张祖锦2023年数学专业真题分类70天之第55天

zhangzujin · 发表于 2023-3-5 13:13:29

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## 张祖锦2023年数学专业真题分类70天之第55天 --- 1243、 5、设 $\displaystyle m,n$ 为正整数, $\displaystyle A$ 为 $\displaystyle n$ 阶实对称矩阵. 证明: $\displaystyle A$ 正定当且仅当对任意的 $\displaystyle n\times m$ 列满秩矩阵 $\displaystyle B$, 均有 $\displaystyle B^\mathrm\{T\} AB$ 正定. (西南大学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)、 $\displaystyle \Rightarrow$: 由 $\displaystyle \mathrm\{rank\} B=m$ 知 $\displaystyle Bx=0$ 只有零解. 从而对 $\displaystyle \forall\ 0\neq x\in\mathbb\{R\}^m$, $\displaystyle y=Bx\neq 0$, \begin\{aligned\} x^\mathrm\{T\}(B^\mathrm\{T\} AB)x=y^\mathrm\{T\} Ay > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle B^\mathrm\{T\} AB$ 正定. (2)、 $\displaystyle \Leftarrow$: 用反证法. 若 $\displaystyle A$ 不是正定的, 则 $\displaystyle A$ 有一个实特征值 $\displaystyle \lambda\_1\leq 0$. 设 $\displaystyle 0\neq \alpha\_1\in\mathbb\{R\}^n$ 为对应的单位特征向量, 将其扩充为 $\displaystyle \mathbb\{R\}^n$ 的一组标准正交基 $\displaystyle \alpha\_1,\cdots,\alpha\_n$. 取 $\displaystyle B=(\alpha\_1,\cdots,\alpha\_m)$, 则 $\displaystyle \mathrm\{rank\} B=m$, 而 $\displaystyle B$ 列满秩, 但 \begin\{aligned\} B^\mathrm\{T\} AB=\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\}\\\\\vdots\\\\\alpha\_m^\mathrm\{T\}\end\{array\}\right)A(\alpha\_1,\cdots,\alpha\_m) =\left(\begin\{array\}\{cccccccccccccccccccc\}\alpha\_1^\mathrm\{T\} A\alpha\_1&\star\\\\ \star&\star\end\{array\}\right) =\left(\begin\{array\}\{cccccccccccccccccccc\}\lambda\_1&\star\\\\ \star&\star\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\} 蕴含 $\displaystyle e\_1^\mathrm\{T\} (B^\mathrm\{T\} AB)e\_1=\lambda\_1\leq 0$. 这与 $\displaystyle B^\mathrm\{T\} AB$ 正定矛盾. 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)/ --- 1244、 6、设 $\displaystyle A,B,X$ 是复数域上的 $\displaystyle n$ 阶矩阵, 且满足 $\displaystyle XA=BX$. 证明: (1)、对任意的复系数多项式 $\displaystyle f(x)$ 均有 $\displaystyle Xf(A)=f(B)X$; (2)、若 $\displaystyle A,B$ 没有相同的特征值, 则 $\displaystyle X=0$. (西南大学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)、由 $\displaystyle XA=BX$, \begin\{aligned\} XA^k=B^kX\Rightarrow XA^\{k+1\}=XA^kA=B^kXA=B^kBX=B^\{k+1\}X \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 $\displaystyle XA^k=B^kX,\forall\ k\geq 0$. 从而对任一复系数多项式 $\displaystyle f(x)$, $\displaystyle Xf(A)=f(B)X$. (2)、设 $\displaystyle f(x)=|xE-A|, g(x)=|xE-B|$. 由 $\displaystyle A,B$ 没有公共特征值知 \begin\{aligned\} &(f,g)=1\Rightarrow \exists\ u,v\in\mathbb\{C\}[x],\mathrm\{ s.t.\} uf+vg=1\\\\ \Rightarrow& E=u(B)f(B)+v(B)g(B)\xlongequal[\tiny\mbox\{Cayley\}]\{\tiny\mbox\{Hamilton-\}\} u(B)f(B)\Rightarrow f(B)\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\} 再由第 1 步知 \begin\{aligned\} f(B)X=Xf(A)\xlongequal[\tiny\mbox\{Cayley\}]\{\tiny\mbox\{Hamilton-\}\} X0=0\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\} 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)/ --- 1245、 7、设 \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&2\cos\theta\\\\ 0&1\end\{array\}\right),\quad B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ 2\cos\theta&-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\} (1)、当 $\displaystyle \theta=0$ 时, 不存在正整数 $\displaystyle n$, 使得 $\displaystyle (AB)^n$ 为单位矩阵; (2)、对任意正整数 $\displaystyle m > 1$, 当 $\displaystyle \theta=\frac\{\pi\}\{m\}$ 时, 使得 $\displaystyle (AB)^n$ 为单位矩阵的最小正整数 $\displaystyle n$ 恰为 $\displaystyle m$. (西南大学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)、当 $\displaystyle \theta=0$ 时, \begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&2\\\\ 0&1\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0\\\\ 2&-1\end\{array\}\right)\Rightarrow C\equiv AB=\left(\begin\{array\}\{cccccccccccccccccccc\}3&-2\\\\ 2&-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\} 易知 $\displaystyle C$ 的特征值为 $\displaystyle 1,1$. 