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张祖锦2023年数学专业真题分类70天之第52天

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发表于 2023-3-5 13:11:53 | 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52天 --- 1174、 (3)、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&3\\\\ 4&5&6\\\\ 7&8&9\end\{array\}\right)$, $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是线性无关的 $\displaystyle 3$ 维列向量, 则 \begin\{aligned\} \mathrm\{rank\}(A\alpha\_1,A\alpha\_2,A\alpha\_3)=\underline\{\ \ \ \ \ \ \ \ \ \ \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (厦门大学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 $\displaystyle A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-1\\\\ 0&1&2\\\\ 0&0&0\end\{array\}\right)$ 知 \begin\{aligned\} \mathrm\{rank\}(A\alpha\_1,A\alpha\_2,A\alpha\_3)=\mathrm\{rank\}\left\[A(\alpha\_1,\alpha\_2,\alpha\_3)\right\]=\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)/ --- 1175、 (7)、 设 $\displaystyle \alpha,\beta$ 是数域 $\displaystyle \mathbb\{F\}$ 上的 $\displaystyle n$ 维列向量, $\displaystyle \beta^\mathrm\{T\}\alpha=3$, 则 $\displaystyle \alpha\beta^\mathrm\{T\}$ 的特征值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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\} (\alpha\beta^\mathrm\{T\})\alpha=(\beta^\mathrm\{T\} \alpha)\alpha=3\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再由题设, $\displaystyle \alpha\neq 0, \beta\neq 0$, $\displaystyle \alpha\beta^\mathrm\{T\}\neq 0\Rightarrow \mathrm\{rank\} (\alpha\beta^\mathrm\{T\})\geq 1$. 但 $\displaystyle \mathrm\{rank\}(\alpha\beta^\mathrm\{T\})\leq \mathrm\{rank\} \alpha=1$. 故 $\displaystyle \alpha\beta^\mathrm\{T\} x=0$ 有两个线性无关的解向量 $\displaystyle \alpha\_2,\cdots,\alpha\_n$. 由于矩阵属于不同特征值的特征向量线性无关知 $\displaystyle P=(\alpha,\alpha\_2,\cdots,\alpha\_n)$ 可逆, 且 \begin\{aligned\} P^\{-1\}(\alpha\beta^\mathrm\{T\})P=\mathrm\{diag\}(3,0,\cdots,0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A$ 的特征值为 $\displaystyle 3$ (单重), $\displaystyle 0$ ($n-1$ 重).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)/ --- 1176、 (8)、 若 $\displaystyle n$ 阶方阵仅有特征值 $\displaystyle 1$ 且只有一个线性无关的特征向量, 则 $\displaystyle A$ 的不变因子为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (厦门大学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) / 设题中矩阵为 $\displaystyle A$, 则 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 的对角元为 $\displaystyle 1$. 又由 $\displaystyle A$ 只有一个线性无关的特征向量知 $\displaystyle J=J\_n(1)$, 而 $\displaystyle A$ 的初等因子为 $\displaystyle (\lambda-1)^n$, 不变因子为 $\displaystyle 1,\cdots,1,(\lambda-1)^n$.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)/ --- 1177、 3、 若 $\displaystyle A$ 是可逆实矩阵, 证明: 存在正交阵 $\displaystyle Q$, 使得 $\displaystyle QA$ 为上三角阵且对角元全为正数. (厦门大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle A=(\alpha\_1,\cdots,\alpha\_n)$, 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). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle Q=\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1^\mathrm\{T\}\\\\\vdots\\\\\beta\_n^\mathrm\{T\}\end\{array\}\right)$, 则 \begin\{aligned\} QA=\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). \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)/ --- 1178、 4、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶复矩阵, $\displaystyle g(x)$ 是 $\displaystyle A$ 的最小多项式, $\displaystyle f(x)$ 是复多项式, 满足 \begin\{aligned\} \left(f(x),g(x)\right)=d(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明: (1)、 $\displaystyle \mathrm\{rank\} f(A)=\mathrm\{rank\} d(A)$; (2)、 $\displaystyle f(A)$ 可逆当且仅当 $\displaystyle d(x)=1$. (厦门大学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\} &(f,g)=d\Rightarrow \exists\ u,v,\mathrm\{ s.t.