张祖锦2023年数学专业真题分类70天之第56天

zhangzujin · 发表于 2023-3-5 13:13:45

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## 张祖锦2023年数学专业真题分类70天之第56天 --- 1266、 (0-7)、用 $\displaystyle J$ 表示元素全为 $\displaystyle 1$ 的 $\displaystyle n$ 阶方阵 ($n\geq 2$), $\displaystyle f(x)=ax^2+bx+c\in\mathbb\{R\}[x]$, 则 $\displaystyle f(J)$ 的全部特征值为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学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 $\displaystyle \mathrm\{rank\} J=1$ 知 $\displaystyle Jx=0$ 的基础解系有 $\displaystyle n-1$ 个向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-1\}$. 设 $\displaystyle \eta\_n=(1,\cdots,1)^\mathrm\{T\}$, 则 $\displaystyle J\eta\_n=n\eta\_n$. 于是 \begin\{aligned\} &P\equiv(\eta\_1,\cdots,\eta\_n)\Rightarrow JP=P\left(\begin\{array\}\{cccccccccccccccccccc\}0\_\{n-1\}&\\\\ &n\end\{array\}\right)\\\\ \Rightarrow& P^\{-1\}f(J)P=\left(\begin\{array\}\{cccccccccccccccccccc\}cE\_\{n-1\}&\\\\ &an^2+bn+c\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\} 故应填 $\displaystyle \underbrace\{c,\cdots,c\}\_\{n-1\},an^2+bn+c$.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)/ --- 1267、 (0-8)、 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ -1&-1&-1&-1\\\\ 1&1&1&1\\\\ 2&2&2&0\end\{array\}\right)$ 的最小多项式为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle A$ 的特征值为 $\displaystyle 1,0,0,0$. 由 $\displaystyle A\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&1&1&0\\\\ 0&0&0&1\\\\ 0&0&0&0\end\{array\}\right)$ 知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 \begin\{aligned\} \mathrm\{rank\} J=\mathrm\{rank\} A=3\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&\\\\ &0&1&\\\\ &&0&\\\\ &&&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A$ 的最小多项式为 $\displaystyle \lambda^3(\lambda-1)$.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)/ --- 1268、 5、 (20 分) 证明如下结论: (0-13)、设 $\displaystyle A$ 是实反对称矩阵, 则 $\displaystyle (E-A)(E+A)^\{-1\}$ 是正交矩阵; (长安大学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\} &\quad (E-A)(E+A)^\{-1\}\left\[(E-A)(E+A)^\{-1\}\right\]^\mathrm\{T\}\\\\ =&(E-A)(E+A)^\{-1\}(E-A)^\{-1\}(E+A)\\\\ =&(E-A)\left\[(E-A)(E+A)\right\]^\{-1\}(E+A)\\\\ =&(E-A)\left\[(E+A)(E-A)\right\]^\{-1\}(E+A) =E. \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)/ --- 1269、 (0-14)、设 $\displaystyle A=(a\_\{ij\})\_\{m\times n\}, B=(b\_\{ij\})\_\{n\times s\}$, 则 \begin\{aligned\} \mathrm\{rank\}(AB)\geq \mathrm\{rank\} A+\mathrm\{rank\} B-n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (长安大学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\\\\ -A&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&B\\\\ A&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&-B\\\\ 0&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&-AB\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\}(B)\leq \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}E&B\\\\ A&0\end\{array\}\right)\\\\ =&\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&-AB\end\{array\}\right)=n+\mathrm\{rank\}(AB). \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)/ --- 1270、 9、 (15 分) 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0&1\\\\ 1&2&0\\\\ -4&0&3\end\{array\}\right)$, 求 $\displaystyle A$ 的若尔当标准形 $\displaystyle J$ 及可逆矩阵 $\displaystyle P$, 使得 $\displaystyle J=P^\{-1\}AP$. (长安大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,1,1$. 