又由 \begin\{aligned\} C-E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1\\\\ 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 $\displaystyle C$ 的 Jordan 标准形为 $\displaystyle \left(\begin\{array\}\{cccccccccccccccccccc\}1&-1\\\\ 0&0\end\{array\}\right)$. 从而存在复可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\{-1\}CP=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1\\\\ 0&1\end\{array\}\right)\Rightarrow P^\{-1\}C^nP=\left(\begin\{array\}\{cccccccccccccccccccc\}1&n\\\\ 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\} 用反证法. 若 $\displaystyle C^n=E$, 则 $\displaystyle E=\left(\begin\{array\}\{cccccccccccccccccccc\}1&n\\\\ 0&1\end\{array\}\right)\Rightarrow n=0$. 这是一个矛盾. 故有结论. (2)、当 $\displaystyle \theta=\frac\{\pi\}\{m\}\ (m > 1)$ 时, \begin\{aligned\} C\equiv AB=\left(\begin\{array\}\{cccccccccccccccccccc\}4\cos^2\frac\{\pi\}\{m\}-1&-2\cos\frac\{\pi\}\{m\}\\\\ 2\cos\frac\{\pi\}\{m\}&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\} 易知 $\displaystyle C$ 的特征值为 $\displaystyle \mathrm\{e\}^\{\pm \frac\{2\pi\}\{m\}\}$, 而存在复可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\{-1\}CP=\mathrm\{diag\}(\mathrm\{e\}^\frac\{2\pi\}\{m\},\mathrm\{e\}^\{-\frac\{2\pi\}\{m\}\}) \Rightarrow P^\{-1\}C^nP=\mathrm\{diag\}(\mathrm\{e\}^\frac\{2n\pi\}\{m\},\mathrm\{e\}^\{-\frac\{2n\pi\}\{m\}\}). \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\} C^n=E\Leftrightarrow \mathrm\{e\}^\frac\{2n\pi\}\{m\}=1\Leftrightarrow \frac\{2n\pi\}\{m\}\in 2\pi\mathbb\{Z\}\Rightarrow \min\_\{n\geq 1; C^n=E\}n=m. \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)/ --- 1246、 (2)、设 $\displaystyle A,B,C$ 为 $\displaystyle 3$ 阶实方阵, 且 $\displaystyle BAA^\mathrm\{T\} =CAA^\mathrm\{T\}$, 则 $\displaystyle \mathrm\{rank\}(BA-CA)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ ($A^\mathrm\{T\}$ 表示矩阵 $\displaystyle A$ 的转置). (西南交通大学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) / (2-1)、先证明方程组 $\displaystyle AX=0$ 与 $\displaystyle A^\mathrm\{T\} AX=0$ 同解. 显然 $\displaystyle AX=0\Rightarrow A^\mathrm\{T\} AX=0$. 反之, \begin\{aligned\} A^\mathrm\{T\} AX=0&\Rightarrow 0=X^\mathrm\{T\} A^\mathrm\{T\} AX=(AX)^\mathrm\{T\} (AX)\Rightarrow AX=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2-2)、回到题目. \begin\{aligned\} &BAA^\mathrm\{T\}=CAA^\mathrm\{T\}\Leftrightarrow(B-C)AA^\mathrm\{T\}=0\\\\ \stackrel\{\mbox\{转置\}\}\{\Leftrightarrow\}&AA^\mathrm\{T\}(B^\mathrm\{T\}-C^\mathrm\{T\})=0\stackrel\{\mbox\{第 i 步\}\}\{\Leftrightarrow\}A^\mathrm\{T\}(B^\mathrm\{T\}-C^\mathrm\{T\})=0\\\\ \stackrel\{\mbox\{转置\}\}\{\Leftrightarrow\}&(B-C)A=0\Leftrightarrow \mathrm\{rank\}\left\[(B-C)A\right\]=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)/ --- 1247、 (3)、设实数域上 $\displaystyle n$ 阶方阵 $\displaystyle A$ 的每行元素之和等于常数 $\displaystyle a$, 则 $\displaystyle A^2+E$ 的每行元素之和等于 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西南交通大学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 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则 \begin\{aligned\} Ae=ae\Rightarrow (A^2+E)e=A(ae)+e=(a^2+1)e. \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 $\displaystyle a^2+1$.