\} uf+vg=d\\\\ \Rightarrow&d(A)=u(A)f(A)+v(A)g(A)=u(A)f(A)\qquad(I)\\\\ \Rightarrow& \mathrm\{rank\} d(A)\leq \mathrm\{rank\} f(A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 另一方面, \begin\{aligned\} d\mid f\Rightarrow \exists\ h,\mathrm\{ s.t.\} f=dh \Rightarrow f(A)=d(A)h(A)\Rightarrow \mathrm\{rank\} f(A)\leq \mathrm\{rank\} d(A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \mathrm\{rank\} f(A)=\mathrm\{rank\} d(A)$. (2)、 (2-1)、 $\displaystyle \Leftarrow$: 由 $\displaystyle (I)$ 知 \begin\{aligned\} E=d(A)=u(A)f(A)\Rightarrow \left\[f(A)\right\]^\{-1\}=u(A), f(A)\mbox\{可逆\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-2)、 $\displaystyle \Rightarrow$: 设 $\displaystyle f(\lambda)=(\lambda-\lambda\_1)\cdots(\lambda-\lambda\_s)$, 其中 $\displaystyle \lambda\_i$ 是 $\displaystyle f$ 的复根. 则 \begin\{aligned\} &0\neq \det f(A) =\det (A-\lambda\_1E)\cdots \det (A-\lambda\_sE)\\\\ \Rightarrow& \forall\ i, \det (A-\lambda\_iE)\neq 0 \Rightarrow \forall\ i, \mbox\{ $\displaystyle \lambda\_i$ 不是 $\displaystyle A$ 的特征值\}\\\\ \Rightarrow&\forall\ i,\mbox\{ $\displaystyle \lambda\_i$ 不是 $\displaystyle g(\lambda)$ 的根 (最小多项式以所有特征值为根)\}\\\\ \Rightarrow&d(\lambda)=\left(g(\lambda),f(\lambda)\right)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 也可用反证法证明如下. 若 $\displaystyle \partial(d)\geq 1$, 则 \begin\{aligned\} \exists\ g\_1, f\_1,\mathrm\{ s.t.\} g=dg\_1, f=df\_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 \begin\{aligned\} 0=g(A)f\_1(A)=d(A)g\_1(A)f\_1(A)=g\_1(A)f(A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle f(A)$ 可逆知 $\displaystyle g\_1(A)=0$, 而找到了次数必 $\displaystyle g$ 更小的 $\displaystyle A$ 的零化多项式, 与 $\displaystyle g$ 是 $\displaystyle A$ 的最小多项式矛盾. 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)/ --- 1179、 5、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶复矩阵, $\displaystyle 0 < \mathrm\{rank\} A < n$, $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^2)$. 证明: 存在课内矩阵 $\displaystyle P$ 及可逆矩阵 $\displaystyle B$ 使得 \begin\{aligned\} A=P\left(\begin\{array\}\{cccccccccccccccccccc\}B&0\\\\ 0&0\end\{array\}\right)P^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [张祖锦注: 本题没有必要假设 $\displaystyle A$ 是复矩阵, 具体见参考解答] (厦门大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \mathrm\{rank\} A=\mathrm\{rank\}(A^2)=r$. 令 \begin\{aligned\} V\_1=\left\\{\alpha\in\mathbb\{F\}^n;A\alpha=0\right\\},\quad V\_2=\left\\{\alpha\in\mathbb\{F\}^n;A^2\alpha=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 $\displaystyle V\_1\subset V\_2$, \begin\{aligned\} \dim V\_1=n-\mathrm\{rank\}(A)=n-\mathrm\{rank\}(A^2)=\dim V\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle V\_1=V\_2$. 再令 \begin\{aligned\} W\_1=\left\\{A\alpha;\alpha\in\mathbb\{F\}^n\right\\},\quad W\_2=\left\\{A^2\alpha;\alpha\in\mathbb\{F\}^n\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle W\_2\subset W\_1$, \begin\{aligned\} \dim W\_1=\mathrm\{rank\}(A)=\mathrm\{rank\}(A^2)=\dim W\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle W\_1=W\_2$. 