由 \begin\{aligned\} A-2E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&0&1\\\\ 0&0&0\end\{array\}\right), A-E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}\\\\ 0&1&\frac\{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 $\displaystyle \mathrm\{rank\}(J-E)=\mathrm\{rank\}(A-E)=2\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&&\\\\ &1&1\\\\ &&1\end\{array\}\right)$. 设 $\displaystyle P=(\alpha\_1,\alpha\_2,\alpha\_3)$, 则 \begin\{aligned\} (A-2E)\alpha\_1=0, (A-E)\alpha\_2=0, (A-E)\alpha\_3=\alpha\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故可取 $\displaystyle \alpha\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\1\\\\0\end\{array\}\right), \alpha\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\2\end\{array\}\right)$. 再由 \begin\{aligned\} (A-E,\alpha\_2)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}&-\frac\{1\}\{2\}\\\\ 0&1&\frac\{1\}\{2\}&-\frac\{1\}\{2\}\\\\ 0&0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知可取 $\displaystyle \alpha\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{2\}\\\\-\frac\{1\}\{2\}\\\\0\end\{array\}\right)$. 因此, $\displaystyle P=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&-\frac\{1\}\{2\}\\\\ 1&-1&-\frac\{1\}\{2\}\\\\ 0&2&0\end\{array\}\right), J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&&\\\\ &1&1\\\\ &&1\end\{array\}\right)$.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)/ --- 1271、 (2)、设 $\displaystyle A$ 是奇数阶的正交矩阵, 且 $\displaystyle A-E$ 可逆, 则 $\displaystyle |A|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (郑州大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设, $\displaystyle |A-E|\neq 0$, 而 $\displaystyle A$ 的实特征值只能为 $\displaystyle -1$. 又由 $\displaystyle A$ 的虚特征值成对出现, 而可设 $\displaystyle A$ 的特征值为 \begin\{aligned\} a\_i\pm b\_i\left(1\leq i\leq s\right), \underbrace\{-1,\cdots,-1\}\_\{n-2s\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 从而 $\displaystyle |A|=\prod\_\{i=1\}^s (a\_i^2+b\_i^2)\cdot (-1)^\{n-2s\}=-1$.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)/ --- 1272、 (4)、设 $\displaystyle \alpha=(a,b,c)^\mathrm\{T\}, \beta=\left(\frac\{1\}\{a\},\frac\{1\}\{b\},\frac\{1\}\{c\}\right)^\mathrm\{T\}$, 则 $\displaystyle (\alpha\beta^\mathrm\{T\})^n=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (郑州大学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 $\displaystyle \beta^\mathrm\{T\} \alpha=3$ 知 \begin\{aligned\} (\alpha\beta^\mathrm\{T\})^n=\alpha (\beta^\mathrm\{T\} \alpha)^\{n-1\}\beta^\mathrm\{T\}=3^\{n-1\}\alpha\beta^\mathrm\{T\}=3^\{n-1\}\left(\begin\{array\}\{cccccccccccccccccccc\}1&\frac\{a\}\{b\}&\frac\{a\}\{c\}\\\\ \frac\{b\}\{a\}&1&\frac\{b\}\{c\}\\\\ \frac\{c\}\{a\}&\frac\{c\}\{b\}&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)/ --- 1273、 (4)、设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}?\end\{array\}\right)$, 求 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$, 并求可逆矩阵 $\displaystyle P$, 使得 $\displaystyle P^\{-1\}AP=J$. [题目有问题, 跟锦数学微信公众号没法做哦.] (郑州大学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.]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)/ --- 1274、 (5)、设 $\displaystyle A,B$ 都是三阶可逆矩阵, $\displaystyle A,B$ 的每行元素之和分别为 $\displaystyle a,b$. 