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)/ --- 1248、 (4)、设 $\displaystyle A$ 为实数域上的奇数阶正交矩阵, 且 $\displaystyle \det A=1$, 则 $\displaystyle \det(E-A)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ ($\det A$ 表示矩阵 $\displaystyle A$ 的行列式). (西南交通大学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\} &|E-A|=|A^\mathrm\{T\}|\cdot|E-A|=|A^\mathrm\{T\}-E|=|(A-E)^\mathrm\{T\}|\\\\ =&|A-E|=(-1)^n|E-A|=-|E-A| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle |E-A|=0$.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)/ --- 1249、 (5)、设 $\displaystyle A$ 为 $\displaystyle n$ 阶实对称矩阵, 且 $\displaystyle A^2-3A+aE=0$. 若 $\displaystyle A$ 是正定矩阵, 则 $\displaystyle a$ 的取值范围是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. [张祖锦注: 最好将题目改为 $\displaystyle A$ 的最小多项式为 $\displaystyle x^2-3x+a$.] (西南交通大学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 $\displaystyle A$ 的特征值, 则由题设知 \begin\{aligned\} \lambda^2-3\lambda+a=0\Rightarrow \lambda=\frac\{3\pm \sqrt\{9-4a\}\}\{2\} > 0\Leftrightarrow 0 < \lambda\leq\frac\{9\}\{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)/ --- 1250、 (2)、 (15 分) 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ -4&4&0\\\\ -2&1&2\end\{array\}\right)$, 求 $\displaystyle A^\{10\}$. (西南交通大学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 $\displaystyle 2,2,2$. 由 \begin\{aligned\} 2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&-\frac\{1\}\{2\}&0\\\\ 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 $\displaystyle A$ 的属于特征值 $\displaystyle 2$ 的特征向量分别为 $\displaystyle \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\0\end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}0\\\\0\\\\1\end\{array\}\right)$. 设 $\displaystyle A$ 的 Jordan 标准形为 $\displaystyle J$, 则 [分析严格上三角部分 $\displaystyle 1$ 的个数即可知道秩] \begin\{aligned\} \mathrm\{rank\}(J-2E)=\mathrm\{rank\}(A-2E)=1\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&\\\\ &2&\\\\ &&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\} 于是 $\displaystyle (J-2E)^2=0\Rightarrow (A-2E)^2=0$. 设 \begin\{aligned\} x^\{10\}=q(x)(x-2)^2+ax+b, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则令 $\displaystyle x=2$, 求导后令 $\displaystyle x=2$ 知 \begin\{aligned\} &2^\{10\}=2a+b, 10\cdot 2^9=a\Rightarrow a=5\cdot 2^\{10\}, b=-9\cdot 2^\{10\}\\\\ \Rightarrow& A^\{10\}=aA+bE=5\cdot 2^\{10\}\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0\\\\ -4&4&0\\\\ -2&1&2\end\{array\}\right)-9\cdot 2^\{10\}E\\\\ &=\left(\begin\{array\}\{cccccccccccccccccccc\}-9\cdot 2^\{10\}&5\cdot 2^\{10\}&0\\\\ -20\cdot 2^\{10\}&11\cdot 2^\{10\}&0\\\\ -10\cdot 2^\{10\}&5\cdot 2^\{10\}&2^\{10\}\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)/ --- 1251、 (5)、 (15 分) 设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&a&1\\\\ 0&0&1&2\end\{array\}\right)$. (1)、已知 $\displaystyle A$ 的一个特征值为 $\displaystyle 3$, 求 $\displaystyle a$ 的值; (2)、求矩阵 $\displaystyle B$, 使得 $\displaystyle (AB)^\mathrm\{T\}(AB)$ 为对角矩阵. (西南交通大学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)、 $\displaystyle 0=|A-3E|=8(2-a)\Rightarrow a=2$. (2)、设 $\displaystyle C=A^\mathrm\{T\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0&0\\\\ 0&1&0&0\\\\ 0&0&5&4\\\\ 0&0&4&5\end\{array\}\right)$, 则可取 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&\\\\ &1&&\\\\ &&1&-\frac\{4\}\{5\}\\\\ &&&1\end\{array\}\right)$ 后, \begin\{aligned\} (AB)^\mathrm\{T\}(AB)=B^\mathrm\{T\} (A^\mathrm\{T\} A)B=B^\mathrm\{T\} CB=\mathrm\{diag\}\left(1,1,5,\frac\{9\}\{5\}\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)/ --- 1252、 5、 (10 分) 设 $\displaystyle A$ 为数域 $\displaystyle \mathbb\{K\}$ 上的 $\displaystyle m\times n$ 矩阵, 且 $\displaystyle \mathrm\{rank\} A=r$, 则对任意一个自然数 $\displaystyle k\ (k\leq r)$, 都存在数域 $\displaystyle \mathbb\{K\}$ 上的 $\displaystyle m\times n$ 矩阵 $\displaystyle B$ 和 $\displaystyle m\times n$ 矩阵 $\displaystyle C$, 满足 $\displaystyle A=B+C$, 其中 $\displaystyle \mathrm\{rank\} B=k, \mathrm\{rank\} C=r-k$. (西南交通大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \mathrm\{rank\} A=r$ 知存在可逆矩阵 $\displaystyle P,Q$ 使得 $\displaystyle A=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_r&0\\\\ 0&0\end\{array\}\right)Q$. 