取 $\displaystyle V\_1=V\_2$ 的一组基 $\displaystyle \eta\_1,\cdots,\eta\_\{n-r\}$, 将其扩充为 $\displaystyle \mathbb\{F\}^n$ 的一组基 $\displaystyle \eta\_1,\cdots,\eta\_n$, 则 $\displaystyle A\eta\_\{n-r+1\},\cdots,A\eta\_n$ 是 $\displaystyle W\_1$ 的一组基, $\displaystyle A^2\eta\_\{n-r+1\},\cdots,A^2\eta\_n$ 是 $\displaystyle W\_2$ 的一组基. 因 $\displaystyle W\_1=W\_2$, 我们得到 \begin\{aligned\} A\eta\_\{n-r+1\},\cdots,A\eta\_n\mbox\{与\}A^2\eta\_\{n-r+1\},\cdots,A^2\eta\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 等价, 而存在可逆矩阵 $\displaystyle B$ 使得 \begin\{aligned\} (A^2\eta\_\{n-r+1\},\cdots,A^2\eta\_n)=(A\eta\_\{n-r+1\},\cdots,A\eta\_n)B. \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\} &\quad \sum\_\{i=1\}^\{n-r\}k\_i\eta\_i+\sum\_\{j=n-r+1\}l\_jA\eta\_j=0\\\\\ &\Rightarrow \sum\_j l\_jA^2\eta\_j=0\quad \left(\mbox\{用\}A\mbox\{作用\}\right)\\\\ &\Rightarrow l\_j=0\Rightarrow k\_i=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle \eta\_1,\cdots,\eta\_\{n-r\},A\eta\_\{n-r+1\},\cdots,A\eta\_n$ 是 $\displaystyle \mathbb\{F\}^n$ 的一组基. 令 \begin\{aligned\} P=(A\eta\_\{n-r+1\},\cdots,A\eta\_n,\eta\_1,\cdots,\eta\_\{n-r\}), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 可逆, 且 \begin\{aligned\} AP=&(A^2\eta\_\{n-r+1\},\cdots,A^2\eta\_n,0,\cdots,0)\\\\ =&(A\eta\_\{n-r+1\},\cdots,A\eta\_n,0,\cdots,0) \left(\begin\{array\}\{cccccccccccccccccccc\}B&0\\\\ 0&0\end\{array\}\right) =P\left(\begin\{array\}\{cccccccccccccccccccc\}B&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\} 于是 $\displaystyle \{A=P\left(\begin\{array\}\{cccccccccccccccccccc\}B&0\\\\ 0&0\end\{array\}\right)P^\{-1\}\}$.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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1180、 6、 $\displaystyle A$ 是 $\displaystyle n$ 阶实正定矩阵, 且非对角元均小于 $\displaystyle 0$. 证明: $\displaystyle A^\{-1\}$ 的所有元素都大于 $\displaystyle 0$. (厦门大学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 $\displaystyle n$ 作数学归纳法. 当 $\displaystyle n=1$ 时, $\displaystyle A=(a)\Rightarrow A^\{-1\}=\left(\frac\{1\}\{a\}\right)$, 结论成立. 假设结论对 $\displaystyle n-1$ 阶矩阵成立, 则对 $\displaystyle n$ 阶正定矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}B&\alpha\\\\ \alpha^\mathrm\{T\}&a\_\{nn\}\end\{array\}\right)$, 写出 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&-a\_\{nn\}^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}B&\alpha\\\\ \alpha^\mathrm\{T\}&a\_\{nn\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -a\_\{nn\}^\{-1\}\alpha^\mathrm\{T\}&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}B-a\_\{nn\}^\{-1\}\alpha\alpha^\mathrm\{T\}&0\\\\ 0&a\_\{nn\}\end\{array\}\right)\quad (I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 后知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}B-a\_\{nn\}^\{-1\}\alpha\alpha^\mathrm\{T\}&0\\\\ 0&a\_\{nn\}\end\{array\}\right)$ 正定. 注意到 $\displaystyle a\_\{nn\}^\{-1\}\alpha\alpha^\mathrm\{T\}$ 的元素都大于 $\displaystyle 0$, 而 $\displaystyle B-a\_\{nn\}^\{-1\}\alpha\alpha^\mathrm\{T\}$ 的非对角元都小于 $\displaystyle 0$. 按归纳假设, $\displaystyle C=(B-a\_\{nn\}^\{-1\}\alpha\alpha^\mathrm\{T\})^\{-1\}$ 的元素均大于 $\displaystyle 0$. 