记 $\displaystyle A^\star, B^\star$ 分别为 $\displaystyle A,B$ 的伴随矩阵. (5-1)、证明 $\displaystyle A^\{-1\}$ 的每行元素之和为 $\displaystyle \frac\{1\}\{a\}$; (5-2)、求 $\displaystyle A^\star B^\star$ 的每行元素之和. (郑州大学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) / (5-1)、设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则 \begin\{aligned\} Ae=ae\Rightarrow \frac\{1\}\{a\}e=A^\{-1\}e\Rightarrow \mbox\{$A^\{-1\}$ 的每行元素之和为 $\displaystyle \frac\{1\}\{a\}$\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (5-2)、 \begin\{aligned\} &A^\star B^\star e=(|A|A^\{-1\})(|B|B^\{-1\})e=|A|\cdot |B|A^\{-1\}B^\{-1\}e\\\\ =&|A|\cdot |B|A^\{-1\}\frac\{1\}\{b\}e =|A|\cdot |B|\frac\{1\}\{b\}\frac\{1\}\{a\}e =\frac\{|A|\cdot |B|\}\{ab\}e. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A^\star B^\star$ 的每行元素之和为 $\displaystyle \frac\{|A|\cdot |B|\}\{ab\}$.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)/ --- 1275、 (6)、设 $\displaystyle A$ 是 $\displaystyle m\times n$ 矩阵, $\displaystyle B$ 是 $\displaystyle k\times n$ 矩阵, 且 $\displaystyle B$ 行满秩. 证明: $\displaystyle AX=0$ 与 $\displaystyle BX=0$ 同解的充要条件是存在 $\displaystyle m\times k$ 列满秩矩阵 $\displaystyle Q$, 使得 $\displaystyle A=QB$. (郑州大学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)、 $\displaystyle \Leftarrow$: $\displaystyle BX=0\Rightarrow AX=QBX=0$. 反之, 由 $\displaystyle \mathrm\{rank\} Q=k$ 知 $\displaystyle QY=0$ 只有零解, 而 \begin\{aligned\} AX=0\Rightarrow QBX=0\Rightarrow 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 $\displaystyle AX=0, BX=0$ 同解. (2)、 $\displaystyle \Rightarrow$: 由 $\displaystyle \mathrm\{rank\} B=k$ 知 $\displaystyle BX=0$ 的基础解系有 $\displaystyle n-k$ 个向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-k\}$. 由 $\displaystyle AX=0, BX=0$ 同解知 \begin\{aligned\} n-\mathrm\{rank\} A=n-k\Rightarrow \mathrm\{rank\} A=k, A(\eta\_1,\cdots,\eta\_\{n-k\})=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}\beta\_1\\\\\vdots\\\\\beta\_k\end\{array\}\right)^\mathrm\{T\}$, 则 $\displaystyle \mathrm\{rank\} B=k$, $\displaystyle BX=0$ 的基础解系有 $\displaystyle n-k$ 个向量 $\displaystyle \eta\_1,\cdots,\eta\_\{n-k\}$. 设 $\displaystyle C=(\eta\_1,\cdots,\eta\_\{n-k\})$, 则 \begin\{aligned\} BC=0\Rightarrow 0=C^\mathrm\{T\} B^\mathrm\{T\}=C^\mathrm\{T\}(\beta\_1^\mathrm\{T\},\cdots,\beta\_\{n-k\}^\mathrm\{T\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle \mathrm\{rank\}(C^\mathrm\{T\})=n-k, \mathrm\{rank\}(\beta\_1^\mathrm\{T\}, \cdots,\beta\_\{n-k\}^\mathrm\{T\})=n-k$ 知 $\displaystyle C^\mathrm\{T\} X=0$ 的基础解系为 $\displaystyle \beta\_1^\mathrm\{T\},\cdots,\beta\_\{n-k\}^\mathrm\{T\}$. 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\vdots\\\\\alpha\_m\end\{array\}\right)$, 则由必要性假设知 \begin\{aligned\} &AC=0\Rightarrow C^\mathrm\{T\} (\alpha\_1^\mathrm\{T\},\cdots,\alpha\_m^\mathrm\{T\})=0 \Rightarrow \alpha\_i^\mathrm\{T\}=\sum\_\{j=1\}^k q\_\{ji\}\beta\_j^\mathrm\{T\}\\\\ \Rightarrow& (\alpha\_1^\mathrm\{T\},\cdots,\alpha\_m^\mathrm\{T\})=(\beta\_1^\mathrm\{T\},\cdots,\beta\_k^\mathrm\{T\})Q^\mathrm\{T\} \Rightarrow A=QB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle \mathrm\{rank\} Q\geq \mathrm\{rank\}(QB)=\mathrm\{rank\} A=k\Rightarrow \mathrm\{rank\} Q=k$ 知 $\displaystyle Q$ 列满秩.