令 \begin\{aligned\} B=P\left(\begin\{array\}\{cccccccccccccccccccc\}E\_k&\\\\ &0\end\{array\}\right)Q, C=P\left(\begin\{array\}\{cccccccccccccccccccc\}0\_\{k\times k\}&&\\\\ &E\_\{r-k\}&\\\\ &&0\end\{array\}\right)Q, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle A=B+C, \mathrm\{rank\} B=k, \mathrm\{rank\} C=r-k$.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)/ --- 1253、 6、 (10 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实方阵, 且 $\displaystyle A^\mathrm\{T\} A$ 的特征值大于 $\displaystyle 1$. 证明: $\displaystyle E-A$ 可逆. (西南交通大学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 $\displaystyle A^\mathrm\{T\} A$ 实对称知存在正交阵 $\displaystyle Q$ 使得 \begin\{aligned\} Q^\mathrm\{T\} (A^\mathrm\{T\} A) Q=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i > 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 用反证法. 若$E-A$ 不可逆, 则 $\displaystyle |E-A|=0$, $\displaystyle (E-A)x=0$ 有非零解 $\displaystyle 0\neq \alpha\in\mathbb\{R\}^n$. 于是 $\displaystyle A\alpha=\alpha$. 设 $\displaystyle \alpha=Q\beta$, 则 $\displaystyle 0\neq\beta=\left(\begin\{array\}\{cccccccccccccccccccc\}b\_1\\\\\vdots\\\\b\_n\end\{array\}\right)\in\mathbb\{R\}^n$, \begin\{aligned\} &\beta^\mathrm\{T\}\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)\beta =\beta^\mathrm\{T\} Q^\mathrm\{T\} A^\mathrm\{T\} AQ\beta =\alpha^\mathrm\{T\} A^\mathrm\{T\} A\alpha\\\\ =&(A\alpha)^\mathrm\{T\} A\alpha =\alpha^\mathrm\{T\} \alpha =\beta^\mathrm\{T\} Q^\mathrm\{T\} Q\beta=\beta^\mathrm\{T\} \beta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 \begin\{aligned\} &\sum\_\{k=1\}^n (\lambda\_k-1)b\_k^2=0\stackrel\{\lambda\_k > 1\}\{\Rightarrow\} (\lambda\_k-1)b\_k^2=0, \forall\ k\Rightarrow b\_k=0, \forall\ k\Rightarrow \beta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 这是一个矛盾. 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)/ --- 1254、 4、设 $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, 满足 \begin\{aligned\} A^2=2022A+2023I. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 求实数 $\displaystyle c\_0, c\_1\ (c\_0 < c\_1)$, 使得 \begin\{aligned\} \mathrm\{rank\}(A+c\_0I)+\mathrm\{rank\}(A+c\_1I)=n, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 并证明之. 这里 $\displaystyle \mathrm\{rank\} A$ 表示矩阵 $\displaystyle A$ 的秩. (湘潭大学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)、先证明一个结论. 设 $\displaystyle A,B$ 分别为 $\displaystyle n$ 阶矩阵且 $\displaystyle AB=0$, 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. \begin\{aligned\} 0=A^2-2022A-2023I=(A-2023I)(A+I).\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 取 $\displaystyle c\_0=-2023, c\_1=1$, 则 \begin\{aligned\} \mathrm\{rank\}(A-2023I)+\mathrm\{rank\}(A+I)\geq&\mathrm\{rank\}\left\[(A-2023I)+(A+I)\right\]\\\\ =&\mathrm\{rank\}(-2024I)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 再由 $\displaystyle (I)$ 及第 1 步知 $\displaystyle \mathrm\{rank\}(A-2023I)+\mathrm\{rank\}(A+I)\leq n$. 