继续由 $\displaystyle (I)$ 知 \begin\{aligned\} A^\{-1\}=&\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -a\_\{nn\}^\{-1\}\alpha^\mathrm\{T\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}C&\\\\ &a\_\{nn\}^\{-1\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-a\_\{nn\}^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}C&-a\_\{nn\}^\{-1\}\alpha C\\\\ -a\_\{nn\}^\{-1\}\alpha C^\{-1\}&a\_\{nn\}^\{-1\}+a\_\{nn\}^\{-2\}\alpha^\mathrm\{T\} C\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\} 由 $\displaystyle C$ 的元素都大于 $\displaystyle 0$, $\displaystyle \alpha$ 的元素都小于 $\displaystyle 0$, 而 $\displaystyle A^\{-1\}$ 的元素都大于 $\displaystyle 0$.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)/ --- 1181、 3、 设 $\displaystyle A,B$ 为 $\displaystyle n$ 阶矩阵且 $\displaystyle A+B=AB$, 求 $\displaystyle A-E$ 的逆矩阵, 并证明 $\displaystyle AB=BA$. (山东大学2023年高等代数与常微分方程考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} &A+B=AB\Rightarrow (A-E)(B-E)=E\\\\ \Rightarrow& A-E\mbox\{可逆, 且\}(A-E)^\{-1\}=B-E\\\\ \Rightarrow& (B-E)(A-E)=E\Rightarrow BA-A-B=0\\\\ \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\} 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)/ --- 1182、 4、 求 $\displaystyle n$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&\cdots&0\\\\ a&1&0&\cdots&0\\\\ a^2&a&1&\cdots&0\\\\ \vdots&\vdots&\vdots&&\vdots\\\\ a^\{n-1\}&a^\{n-2\}&a^\{n-3\}&\cdots&1\end\{array\}\right)$ 的逆矩阵. (山东大学2023年高等代数与常微分方程考研试题) [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 $\displaystyle N=\left(\begin\{array\}\{cccccccccccccccccccc\}0&&&\\\\ 1&\ddots&&\\\\ &\ddots&\ddots&\\\\ &&1&0\end\{array\}\right)$, 则 \begin\{aligned\} A=E+aN+a^2N^2+\cdots+a^\{n-1\}N^\{n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle (E-aN)A=E$ 知 $\displaystyle A^\{-1\}=E-aN=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ -a&\ddots&&\\\\ &\ddots&\ddots&\\\\ &&-a&1\end\{array\}\right)$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1183、 5、 设 $\displaystyle A,B$ 均为 $\displaystyle n$ 阶正定矩阵, 证明 $\displaystyle AB$ 正定的充要条件为 $\displaystyle AB=BA$. (山东大学2023年高等代数与常微分方程考研试题) [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\} BA=B^\mathrm\{T\} A^\mathrm\{T\} =(AB)^\mathrm\{T\}=AB . \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 $\displaystyle AB=BA$ 知 \begin\{aligned\} (AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\} =BA=AB, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 $\displaystyle AB$ 为实对称矩阵. 进一步, \begin\{aligned\} A,B\mbox\{正定\}&\Rightarrow \exists\ \mbox\{可逆\}P,Q,\mathrm\{ s.t.\} A=P^\mathrm\{T\} P, B=Q^\mathrm\{T\} Q\\\\ &\Rightarrow AB=P^\mathrm\{T\} P Q^\mathrm\{T\} Q\\\\ & \Rightarrow \left(P^\mathrm\{T\}\right)^\{-1\} ABP^\mathrm\{T\} =PQ^\mathrm\{T\} QP^\mathrm\{T\}=(QP^\mathrm\{T\})^\mathrm\{T\} QP^\mathrm\{T\}\\\\ &\Rightarrow \left(P^\mathrm\{T\}\right)^\{-1\} ABP^\mathrm\{T\}\mbox\{正定\}\left(\mbox\{$QP^\mathrm\{T\}$ 可逆\}\right)\\\\ &\Rightarrow \left(P^\mathrm\{T\}\right)^\{-1\} ABP^\mathrm\{T\}\mbox\{的特征值均大于 $\displaystyle 0$\}\\\\ &\Rightarrow AB\left(\sim \left(P^\mathrm\{T\}\right)^\{-1\} ABP^\mathrm\{T\}\right) \mbox\{的特征值均大于 $\displaystyle 0$\}\\\\ &\Rightarrow AB\mbox\{正定\}. \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)/ --- 1184、 4、 (15 分) 矩阵 $\displaystyle A$ 为 $\displaystyle 3$ 阶实对称矩阵, 它的秩为 $\displaystyle 2$, 且 $\displaystyle -2$ 是其二重特征值, \begin\{aligned\} (1,0,0)^\mathrm\{T\}, (2,1,1)^\mathrm\{T\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 为 $\displaystyle A$ 的属于 $\displaystyle -2$ 的特征向量. 