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)/ --- 1276、 (4)、已知 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ 1&0&0&0\end\{array\}\right)$, 则 $\displaystyle A^4=\underline\{\ \ \ \ \ \ \ \ \ \ \}$, 方阵 $\displaystyle A$ 的特征值分别为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学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) / $\displaystyle A^4=E\_4$, $\displaystyle A$ 的特征值为 $\displaystyle 4$ 次单位根 $\displaystyle 1,\mathrm\{ i\}, -1,-\mathrm\{ i\}$.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)/ --- 1277、 (7)、方阵 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&1\\\\ 2&-3&2\\\\ 3&-6&4\end\{array\}\right)$ 的 Jordan 标准形为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (中国科学技术大学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 $\displaystyle A$ 的特征值全为 $\displaystyle 1$. 又由 \begin\{aligned\} A-E\to \left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&1\\\\ 0&0&0\\\\ 0&0&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 $\displaystyle \mathrm\{rank\}(J-E)=1\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\\\\ &1&\\\\ &&1\end\{array\}\right)$.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)/ --- 1278、 3、设 $\displaystyle f$ 是一元多项式, $\displaystyle B$ 是 $\displaystyle n$ 阶方阵. (1)、设 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 是 $\displaystyle B$ 的所有特征值, 证明: $\displaystyle f(\lambda\_1),\cdots, f(\lambda\_n)$ 是 $\displaystyle f(B)$ 的所有特征值. (2)、设 $\displaystyle f$ 是 $\displaystyle A$ 的特征多项式. 证明: $\displaystyle A,B$ 有公共特征值的充要条件是 $\displaystyle f(B)$ 是奇异的. (中国科学院大学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 Jordan 标准形理论知存在复可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} &P^\{-1\}BP=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star&\star\\\\ &\ddots&\star\\\\ &&\lambda\_n\end\{array\}\right)\\\\ \Rightarrow&P^\{-1\}f(B)P=\left(\begin\{array\}\{cccccccccccccccccccc\}f(\lambda\_1)&\star&\star\\\\ &\ddots&\star\\\\ &&f(\lambda\_n)\end\{array\}\right)\\\\ \Rightarrow&|\lambda E\_n-f(B)|=\prod\_\{i=1\}^n |\lambda -f(\lambda\_i)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle f(\lambda\_1),\cdots, f(\lambda\_n)$ 是 $\displaystyle f(B)$ 的所有特征值. (2)、只要证明 $\displaystyle A,B$ 没有公共特征值的充要条件是 $\displaystyle f(B)$ 是非奇异的. (2-1)、 $\displaystyle \Rightarrow$: 设 $\displaystyle g$ 是 $\displaystyle B$ 的特征多项式, 则 $\displaystyle A,B$ 没有公共特征值等价于 \begin\{aligned\} &(f,g)=1\Leftrightarrow \exists\ u,v,\mathrm\{ s.t.\} uf+vg=1\\\\ \Leftrightarrow&E\_n=u(B)f(B)+v(B)g(B)=u(B)f(B) \Rightarrow f(B)\mbox\{非奇异\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-2)、 $\displaystyle \Leftarrow$: 反之, 若 $\displaystyle f(B)$ 非奇异, 先证 $\displaystyle AX=XB$ 只有零解. 事实上, 若 $\displaystyle X$ 是一个解, 则 $\displaystyle A^kX=XB^k, \forall\ k\geq 1$, \begin\{aligned\} 0=f(A)X=Xf(B)\stackrel\{f(B)\mbox\{非奇异\}\}\{\Rightarrow\}X=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再用反证法证 $\displaystyle A,B$ 没有公共特征值. 若有, 比如 $\displaystyle \lambda\_0$, 则由 $\displaystyle |\lambda E-B|=|\lambda E-B^\mathrm\{T\}|$ 知 $\displaystyle \lambda\_0$ 也是 $\displaystyle B^\mathrm\{T\}$ 的特征值, 而 \begin\{aligned\} \exists\ \alpha\neq 0,\ \beta\neq 0,\mathrm\{ s.t.