故 \begin\{aligned\} \mathrm\{rank\}(A-2023I)+\mathrm\{rank\}(A+I)=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)/ --- 1255、 5、设 $\displaystyle A,B$ 是 $\displaystyle n$ 阶正定矩阵. (1)、证明: $\displaystyle A^\star$ 是正定矩阵; (2)、 $\displaystyle AB=BA$ 是 $\displaystyle AB$ 为正定矩阵的什么条件, 并证明你的结论. (湘潭大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、由 $\displaystyle A$ 正定知存在正交阵 $\displaystyle P$ 使得 \begin\{aligned\} &P^\mathrm\{T\} AP=P^\{-1\}AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n), \lambda\_i > 0\left(\Rightarrow |A|=\prod\_\{i=1\}^n \lambda\_i > 0\right)\\\\ \Rightarrow& P^\{-1\}A^\{-1\}P=\mathrm\{diag\}\left(\frac\{1\}\{\lambda\_1\},\cdots,\frac\{1\}\{\lambda\_n\}\right)\\\\ \Rightarrow& P^\{-1\}A^\star P=P^\{-1\}|A|A^\{-1\}P=\mathrm\{diag\}\left(\frac\{|A|\}\{\lambda\_1\},\cdots,\frac\{|A|\}\{\lambda\_n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle A^\{-1\}$ 正定. (2)、充要条件. (2-1)、 $\displaystyle \Rightarrow$: 由 $\displaystyle (AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\} =BA$ 知 $\displaystyle AB$ 是实对称矩阵. 设 $\displaystyle \lambda\in\mathbb\{R\}$ 是 $\displaystyle AB$ 的任一特征值, $\displaystyle 0\neq\alpha\in\mathbb\{R\}^n$ 为对应的特征向量, 则 \begin\{aligned\} &AB\alpha=\lambda \alpha\Rightarrow B\alpha=\lambda A^\{-1\}\alpha\\\\ \Rightarrow& 0 < \alpha^\mathrm\{T\} B\alpha=\lambda\alpha^\mathrm\{T\} A^\{-1\}\alpha\Rightarrow \lambda=\frac\{\alpha^\mathrm\{T\} B\alpha\}\{\alpha^\mathrm\{T\} A^\{-1\}\alpha\}\stackrel\{\mbox\{题设及第 1 步\}\}\{ > \}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 既然实对称矩阵 $\displaystyle AB$ 的特征值都大于 $\displaystyle 0$, 那么 $\displaystyle AB$ 就是正定的. (2-2)、 $\displaystyle \Leftarrow$: 若 $\displaystyle AB$ 正定, 则 $\displaystyle AB$ 实对称, 而 $\displaystyle AB=(AB)^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\}=BA$.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)/ --- 1256、 7、设 $\displaystyle n$ 阶方阵 $\displaystyle A,B$ 满足 $\displaystyle AB=A-B$, 证明: (1)、 $\displaystyle \lambda=1$ 不是 $\displaystyle B$ 的特征值; (2)、若 $\displaystyle B$ 相似于对角矩阵, 则有可逆矩阵 $\displaystyle T$, 使得 $\displaystyle T^\{-1\}AT, T^\{-1\}BT$ 均为对角矩阵. (湘潭大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 \begin\{aligned\} &AB-A+B=0\Rightarrow (A+I)(B-I)=-I\\\\ \Rightarrow& |B-I|\neq 0\Rightarrow \mbox\{$1$ 不是 $\displaystyle B$ 的特征值\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle AB=A-B\Rightarrow A(I-B)=B\Rightarrow A=(I-B)^\{-1\}B$. 由题设, 存在可逆矩阵 $\displaystyle T$ 使得 \begin\{aligned\} &T^\{-1\}BT=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)\\\\ \Rightarrow& T^\{-1\}(I-B)T=\mathrm\{diag\}(1-\lambda\_1,\cdots,1-\lambda\_n)\\\\ \Rightarrow&T^\{-1\}(I-B)^\{-1\}T=\mathrm\{diag\}\left(\frac\{1\}\{1-\lambda\_1\},\cdots,\frac\{1\}\{1-\lambda\_n\}\right)\\\\ \Rightarrow& T^\{-1\}(I-B)^\{-1\}BT=T^\{-1\}(I-B)^\{-1\}T\cdot T^\{-1\}BT\\\\ &=\mathrm\{diag\}\left(\frac\{\lambda\_1\}\{1-\lambda\_1\},\cdots,\frac\{\lambda\_n\}\{1-\lambda\_n\}\right). \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 $\displaystyle T^\{-1\}AT, T^\{-1\}BT$ 均为对角矩阵.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)/ --- 1257、 8、已知方阵 $\displaystyle A$ 的最小多项式为 $\displaystyle m(\lambda)=\lambda^3+\lambda^2-\lambda-1$. (1)、证明: $\displaystyle A$ 可逆, $\displaystyle A+E, A-E$ 均不可逆; (2)、如果 $\displaystyle A$ 的特征多项式为 $\displaystyle f(\lambda)=(\lambda-1)^2(\lambda+1)^3$, 求 $\displaystyle A$ 的 Jordan 标准形. (湘潭大学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)、 $\displaystyle m(x)=(x+1)^2(x-1)$. 设 $\displaystyle \lambda$ 是 $\displaystyle A$ 的特征值, 则 $\displaystyle 0=m(\lambda)\Rightarrow \lambda=-1\mbox\{或\} 1$. 既然 $\displaystyle A$ 的特征值都非零, 而 $\displaystyle |A|$ 作为 $\displaystyle A$ 的所有特征值的乘积, 也非零. 故 $\displaystyle A$ 可逆. 