求矩阵 $\displaystyle A$. (山西大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由实对称矩阵属于不同特征值的特征向量正交知 $\displaystyle A$ 的属于特征值 $\displaystyle 2$ 的特征向量 $\displaystyle x$ 满足 \begin\{aligned\} Ax=0, A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 2&1&1\end\{array\}\right)\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故可取 $\displaystyle x=(0,-1,1)^\mathrm\{T\}$. 于是 \begin\{aligned\} P\equiv \left(\begin\{array\}\{cccccccccccccccccccc\}0&1&2\\\\ -1&0&1\\\\ 1&0&1\end\{array\}\right)\Rightarrow AP=P\mathrm\{diag\}(2,-2,-2)\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}-2&0&0\\\\ 0&0&-2\\\\ 0&-2&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\} 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)/ --- 1185、 6、 (15 分) 求 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&0&0\\\\ 3&-4&0&0\\\\ -7&6&2&1\\\\ 7&6&1&0\end\{array\}\right)$ 的不变因子, 初等因子, 若当标准形和有理标准形. [题目有问题, 跟锦数学微信公众号没法做哦. 毕竟算出来有根号, 考试一般不会出现] (山西大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目有问题, 跟锦数学微信公众号没法做哦. 毕竟算出来有根号, 考试一般不会出现]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)/ --- 1186、 10、 (15 分) 证明下列结论. (1)、 $\displaystyle A$ 为 $\displaystyle n$ 阶实矩阵, 则 $\displaystyle A$ 为反对称矩阵的充分必要条件为 $\displaystyle AA^\mathrm\{T\}=-A^2$. (山西大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \Rightarrow$: $\displaystyle AA^\mathrm\{T\}=A(-A)=-A^2$. $\displaystyle \Leftarrow$: 由 \begin\{aligned\} &\mathrm\{tr\}\left\[(A^\mathrm\{T\}+A)^\mathrm\{T\}(A^\mathrm\{T\}+A)\right\] =\mathrm\{tr\}\left\[(A+A^\mathrm\{T\})(A^\mathrm\{T\}+A)\right\]\\\\ =&\mathrm\{tr\}\left\[AA^\mathrm\{T\}+A^2+(A^2)^\mathrm\{T\}+A^\mathrm\{T\} A\right\] =\mathrm\{tr\}\left\[(A^2)^\mathrm\{T\}+A^\mathrm\{T\} A\right\]\\\\ \stackrel\{\mbox\{转置\}\}\{=\}&\mathrm\{tr\}(A^2+A^\mathrm\{T\} A) =\mathrm\{tr\}(A^2)+\mathrm\{tr\}(A^\mathrm\{T\} A)\\\\ =&\mathrm\{tr\}(A^2)+\mathrm\{tr\}(AA^\mathrm\{T\})=\mathrm\{tr\}(A^2+AA^\mathrm\{T\})=\mathrm\{tr\} 0=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A^\mathrm\{T\}+A$ 的所有元素的平方和为 $\displaystyle 0$, 而 $\displaystyle A^\mathrm\{T\}+A=0$, $\displaystyle A$ 反对称.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)/ --- 1187、 (2)、 $\displaystyle A$ 为正定矩阵, 则存在正定矩阵 $\displaystyle B$, 使得 $\displaystyle A=B^2$. (山西大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 正定知存在正交阵 $\displaystyle P$ 使得 \begin\{aligned\} A=P^\mathrm\{T\} \mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)P, \lambda\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle B=P^\mathrm\{T\} \mathrm\{diag\}\left(\sqrt\{\lambda\_1\},\cdots,\sqrt\{\lambda\_n\}\right)P$, 则 $\displaystyle B$ 正定, 且 $\displaystyle A=B^2$.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)/ --- 1188、 4、 (15 分) 设 $\displaystyle A=(a\_\{ij\})\in\mathbb\{R\}^\{n\times n\}$, 已知 \begin\{aligned\} a\_\{ii\} > 0, a\_\{ij\} < 0\left(i\neq j\right), \sum\_\{j=1\}^n a\_\{ij\}=0\left(1\leq i\leq n, 1\leq i\neq j\leq n\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明: $\displaystyle \mathrm\{rank\} A=n-1$. (陕西师范大学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)、 我们先证明一个结论. 