆知 $\displaystyle \alpha\_1,\cdots,\alpha\_n$ 线性无关. 将它们施行 Gram-Schmidt 标准正交化过程, \begin\{aligned\} \beta\_1&=\frac\{\alpha\_1\}\{|\alpha\_1|\}=\frac\{\alpha\_1\}\{k\_1\},\\\\ \beta\_2&=\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{|\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1|\}=\frac\{\alpha\_2-(\alpha\_2,\beta\_1)\beta\_1\}\{k\_2\},\cdots,\\\\ \beta\_n&=\frac\{\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\}\{|\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}|\}\\\\ &=\frac\{\alpha\_n-(\alpha\_n,\beta\_1)\beta\_1-\cdots-(\alpha\_n,\beta\_\{n-1\})\beta\_\{n-1\}\}\{k\_n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 此即 \begin\{aligned\} A=(\alpha\_1,\cdots,\alpha\_n)=(\beta\_1,\cdots,\beta\_n) \left(\begin\{array\}\{cccccccccccccccccccc\}k\_1&(\alpha\_2,\beta\_1)&\cdots&(\alpha\_n,\beta\_1)\\\\ &k\_2&\cdots&(\alpha\_n,\beta\_2)\\\\ &&\ddots&\vdots\\\\ &&&k\_n\end\{array\}\right)\equiv QT. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle Q$ 正交, $\displaystyle T$ 上三角, 且对角元大于 $\displaystyle 0$. (2)、 唯一性. 设 $\displaystyle A=QT=Q\_1T\_1$, 其中 $\displaystyle Q\_1$ 正交, $\displaystyle T\_1$ 上三角, 且对角元大于 $\displaystyle 0$, 则 \begin\{aligned\} Q\_1^\{-1\}Q=T\_1T^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 注意到正交 (上三角) 矩阵的逆还是正交 (上三角) 矩阵, 正交 (上三角) 矩阵的乘积还是正交 (上三角) 矩阵, 我们知上式左端是正交阵, 右端是上三角矩阵, 且对角元大于 $\displaystyle 0$. 作为正交阵的 $\displaystyle T\_1T^\{-1\}$ 的第一列是单位向量, 而它只有第一个元素非零且为正数, 而 $\displaystyle =1$. 从而 $\displaystyle T\_1T^\{-1\}$ 的第一行除了第一个元素外全为 $\displaystyle 0$. 继续看 $\displaystyle T\_1T^\{-1\}$ 的第二列. 除了第二个元素非零且为正数. 由于第二列为单位为单位向量, 而它的第二元素也 $\displaystyle =1$. 如此这般即知 $\displaystyle T\_1T^\{-1\}=E\_n\Rightarrow T\_1=T\Rightarrow Q\_1=Q$. 唯一性证得.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)/ --- 1308、 6、 (12 分) 证明: $\displaystyle \mathrm\{rank\} A+\mathrm\{rank\}(A-E)=n$ 的充要条件是 $\displaystyle A=A^2$, 其中 $\displaystyle A\in M\_n(\mathbb\{K\})$. (重庆大学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&\\\\ &A-E\end\{array\}\right)\to&\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ &A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ -A&-E\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}A-A^2&\\\\ -A&-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A-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\}(A)+\mathrm\{rank\}(A-E)=\mathrm\{rank\}(A-A^2)+n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} \mathrm\{rank\}(A)+\mathrm\{rank\}(A-E)=n&\Leftrightarrow \mathrm\{rank\}(A-A^2)=0\\\\ &\Leftrightarrow A-A^2=0\Leftrightarrow A=A^2. \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)/ --- 1309、 7、 (12 分) 设 $\displaystyle A$ 为 $\displaystyle n$ 阶方阵, 且 $\displaystyle A^2=A$. 证明: \begin\{aligned\} (A+E)^k=E+(2^k-1)A, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle E$ 为单位矩阵. (重庆师范大学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) / 对 $\displaystyle k$ 作数学归纳法. 当 $\displaystyle k=1$ 时结论自然成立. 若结论对 $\displaystyle k$ 成立, 则 \begin\{aligned\} &(A+E)^\{k+1\}=(A+E)^k(A+E)\xlongequal\{\tiny\mbox\{归纳假设\}\} [E+(2^k-1)A] (A+E)\\\\ =&A+E+(2^k-1)A^2+(2^k-1)A \xlongequal\{\tiny\mbox\{题设\}\} A+E+(2^k-1)A+(2^k-1)A\\\\ =&E+(2^\{k+1\}-1)A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 归纳步证毕.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)/ --- 1310、 11、 (15 分) 设 $\displaystyle f\_i(x),g\_i(x)$ 为数域 $\displaystyle \mathbb\{P\}$ 上的多项式, 且 \begin\{aligned\} \left(f\_1(x)\cdots f\_n(x),g\_1(x)\cdots g\_n(x)\right)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再设 $\displaystyle \lambda E-A$ 与 $\displaystyle \lambda E-B$ 分别与 \begin\{aligned\} \mathrm\{diag\}(f\_1g\_1,\cdots,f\_ng\_n), \mathrm\{diag\}\left(f\_\{i\_1\}g\_\{i\_1\},\cdots,f\_\{i\_n\}g\_\{i\_n\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 等价, 其中 $\displaystyle i\_1\cdots i\_n$ 是 $\displaystyle 12\cdots n$ 的一个排列. 证明: $\displaystyle A,B$ 相似. (重庆师范大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle P$ 是 $\displaystyle (j,i\_j)$ 元为 $\displaystyle 1$, $\displaystyle 1\leq j\leq n$, 其余元为 $\displaystyle 0$ 的置换矩阵, 则 \begin\{aligned\} P^\mathrm\{T\} \mathrm\{diag\}(f\_1g\_1,\cdots,f\_ng\_n)P=\mathrm\{diag\}\left(f\_\{i\_1\}g\_\{i\_1\},\cdots,f\_\{i\_n\}g\_\{i\_n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 与题设联合即知 $\displaystyle \lambda E-A, \lambda E-B$ 等价, 从而 $\displaystyle A,B$ 相似.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)/ --- 1311、 (5)、 实二次型 \begin\{aligned\} f(x\_1,x\_2,x\_3)=2x\_1x\_2+2x\_1x\_3-6x\_2x\_3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的规范形为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (安徽大学2023年高等代数考研试题) [二次型 ] [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) / $\displaystyle f$ 的矩阵为 \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&1\\\\ 1&0&-3\\\\ 1&-3&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\} 则 \begin\{aligned\} P\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0\\\\ 1&-1&0\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_1^\mathrm\{T\} AP&=A\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&-2\\\\ 0&-2&4\\\\ -2&4&0\end\{array\}\right),\\\\ P\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&2\\\\ 0&0&1\end\{array\}\right)\Rightarrow P\_2^\mathrm\{T\} A\_1P\_2&=A\_2=\mathrm\{diag\}(2,-2,6),\\\\ P\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&\\\\ &&1\\\\ &1&\end\{array\}\right)\Rightarrow P\_3^\mathrm\{T\} A\_2P\_3&=\mathrm\{diag\}(2,6,-2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle f$ 的规范形为 $\displaystyle z\_1^2+z\_2^2-z\_3^2$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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