由最小多项式的定义知 \begin\{aligned\} 0=m(A)=(A+E)^2(A-E). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 往用反证法证明 $\displaystyle A+E$ 不可逆. 若 $\displaystyle A+E$ 可逆, 则上式蕴含 $\displaystyle A-E=0$, $\displaystyle A$ 有零化多项式 $\displaystyle x-1$, 次数比 $\displaystyle m(x)$ 的小. 这与最小多项式的定义矛盾. 故有结论. 类似的可证 $\displaystyle A-E$ 不可逆. (2)、由 $\displaystyle m(x)$ 的形式知 $\displaystyle A$ 的 Jordan 标准形有 Jordan 块 $\displaystyle J\_2(-1) J\_1(1)$, 且以 $\displaystyle -1$ 为对角元的 Jordan 块的阶数 $\displaystyle \leq 2$, 以 $\displaystyle 1$ 为对角元的 Jordan 块都是一阶的. 再由 $\displaystyle f(x)$ 的形式知 $\displaystyle A$ 是 $\displaystyle 5$ 阶方阵. 故 $\displaystyle J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&&&&\\\\ &1&&&\\\\ &&-1&&\\\\ &&&-1&1\\\\ &&&&-1\end\{array\}\right)$.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)/ --- 1258、 10、设 $\displaystyle A$ 为 $\displaystyle n$ 阶实对称矩阵, 且 $\displaystyle |A|\neq 0$, 证明: $\displaystyle A$ 为正定矩阵的充要条件是对所有正定矩阵 $\displaystyle B$, 恒有 $\displaystyle \mathrm\{tr\}(AB) > 0$. 这里 $\displaystyle \mathrm\{tr\} A$ 表示矩阵 $\displaystyle A$ 的迹, 即 $\displaystyle A$ 的主对角元素之和. [张祖锦注: 给出的参考解答不要条件’且 $\displaystyle |A|\neq 0$‘!] (湘潭大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle A$ 实对称知存在正交阵 $\displaystyle P$ 使得 $\displaystyle P^\mathrm\{T\} AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)$, 其中 $\displaystyle \lambda\_i\in\mathbb\{R\}$. (1)、 $\displaystyle \Rightarrow$: 若 $\displaystyle A$ 正定, 则 $\displaystyle \lambda\_i > 0$. 进而 \begin\{aligned\} \mathrm\{tr\}(AB)=&\mathrm\{tr\}(P^\{-1\}APP^\{-1\}BP)\\\\ =&\mathrm\{tr\}(\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n)\tilde\{B\})\left(\tilde\{B\}=P^\{-1\}BP\mbox\{正定\}\right)\\\\ =&\sum\_\{i=1\}^n \lambda\_i\tilde\{b\}\_\{ii\} > 0\left(\mbox\{$\tilde\{B\}$ 正定 $\displaystyle \Rightarrow e\_i^\mathrm\{T\} \tilde\{B\}e\_i=\tilde\{b\}\_\{ii\} > 0$\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 $\displaystyle \Leftarrow$: 若对任意的正定矩阵 $\displaystyle B$, 有 $\displaystyle \mathrm\{tr\}(AB) > 0$, 则 \begin\{aligned\} 0 < &\mathrm\{tr\}(AB)=\mathrm\{tr\}(P^\{-1\}APP^\{-1\}BP)=\sum\_\{i=1\}^n \lambda\_i\tilde\{b\}\_\{ii\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle \tilde\{b\}\_\{ii\}$ 为正定矩阵 $\displaystyle \tilde\{B\}=P^\{-1\}BP$ 的第 $\displaystyle i$ 个对角元. 由 $\displaystyle B$ 的任意性知 $\displaystyle \tilde\{B\}$ 的任意性. 往用反证法证明各 $\displaystyle \lambda\_i > 0$. 若某些 $\displaystyle \lambda\_i\leq 0$, 则取 \begin\{aligned\} \tilde\{b\}\_\{ii\}=\left\\{\begin\{array\}\{llllllllllll\}1,&\lambda\_i=0,\\\\ 1,&\lambda\_i > 0,\\\\ \frac\{1\}\{-\lambda\_i\}\sum\_\{j: \lambda\_j > 0\}\lambda\_j,&\lambda\_i < 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\} 后 \begin\{aligned\} 0 < \sum\_\{i=1\}^n \lambda\_i\tilde\{b\}\_\{ii\} =\sum\_\{j: \lambda\_j > 0\}\lambda\_j\cdot 1 +\sum\_\{i: \lambda\_i < 0\}\lambda\_i\cdot \frac\{1\}\{-\lambda\_i\}\sum\_\{j: \lambda\_j > 0\}\lambda\_j\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (若只有一个 $\displaystyle \lambda\_i < 0$, 则 $\displaystyle =0$, 否则 $\displaystyle < 0$) 这是一个矛盾. 故有结论: 各 $\displaystyle \lambda\_i > 0$, 而 $\displaystyle A$ 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)/ --- 1259、 6、 (15 分) 设 $\displaystyle A$ 为 $\displaystyle n\times n$ 实矩阵, 证明: $\displaystyle \mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} A$. (新疆大学2023年高等代数考研试题) [矩阵 ] [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 \begin\{aligned\} V\_1&=\left\\{\alpha\in\mathbb\{R\}^n; A^\mathrm\{T\} A\alpha=0\right\\},\\\\ V\_2&=\left\\{\alpha\in\mathbb\{R\}^n; A\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\} 则显然有 $\displaystyle V\_2\subset V\_1$. 