设矩阵 $\displaystyle A=(a\_\{ij\})$ 是一个 $\displaystyle n$ 级实矩阵, 对任意的 $\displaystyle i=1,2,\cdots,n$, 均有 \begin\{aligned\} a\_\{ii\} > \sum\_\{j\neq i\} |a\_\{ij\}|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle \det A > 0$. 事实上, 设 $\displaystyle \lambda$ 是 $\displaystyle A$ 的实特征值, 则 \begin\{aligned\} &\exists\ x\neq 0,\mathrm\{ s.t.\} Ax=\lambda x\\\\ \Rightarrow&\sum\_\{j=1\}^n a\_\{kj\}x\_j=\lambda x\_k\left(|x\_k|=\max\_\{1\leq i\leq n\}|x\_i| > 0\right)\\\\ \Rightarrow& (\lambda-a\_\{kk\})x\_k=\sum\_\{j\neq k\} a\_\{kj\}x\_j\\\\ \Rightarrow& |\lambda-a\_\{kk\}|\leq \sum\_\{j\neq k\} |a\_\{kj\}|\frac\{|x\_j|\}\{|x\_k|\} \leq \sum\_\{j\neq k\} |a\_\{kj\}|\\\\ \Rightarrow&\lambda \geq a\_\{kk\}-\sum\_\{j\neq k\} |a\_\{kj\}| > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又 $\displaystyle A$ 是实矩阵, 其虚特征值成对出现, 而可设它的全部特征值为 \begin\{aligned\} \lambda\_1,\cdots,\lambda\_s, a\_1+b\_1\mathrm\{ i\}, a\_1-b\_1\mathrm\{ i\}, \cdots, a\_\{n-s\}+b\_\{n-s\}\mathrm\{ i\}, a\_\{n-s\}-b\_\{n-s\}\mathrm\{ i\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle \lambda\_i\in\mathbb\{R\}, a\_j,b\_j\in\mathbb\{R\}, b\_j\neq 0$. 如此, \begin\{aligned\} |A|=\prod\_\{i=1\}^s \lambda\_i\cdot \prod\_\{j=1\}^\{n-s\} (a\_i^2+b\_i^2) > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 回到题目. 将 $\displaystyle A$ 的第 $\displaystyle i$ ($1\leq i\leq n-1$) 加到第 $\displaystyle n$ 行, 可将 $\displaystyle A$ 的第 $\displaystyle n$ 行化为 $\displaystyle 0$. 故 $\displaystyle \mathrm\{rank\} A\leq n-1$. 考虑 $\displaystyle A$ 的 $\displaystyle n-1$ 阶顺序主子阵 \begin\{aligned\} B=\left(\begin\{array\}\{cccccccccccccccccccc\}a\_\{11\}&\cdots&a\_\{1,n-1\}\\\\ \vdots&&\vdots\\\\ a\_\{n-1,1\}&\cdots&a\_\{n-1,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\} 则 \begin\{aligned\} 0 < a\_\{kk\}=-\sum\_\{i=1, i\neq k\}^n a\_\{ik\} > \sum\_\{i=1,i\neq k\}^\{n-1\}a\_\{ik\}, 1\leq k\leq n-1. \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 步结果知 $\displaystyle 0 < |B^\mathrm\{T\}|=|B|$. 故 $\displaystyle \mathrm\{rank\} A=n-1$.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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1189、 8、 (15 分) 有一个 $\displaystyle 6$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}a&-b&&&&\\\\ b&a&1&&&\\\\ &&a&-b&&\\\\ &&b&a&1&&\\\\ &&&&a&-b\\\\ &&&&b&a\end\{array\}\right)$, 其中 $\displaystyle a,b\in\mathbb\{R\}$, 且 $\displaystyle b\neq 0$. 求 $\displaystyle \lambda E-A$ 的不变因子与初等因子以及 $\displaystyle A$ 的若尔当标准形. (陕西师范大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \lambda E-A=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda-a&b&&&&\\\\ -b&\lambda-a&-1&&&\\\\ &&\lambda-a&b&&\\\\ &&-b&\lambda-a&-1&&\\\\ &&&&\lambda-a&b\\\\ &&&&-b&\lambda-a\end\{array\}\right)$ 的右上角有一个 $\displaystyle 5$ 阶子式 $\displaystyle =b^3\neq 0$. 设 $\displaystyle |\lambda E-A|=D\_6$, 则 \begin\{aligned\} D\_6=(\lambda-a)^2D\_4+b^2D\_4=\cdots=[(\lambda-a)^2+b^2]^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A$ 的不变因子为 $\displaystyle 1,1,1,1,1, [(\lambda-a)^2+b^2]^3=[\lambda-(a+b\mathrm\{ i\})]^3[\lambda-(a-b\mathrm\{ i\})]^3$, 初等因子为 $\displaystyle [\lambda-(a+b\mathrm\{ i\})]^3,[\lambda-(a-b\mathrm\{ i\})]^3$, Jordan 标准形为 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}a+b\mathrm\{ i\}&1&&&&\\\\ &a+b\mathrm\{ i\}&1&&&\\\\ &&a+b\mathrm\{ i\}&&&\\\\ &&&a-b\mathrm\{ i\}&1&\\\\ &&&&a-b\mathrm\{ i\}&1\\\\ &&&&&a-b\mathrm\{ i\}\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)/ --- 1190、 3、 已知 $\displaystyle A,B$ 为 $\displaystyle n$ 阶矩阵, $\displaystyle AB=BA$. 