\} A\alpha=\lambda\_0\alpha,\ B^\mathrm\{T\} \beta=\lambda\_0\beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 取 $\displaystyle X=\alpha\beta^\mathrm\{T\}\neq 0$, 则 \begin\{aligned\} AX=&A\alpha \beta^\mathrm\{T\}=\lambda \alpha \beta^\mathrm\{T\},\\\\ XB=&\alpha\beta^\mathrm\{T\} B =\alpha(B^\mathrm\{T\} \beta)^\mathrm\{T\} =\alpha (\lambda \beta)^\mathrm\{T\} =\lambda \alpha\beta^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 因此, $\displaystyle AX=XB$ 有非零解 $\displaystyle X$. 这与已证矛盾.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)/ --- 1279、 7、设实反对称矩阵 $\displaystyle A$ 是可逆的, $\displaystyle \alpha\in\mathbb\{R\}^n$, 试求 $\displaystyle \mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}A&\alpha\\\\ \alpha^\mathrm\{T\}&0\end\{array\}\right)$. (中国科学院大学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 $\displaystyle A$ 是可逆的反对称矩阵知 $\displaystyle A^\{-1\}$ 也是反对称矩阵, 而 $\displaystyle \alpha^\mathrm\{T\} A^\{-1\}\alpha=0$. 再由 \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&0\\\\ 0&-\alpha^\mathrm\{T\} A^\{-1\}\alpha\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A&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 $\displaystyle \mathrm\{rank\} B=\mathrm\{rank\} A=n$.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)/ --- 1280、 9、设 $\displaystyle A$ 是 $\displaystyle n$ 阶实矩阵. 证明: 存在正交矩阵 $\displaystyle Q$, 使得 $\displaystyle Q^\mathrm\{T\} AQ$ 为对角矩阵的充要条件是 $\displaystyle A$ 的特征值均为实数且 $\displaystyle A^\mathrm\{T\} A=AA^\mathrm\{T\}$. (中国科学院大学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)、 $\displaystyle \Rightarrow$: 由必要性假设知 \begin\{aligned\} Q^\mathrm\{T\} AQ=\varLambda =\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 $\displaystyle A$ 的特征值 $\displaystyle \lambda\_1,\cdots,\lambda\_n$ 都是实数, 且 \begin\{aligned\} Q^\mathrm\{T\} A^\mathrm\{T\} Q=\varLambda\Rightarrow A^\mathrm\{T\} A&=Q\varLambda Q^\mathrm\{T\} Q\varLambda Q^\mathrm\{T\} =Q\varLambda^2Q^\mathrm\{T\}\\\\ &=Q\varLambda Q^\mathrm\{T\} Q\varLambda Q^\mathrm\{T\} =AA^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle \Leftarrow$: (2-1)、先利用 $\displaystyle A$ 的特征值都是实数证明 $\displaystyle A$ 正交相似与上三角矩阵. 对 $\displaystyle n$ 作数学归纳法. 取 $\displaystyle A$ 的属于实特征值 $\displaystyle \lambda\_1$ 的单位特征向量 $\displaystyle \eta\_1$, 并将其扩充为 $\displaystyle \mathbb\{R\}^n$ 的一组标准正交基 $\displaystyle \eta\_1,\cdots,\eta\_n$, 并设 $\displaystyle P=(\eta\_1,\cdots,\eta\_n)$, 则 \begin\{aligned\} AP=P\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star \\\\ 0&A\_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 $\displaystyle A\_1$ 的特征值就是 $\displaystyle A$ 除了 $\displaystyle \lambda\_1$ 之外的特征值, 都是实数. 据归纳假设, 存在正交阵 $\displaystyle Q$ 使得 $\displaystyle Q^\mathrm\{T\} A\_1Q$ 为上三角矩阵. 令 $\displaystyle R=P\mathrm\{diag\}(1,Q)$, 则 $\displaystyle R$ 正交, 且 $\displaystyle R^\mathrm\{T\} AR$ 为上三角矩阵. (2-2)、由第 1 步知 \begin\{aligned\} R^\mathrm\{T\} AR=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star&\star\\\\ &\ddots&\star\\\\ &&\lambda\_n\end\{array\}\right)\Rightarrow R^\mathrm\{T\} A^\mathrm\{T\} R=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_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\} 于是 $\displaystyle A^\mathrm\{T\} A=AA^\mathrm\{T\}$ \begin\{aligned\} \Rightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_n\end\{array\}\right) \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star&\star\\\\ &\ddots&\star\\\\ &&\lambda\_n\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star&\star\\\\ &\ddots&\star\\\\ &&\lambda\_n\end\{array\}\right) \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_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\} 比较 $\displaystyle (1,1)$ 元即知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_n\end\{array\}\right)$ 的 $\displaystyle (1,j)$ ($j\geq 2$) 元都是 $\displaystyle 0$. 继续比较 $\displaystyle (2,2)$ 元知 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_n\end\{array\}\right)$ 的 $\displaystyle (2,j)$ ($j\geq 3$) 元都是 $\displaystyle 0$. 等等. 如此这般, $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&&\\\\ \star&\ddots&\\\\ \star&\star&\lambda\_n\end\{array\}\right)=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)$. 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)/ --- 1281、 (5)、合同于单位矩阵的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. A. $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&0\\\\ 1&0&1\end\{array\}\right)$ B. $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&1&1\\\\ 1&1&1\end\{array\}\right)$ C. $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&2&1\\\\ 2&7&1\\\\ 1&1&8\end\{array\}\right)$ D. $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}2&-1&2\\\\ -1&3&-\frac\{3\}\{2\}\\\\ 2&-\frac\{3\}\{2\}&-4\end\{array\}\right)$ (中国矿业大学(北京)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) / $\displaystyle C$. $\displaystyle A$ 第 $\displaystyle 1,3$ 行相同, 而 $\displaystyle \mathrm\{rank\} A=2$, $\displaystyle \mathrm\{rank\} B=1$. 于是 $\displaystyle A,B$ 不可逆, 而不正定. $\displaystyle C$ 的顺序主子式都大于 $\displaystyle 0$, 而 $\displaystyle C$ 正定. $\displaystyle D$ 中 $\displaystyle 0 > -4=d\_\{33\}=e\_3^\mathrm\{T\} De\_3$ 而 $\displaystyle D$ 不正定.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)/ --- 1282、 (8)、有关正交矩阵说法正确的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. A. 特征值全为实数 B. 特征值的模一定等于 $\displaystyle 1$ C. 特征值两两互异 D. 线性无关的特征向量正交 (中国矿业大学(北京)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) / $\displaystyle B$.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)/ --- 1283、 5、已知 $\displaystyle A,B$ 均为正定矩阵, 证明: $\displaystyle AB$ 也是正定矩阵当且仅当 $\displaystyle AB=BA$. (中国矿业大学(北京)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, 正定矩阵的乘积正定的充要条件是这两个矩阵可交换. (1)、 $\displaystyle \Rightarrow$: \begin\{aligned\} AB=(AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\}=BA . \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 $\displaystyle A$ 正定知存在正交阵 $\displaystyle Q$ 使得 \begin\{aligned\} Q^\mathrm\{T\} AQ=\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle \varLambda\_i$ 是以 $\displaystyle \lambda\_i > 0$ 为对角元的对角阵, 且各 $\displaystyle \lambda\_i$ 互异. 