又由 \begin\{aligned\} A^\mathrm\{T\} A\alpha =0&\Rightarrow (A\alpha)^\mathrm\{T\} A\alpha=0\Rightarrow A\alpha=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle V\_1\subset V\_2$. 故 \begin\{aligned\} V\_1=V\_2\Rightarrow&\dim V\_1=\dim V\_2\\\\ \Rightarrow&\mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} A . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 于是 $\displaystyle \mathrm\{rank\}(A^\mathrm\{T\} A)=\mathrm\{rank\} A$.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)/ --- 1260、 9、 (20 分) 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-2&0\\\\ -2&1&-2\\\\ 0&-2&0\end\{array\}\right)$, 求正交矩阵 $\displaystyle T$, 使得 $\displaystyle T^\{-1\}AT$ 成对角形. (新疆大学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 $\displaystyle 4,1,-2$. 由 \begin\{aligned\} 4E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-1\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right), E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&1\\\\ 0&1&\frac\{1\}\{2\}\\\\ 0&0&0 \end\{array\}\right), -2E-A\to\left(\begin\{array\}\{cccccccccccccccccccc\} 1&0&-\frac\{1\}\{2\}\\\\ 0&1&-1\\\\ 0&0&0 \end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle A$ 的属于特征值 $\displaystyle 4,1,-2$ 的特征向量分别为 \begin\{aligned\} \xi\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1 \end\{array\}\right), \xi\_2=\left(\begin\{array\}\{cccccccccccccccccccc\}-2\\\\-1\\\\2 \end\{array\}\right), \xi\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\2\\\\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\} 将 $\displaystyle \xi\_1,\xi\_2,\xi\_3$ 标准正交化为 $\displaystyle \eta\_1,\eta\_2,\eta\_3$. 令 \begin\{aligned\} T=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\} \frac\{2\}\{3\}&-\frac\{2\}\{3\}&\frac\{1\}\{3\}\\\\ -\frac\{2\}\{3\}&-\frac\{1\}\{3\}&\frac\{2\}\{3\}\\\\ \frac\{1\}\{3\}&\frac\{2\}\{3\}&\frac\{2\}\{3\} \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 $\displaystyle T$ 正交, 且 \begin\{aligned\} T^\mathrm\{T\} AT=T^\{-1\}AT=\mathrm\{diag\}\left(4,1,-2\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)/ --- 1261、 10、 (15 分) 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&2&0\\\\ 0&2&0\\\\ -2&-2&1\end\{array\}\right)$. (1)、 (5 分) 求 $\displaystyle A$ 的不变因子; (2)、 (3 分) 求 $\displaystyle A$ 的初等因子; (3)、 (5 分) 求 $\displaystyle A$ 的极小多项式; (4)、 (2 分) 求 $\displaystyle A$ 的若尔当标准形. (新疆大学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 $\displaystyle A-E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&0\\\\ 0&1&0\\\\ 0&0&0\end\{array\}\right)$ 知 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 [如果严格上三角部分没有 $\displaystyle 1$, 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\}2&&\\\\ &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\} 从而 $\displaystyle A$ 的极小多项式为 $\displaystyle (\lambda-2)(\lambda-1)^2$, 初等因子为 $\displaystyle \lambda-2,(\lambda-1)^2$, 不变因子为 $\displaystyle 1,1,(\lambda-2)(\lambda-1)^2$.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)/ --- 1262、 6、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶实方阵, $\displaystyle B=AA^\mathrm\{T\}$. 证明: 对一切正整数 $\displaystyle m$, 有 \begin\{aligned\} \mathrm\{rank\} A=\mathrm\{rank\}(B^m)=\mathrm\{rank\}(B^mA). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (云南大学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)、先证明一个结论. 设 $\displaystyle B\in M\_n(\mathbb\{R\})$ 是半正定对称矩阵, $\displaystyle x\in\mathbb\{R\}^n$. 