证明: \begin\{aligned\} \mathrm\{rank\}(A+B)+\mathrm\{rank\}(AB)\leq \mathrm\{rank\} A+\mathrm\{rank\} B . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (上海财经大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 取定 $\displaystyle \mathbb\{F\}^n$ 的一组基 $\displaystyle \varepsilon\_1,\cdots,\varepsilon\_n$, 并设线性变换 $\displaystyle \mathscr\{A\}, \mathscr\{B\}: \mathbb\{F\}^n\to\mathbb\{F\}^n$ 在该基下的矩阵分别为 $\displaystyle A,B$, 则 \begin\{aligned\} \mathscr\{A\}(\varepsilon\_1,\cdots,\varepsilon\_n)&=(\varepsilon\_1,\cdots,\varepsilon\_n)A,\\\\ \mathscr\{B\}(\varepsilon\_1,\cdots,\varepsilon\_n)&=(\varepsilon\_1,\cdots,\varepsilon\_n)B. \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\} AB=BA&\Rightarrow \mathscr\{A\}\mathscr\{B\}=\mathscr\{B\}\mathscr\{A\}\\\\ &\Rightarrow \mathrm\{im\}(\mathscr\{A\}\mathscr\{B\})=\mathrm\{im\}(\mathscr\{B\}\mathscr\{A\}) \subset \mathrm\{im\} \mathscr\{A\}\cap \mathrm\{im\} \mathscr\{B\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 据维数公式即知 \begin\{aligned\} \mathrm\{rank\}(AB)+\mathrm\{rank\}(A+B) &=\dim\mathrm\{im\}(\mathscr\{A\}\mathscr\{B\})+\dim \mathrm\{im\}(\mathscr\{A\}+\mathscr\{B\})\\\\ &\leq \dim (\mathrm\{im\} \mathscr\{A\}\cap \mathrm\{im\}\mathscr\{B\})+\dim\mathrm\{im\}(\mathscr\{A\}+\mathscr\{B\})\\\\ &=\dim \mathrm\{im\}\mathscr\{A\}+\dim \mathrm\{im\}\mathscr\{B\}\\\\ &=\mathrm\{rank\}(A)+\mathrm\{rank\}(B). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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)/ --- 1191、 5、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, 满足 $\displaystyle E\_n-AA^\mathrm\{T\}$ 可逆. 证明: (1)、 $\displaystyle E\_n-A^\mathrm\{T\} A$ 可逆; (2)、 $\displaystyle A^\mathrm\{T\}(E\_n-AA^\mathrm\{T\})^\{-1\}=(E\_n-A^\mathrm\{T\} A)^\{-1\}A^\mathrm\{T\}$. (上海财经大学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)、 我们有如下常用公式: 由 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&E+BA\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccccccccccccc\}E&-A\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&A\\\\ -B&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ B&E\end\{array\}\right)=&\left(\begin\{array\}\{cccccccccccccccccccc\}E+AB&0\\\\ 0&E\end\{array\}\right)\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 即知 \begin\{aligned\} |E+BA|=|E+AB|. \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 $\displaystyle |E\_n-A^\mathrm\{T\} A|=|E\_n-AA^\mathrm\{T\}|\neq 0$, 而 $\displaystyle E\_n-A^\mathrm\{T\} A$ 可逆. \begin\{aligned\} A^\mathrm\{T\}(E\_n-AA^\mathrm\{T\})=A^\mathrm\{T\}-A^\mathrm\{T\} A\cdot A^\mathrm\{T\}=(E\_n-A^\mathrm\{T\} A)A^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 左乘 $\displaystyle (E\_n-A^\mathrm\{T\} A)^\{-1\}$, 右乘 $\displaystyle (E\_n-AA^\mathrm\{T\})^\{-1\}$ 即得 \begin\{aligned\} A^\mathrm\{T\}(E\_n-AA^\mathrm\{T\})^\{-1\}=(E\_n-A^\mathrm\{T\} A)^\{-1\}A^\mathrm\{T\}. \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)/ --- 1192、 6、 设 $\displaystyle A,B$ 是 $\displaystyle n$ 阶正定矩阵. 