于是 \begin\{aligned\} AB=BA&\Leftrightarrow Q^\mathrm\{T\} AQ\cdot Q^\mathrm\{T\} BQ=Q^\mathrm\{T\} BQ\cdot Q^\mathrm\{T\} AQ\\\\ &\Leftrightarrow \mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s)\tilde\{B\}=\tilde\{B\}\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s)\left(\tilde\{B\}=Q^\mathrm\{T\} BQ\right)\\\\ &\Leftrightarrow \varLambda\_i\tilde\{B\}\_\{ij\} =\tilde\{B\}\_\{ij\}\varLambda\_j\left(\tilde\{B\}=\left(\begin\{array\}\{cccccccccccccccccccc\}\tilde\{B\}\_\{11\}&\cdots&\tilde\{B\}\_\{1s\}\\\\ \vdots&\ddots&\vdots\\\\ \tilde\{B\}\_\{s1\}&\cdots&\tilde\{B\}\_\{ss\}\end\{array\}\right)\right)\\\\ &\Leftrightarrow \tilde\{B\}\_\{ij\}=0\left(i\neq j\right)\\\\ &\Leftrightarrow \tilde\{B\}=\mathrm\{diag\}(\tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 又由 $\displaystyle B$ 正定知 $\displaystyle \tilde\{B\}=Q^\mathrm\{T\} BQ$ 正定, $\displaystyle \tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}$ 也正定. 于是存在正交阵 $\displaystyle R\_i$, 使得 \begin\{aligned\} R\_i^\mathrm\{T\} \tilde\{B\}\_\{ii\}R\_i=D\_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 $\displaystyle D\_i$ 为对角阵. 取 \begin\{aligned\} P=Q\mathrm\{diag\}(R\_1,\cdots,R\_s), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P$ 正交, 且 \begin\{aligned\} P^\mathrm\{T\} AP&=\mathrm\{diag\}(R\_1^\mathrm\{T\},\cdots,R\_s^\mathrm\{T\})\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s) \mathrm\{diag\}(R\_1,\cdots,R\_s)\\\\ &=\mathrm\{diag\}(\varLambda\_1,\cdots,\varLambda\_s),\\\\ P^\mathrm\{T\} BP&=\mathrm\{diag\}(R\_1^\mathrm\{T\},\cdots,R\_s^\mathrm\{T\})\mathrm\{diag\}(\tilde\{B\}\_\{11\},\cdots,\tilde\{B\}\_\{ss\}) \mathrm\{diag\}(R\_1,\cdots,R\_s)\\\\ &=\mathrm\{diag\}(D\_1,\cdots,D\_s). \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^\mathrm\{T\} ABP=\mathrm\{diag\}(\lambda\_1\mu\_1,\cdots,\lambda\_n\mu\_n), \lambda\_i\mu\_i > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle AB$ 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)/ --- 1284、 6、 $\displaystyle A,B$ 为 $\displaystyle n$ 阶方阵, 且 $\displaystyle (A+B)^\{-1\}, (A-B)^\{-1\}$ 都存在, 说明 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B&A\end\{array\}\right)$ 可逆, 并求其逆. (中国矿业大学(北京)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\\\\ E&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B&A\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -E&E\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A-B&B\\\\ 0&A+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\} \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}E&-B(A+B)^\{-1\}\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}A-B&B\\\\ 0&A+B\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}A-B&\\\\ &A+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 $\displaystyle 0\neq |A-B|\cdot |A+B|=\left|\begin\{array\}\{cccccccccc\}A&B\\\\ B&A\end\{array\}\right|$, 而 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B&A\end\{array\}\right)$ 可逆, 且 \begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A&B\\\\ B&A\end\{array\}\right)^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ -E&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}(A-B)^\{-1\}&\\\\ &(A+B)^\{-1\}\end\{array\}\right)\\\\ &\left(\begin\{array\}\{cccccccccccccccccccc\}E&-B(A+B)^\{-1\}\\\\ 0&E\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ E&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\} 经过化简后, 得所求矩阵的逆为 \begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}\boxed\{\begin\{array\}\{c\}(A-B)^\{-1\}\\\\-(A-B)^\{-1\}B(A+B)^\{-1\}\end\{array\}\}&-(A-B)^\{-1\}B(A+B)^\{-1\}\\\\ \boxed\{\begin\{array\}\{c\}(A-B)^\{-1\}+(A+B)^\{-1\}\\\\-(A-B)^\{-1\}B(A+B)^\{-1\}\end\{array\}\}&\boxed\{\begin\{array\}\{c\}(A+B)^\{-1\}\\\\-(A-B)^\{-1\}B(A+B)^\{-1\}\end\{array\}\}\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)/ --- 1285、 (2)、设 $\displaystyle A,B$ 为 $\displaystyle n$ 阶矩阵, 且 $\displaystyle |A|=3, |B|=6, |A^\{-1\}+B|=8$, 则 $\displaystyle |A+B^\{-1\}|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / \begin\{aligned\} &|A+B^\{-1\}|=|A(A^\{-1\}+B)B^\{-1\}|\\\\ =&|A|\cdot |A^\{-1\}+B|\cdot |B|^\{-1\}=3\cdot 8\cdot \frac\{1\}\{6\}=4. \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)/ --- 1286、 (4)、下列关于方阵 $\displaystyle A,B$ 以及伴随矩阵 $\displaystyle A^\star, B^\star$ 说法正确的有 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. A. 若 $\displaystyle A$ 与 $\displaystyle B$ 相似, 则 $\displaystyle A^\star$ 与 $\displaystyle B^\star$ 也相似. B. 若 $\displaystyle A$ 与 $\displaystyle B$ 相似, 则 $\displaystyle A^\star$ 与 $\displaystyle B^\star$ 未必相似. C. 若 $\displaystyle A$ 与 $\displaystyle B$ 合同, 则 $\displaystyle A^\star$ 与 $\displaystyle B^\star$ 也合同. D. 若 $\displaystyle A$ 与 $\displaystyle B$ 合同, 则 $\displaystyle A^\star$ 与 $\displaystyle B^\star$ 未必合同. (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) / $\displaystyle AC$. 若 $\displaystyle A,B$ 相似, 则存在可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\{-1\}AP=B\Rightarrow P^\star A^\star (P^\star)^\{-1\}=B^\star. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle A,B$ 合同, 则存在可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\mathrm\{T\} AP=B\Rightarrow P^\star A^\star P^\{-\mathrm\{T\} \star\}=B^\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)/ --- 1287、 (7)、设 $\displaystyle A$ 是 $\displaystyle 3$ 阶矩阵, $\displaystyle A$ 的秩为 $\displaystyle 2$, 且 \begin\{aligned\} A\left(\begin\{array\}\{cccccccccccccccccccc\}1&3\\\\ -1&0\\\\ 2&2\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&6\\\\ 1&0\\\\ -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\} 则 $\displaystyle A$ 的迹等于 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / 设 $\displaystyle \alpha=(1,-1,2)^\mathrm\{T\},\beta=(3,0,2)^\mathrm\{T\}$, 则由题设, \begin\{aligned\} A\alpha=-\alpha, A\beta=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 联合 $\displaystyle \mathrm\{rank\} A=2$ 知 $\displaystyle A$ 的特征值为 $\displaystyle 0,-1,2$, 而 $\displaystyle \mathrm\{tr\} A=0-1+2=1$.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)/ --- 1288、 (10)、设 $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, 满足 $\displaystyle |A|=6$, 且 $\displaystyle A+A^\star=5E\_n$, 其中 $\displaystyle A^\star$ 表示 $\displaystyle A$ 的伴随矩阵, $\displaystyle E\_n$ 是 $\displaystyle n$ 阶单位矩阵, 则 $\displaystyle A$ 的最小多项式是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / 由题设, \begin\{aligned\} &5A=A\cdot 5E=A(A+A^\star)=A^2+|A|E=A^2+6E\\\\ \Rightarrow& A^2-5A+6E=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A$ 的最小多项式 $\displaystyle m(x)\mid (x^2-5x+6)\Rightarrow m(x)=x-2\mbox\{或\} x-3\mbox\{或\} x^2-5x+6$. (10-1)、若 $\displaystyle m(x)=x-2$, 则 $\displaystyle A=2E\Rightarrow |A|=2^n\neq 6$, 矛盾. (10-2)、若 $\displaystyle m(x)=x-3$, 则 $\displaystyle A=3E\Rightarrow |A|=3^n\neq 6$, 矛盾. 故 $\displaystyle m(x)=x^2-5x+6$.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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