则 $\displaystyle x^\mathrm\{T\} Bx=0$ 等价于 $\displaystyle Bx=0$. $\displaystyle \Leftarrow$: 显然成立. $\displaystyle \Rightarrow$: 由 $\displaystyle B$ 半正定知存在实矩阵 $\displaystyle C$ 使得 $\displaystyle B=C^\mathrm\{T\} C$, 而 \begin\{aligned\} &x^\mathrm\{T\} Bx=0\Rightarrow 0=x^\mathrm\{T\} C^\mathrm\{T\} Cx\stackrel\{y=Cx\}\{=\}y^\mathrm\{T\} y\\\\ \Rightarrow& 0=y=Cx\Rightarrow Bx=C^\mathrm\{T\} Cx=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. 显然 $\displaystyle A^\mathrm\{T\} x=0\Rightarrow Bx=0\Rightarrow B^mx=0$. 反之, \begin\{aligned\} &B^mx=0\Rightarrow AA^\mathrm\{T\} B^\{m-2\}AA^\mathrm\{T\} x=0 \Rightarrow x^\mathrm\{T\} AA^\mathrm\{T\} B^\{m-2\}AA^\mathrm\{T\} x=0\\\\ \overset\{\tiny\mbox\{第1步\}\}\{\Longrightarrow\}& B^\{m-2\}AA^\mathrm\{T\} x=0\Rightarrow B^\{m-1\}X=0\Rightarrow \cdots \Rightarrow Bx=0\\\\ \Rightarrow&AA^\mathrm\{T\} x=0\Rightarrow x^\mathrm\{T\} AA^\mathrm\{T\} x=0\stackrel\{y=A^\mathrm\{T\} x\}\{\Rightarrow\}y^\mathrm\{T\} y=0\Rightarrow 0=y=A^\mathrm\{T\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} &\left\\{x; A^\mathrm\{T\} x=0\right\\}=\left\\{x; B^m x=0\right\\}\qquad(I)\\\\ \Rightarrow&\dim\left\\{x; A^\mathrm\{T\} x=0\right\\}=\dim\left\\{x; B^m=0\right\\}\\\\ \Rightarrow&n-\mathrm\{rank\}(A^\mathrm\{T\})=n-\mathrm\{rank\}(B^m)\Rightarrow \mathrm\{rank\} A=\mathrm\{rank\}(B^m). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (3)、易知 $\displaystyle Ax=0\Rightarrow B^mAx=0$. 反之, \begin\{aligned\} B^m\underline\{Ax\}=0\stackrel\{(I)\}\{\Rightarrow\}A^\mathrm\{T\}\underline\{Ax\}=0\Rightarrow 0=x^\mathrm\{T\} A^\mathrm\{T\} Ax\stackrel\{y=Ax\}\{=\}y^\mathrm\{T\} y \Rightarrow 0=y=Ax. \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\\{x; A x=0\right\\}=\left\\{x; B^mA x=0\right\\}\qquad(I)\\\\ \Rightarrow&\dim\left\\{x; A x=0\right\\}=\dim\left\\{x; B^m A=0\right\\}\\\\ \Rightarrow&n-\mathrm\{rank\}(A)=n-\mathrm\{rank\}(B^m A)\Rightarrow \mathrm\{rank\} A=\mathrm\{rank\}(B^m A). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 1263、 8、 (15 分) 设 $\displaystyle A$ 是 $\displaystyle n$ 阶复方阵. 证明: 如果存在正整数 $\displaystyle m$, 使得 $\displaystyle A^m=E$, 则 $\displaystyle A$ 一定相似于一个主对角线上元素都是 $\displaystyle m$ 次单位根的对角阵, 这里 $\displaystyle E$ 是 $\displaystyle n$ 阶单位阵. (云南大学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) / 由题设, $\displaystyle x^m-1$ 是 $\displaystyle A$ 的零化多项式, 从而 $\displaystyle A$ 的最小多项式 $\displaystyle m(x)\mid (x^m-1)$. 这表明 $\displaystyle m(x)$ 没有重根, 而 $\displaystyle A$ 可对角化, 即存在复可逆矩阵 $\displaystyle P$ 使得 \begin\{aligned\} P^\{-1\}AP=\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_n),\quad \lambda\_i^m=1. \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)/ --- 1264、 (0-5)、设 $\displaystyle 3$ 阶方阵 $\displaystyle A$ 的特征值为 $\displaystyle 1,2,3$, 则 $\displaystyle |A^\star|=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / $\displaystyle |A|=1\cdot 2\cdot 3=6\Rightarrow |A^\star|=|A|^\{n-1\}=6^2=36$.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)/ --- 1265、 (0-6)、已知 $\displaystyle A^2=\left(\begin\{array\}\{cccccccccccccccccccc\}-3&1&0\\\\ -7&2&0\\\\ 0&0&4\end\{array\}\right)$, $\displaystyle A^3=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&&\\\\ &-1&\\\\ &&8\end\{array\}\right)$, 则 $\displaystyle A=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (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) / \begin\{aligned\} A=(A^2)^\{-1\}A^3=\left(\begin\{array\}\{cccccccccccccccccccc\}-2&1&0\\\\ -7&3&0\\\\ 0&0&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\} 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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