证明: (1)、 $\displaystyle A+2023B$ 正定; (2)、 $\displaystyle |A+2023B| > |A|$. (上海财经大学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)、 $\displaystyle A+2023B$ 实对称, 且对 $\displaystyle \forall\ 0\neq x\in\mathbb\{R\}^n$, \begin\{aligned\} x^\mathrm\{T\} Ax > 0, x^\mathrm\{T\} Bx > 0\Rightarrow x^\mathrm\{T\}(A+2023B)x=x^\mathrm\{T\} Ax+2023x^\mathrm\{T\} Bx > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A+2023B$ 正定. (2)、 由 $\displaystyle B$ 正定知存在可逆阵 $\displaystyle P$, 使得 \begin\{aligned\} P^\mathrm\{T\} BP=E\Rightarrow |P^\mathrm\{T\} P|=\frac\{1\}\{|B|\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 $\displaystyle P^\mathrm\{T\} AP$ 仍半正定, 存在正交阵 $\displaystyle Q$, 使得 \begin\{aligned\} Q^\mathrm\{T\} P^\mathrm\{T\} APQ=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} &|Q^\mathrm\{T\} P^\mathrm\{T\} (A+2023B)PQ|=|\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)+2023E|\\\\ &=\prod\_\{i=1\}^n(2023+\lambda\_i) > \lambda\_1\cdots \lambda\_n=|P^\mathrm\{T\} AP|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 化简即知 $\displaystyle |A+2023B| > |A|$.跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1193、 (3)、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶反对称矩阵, 且为正交矩阵, 则 $\displaystyle A^2=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (上海大学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 $\displaystyle A$ 是 $\displaystyle n$ 阶反对称矩阵知 $\displaystyle A$ 的特征值为 $\displaystyle 0$ 或纯虚数. 又由 $\displaystyle A$ 为正交矩阵知 $\displaystyle A$ 的特征值的模长为 $\displaystyle 1$. 联合 $\displaystyle A$ 的虚特征值成对出现知 $\displaystyle A$ 的特征值为 $\displaystyle \pm \mathrm\{ i\}$ (重数相同). 于是实对称矩阵 $\displaystyle A^2$ 的特征值都是 $\displaystyle -1$. 从而存在正交阵 $\displaystyle P$ 使得 \begin\{aligned\} A^2=P\mathrm\{diag\}(-1,\cdots,-1)P^\mathrm\{T\}=-E\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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)/ --- 1194、 (5)、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶正定矩阵, $\displaystyle \alpha$ 为非零 $\displaystyle n$ 维列向量, 则实对称矩阵 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&0\end\{array\}\right)$ 的正负惯性指数差为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (上海大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -\alpha^\mathrm\{T\} A^\{-1\}&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-A^\{-1\}\alpha\\\\ 0&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A&\\\\ &-\alpha^\mathrm\{T\} A^\{-1\}\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\} 知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&0\end\{array\}\right)$ 的正负惯性指数差为 $\displaystyle n-1$.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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1195、 (2)、 若实对称矩阵 $\displaystyle A,B$ 相似, 则 $\displaystyle A,B$ 未必合同; 反之, 若 $\displaystyle A,B$ 合同, 则 $\displaystyle A,B$ 一定相似. $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ (上海大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \times$. 若实对称矩阵 $\displaystyle A,B$ 相似, 则它们正交相似, 而 $\displaystyle A,B$ 未必合同.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)/ --- 1196、 (3)、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, 若 $\displaystyle A^3=A$, 则 $\displaystyle A$ 与对角矩阵相似. $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ (上海大学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) / $\displaystyle \surd$. $\displaystyle A$ 的最小多项式整除 $\displaystyle \lambda^3-\lambda$, 而没有重根, 可对角化.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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