## 张祖锦2023年数学专业真题分类70天之第54天
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1220、 4、 证明: 任意秩为 $\displaystyle r$ 的实对称矩阵, 可以分解为 $\displaystyle r$ 个秩为 $\displaystyle 1$ 的实对称矩阵之和. (天津大学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 $\displaystyle A$ 是秩为 $\displaystyle r$ 的实对称矩阵, 则存在正交阵 $\displaystyle P$ 使得
\begin\{aligned\} A=P\mathrm\{diag\}(\lambda\_1,\cdots,\lambda\_r,0,\cdots,0)P^\mathrm\{T\}, 0\neq \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\}
令 $\displaystyle B\_i=\lambda\_i PE\_\{ii\}P^\mathrm\{T\}$, 则 $\displaystyle B\_i$ 是秩为 $\displaystyle 1$ 的实对称矩阵, 且
\begin\{aligned\} A=B\_1+\cdots+B\_r. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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1221、 2、 实矩阵
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2\\\\ 2&a\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}4&b\\\\ 3&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\}
证明: (1)、 矩阵方程 $\displaystyle AX=B$ 有解但 $\displaystyle BY=A$ 无解的充要条件是 $\displaystyle a\neq 2,b=\frac\{4\}\{3\}$; (2)、 $\displaystyle A$ 相似于 $\displaystyle B$ 的充要条件是 $\displaystyle a=3,b=\frac\{2\}\{3\}$; (3)、 $\displaystyle A$ 合同于 $\displaystyle B$ 的充要条件是 $\displaystyle a < 2, b=3$. (武汉大学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)、 由线性方程组理论知
\begin\{aligned\} AX=B\mbox\{有解\}\Leftrightarrow \mathrm\{rank\}(A,B)=\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\}
对 $\displaystyle (A,B)$ 施行初等行变换知
\begin\{aligned\} (A,B)&=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&4&b\\\\ 2&a&3&1\end\{array\}\right)\rightarrow \left(\begin\{array\}\{cccccccccccccccccccc\}2&2&4&b\\\\ 0&a-2&-1&1-b\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当且仅当 $\displaystyle a\neq 2$ 时, $\displaystyle \mathrm\{rank\}(A)=2=\mathrm\{rank\}(A,B)$. 同理, 由
\begin\{aligned\} (B,A)&=\left(\begin\{array\}\{cccccccccccccccccccc\}4&b&2&2\\\\ 3&1&2&a\end\{array\}\right)\rightarrow\left(\begin\{array\}\{cccccccccccccccccccc\}4&b&2&2\\\\ 0&1-\frac\{3b\}\{4\}&\frac\{1\}\{2\}&a-\frac\{3\}\{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\}
知 $\displaystyle BY=A$ 无解 $\displaystyle \Leftrightarrow \mathrm\{rank\}(B)\neq \mathrm\{rank\}(B,A)\Leftrightarrow 1-\frac\{3b\}\{4\}=0\Leftrightarrow b=\frac\{4\}\{3\}$. (2)、 (2-1)、 $\displaystyle \Rightarrow$:
\begin\{aligned\} A\sim B&\Rightarrow \mathrm\{tr\} A=\mathrm\{tr\} B, |A|=|B|\Rightarrow a=3, b=\frac\{2\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 $\displaystyle \Leftarrow$:
\begin\{aligned\} a=3, b=\frac\{2\}\{3\}&\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2\\\\ 2&3\end\{array\}\right), B=\left(\begin\{array\}\{cccccccccccccccccccc\}4&\frac\{2\}\{3\}\\\\ 3&1\end\{array\}\right)\\\\ &\Rightarrow A,B\mbox\{的特征多项式为 $\displaystyle \lambda^2-5\lambda+2=0$, 无重根\}\\\\ &\Rightarrow A,B\mbox\{相似于同一对角阵\}\Rightarrow A\sim B . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 (3-1)、 $\displaystyle \Rightarrow$: 若 $\displaystyle A,B$ 合同, 则 $\displaystyle B$ 也是对称矩阵, 故 $\displaystyle b=3$. 由
\begin\{aligned\} B=\left(\begin\{array\}\{cccccccccccccccccccc\}4&3\\\\ 3&1\end\{array\}\right)&\Rightarrow \mbox\{ $\displaystyle B$ 的特征值为 $\displaystyle \frac\{5+3\sqrt\{5\}\}\{2\},\frac\{5-3\sqrt\{5\}\}\{2\}$\}\\\\ &\Rightarrow \mbox\{ $\displaystyle B$ 的正负惯性指数均为 $\displaystyle 1$\}\\\\ &\Rightarrow\mbox\{ $\displaystyle A$ 的正负惯性指数均为 $\displaystyle 1$\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2\\\\ 2&a\end\{array\}\right)&\Rightarrow A\mbox\{的特征值为 $\displaystyle \frac\{2+a\pm\sqrt\{20-4a+a^2\}\}\{2\}$\} \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\} 20-4a+a^2 > 0, 2+a < \sqrt\{20-4a+a^2\}\Leftrightarrow a < 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3-2)、 $\displaystyle \Leftarrow$: 若 $\displaystyle a < 2, b=3$, 则由必要性的论证知 $\displaystyle A,B$ 合同.跟锦数学微信公众号. [在线资料](
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1222、 7、 (20 分) 设 $\displaystyle A$ 是 $\displaystyle m$ 阶方阵, $\displaystyle B$ 是 $\displaystyle n$ 阶方阵, $\displaystyle Q$ 是 $\displaystyle m\times n$ 矩阵, $\displaystyle f\_A(x), f\_B(x)$ 是 $\displaystyle A,B$ 的特征多项式. (1)、 证明: $\displaystyle f\_B(A)$ 可逆的充要条件是 $\displaystyle \left(f\_A(x),f\_B(x)\right)=1$; (2)、 证明: 若 $\displaystyle \left(f\_A(x),f\_B(x)\right)=1$, 且 $\displaystyle AQ=QB$, 则 $\displaystyle Q=0$. (武汉大学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)、 (1-1)、 $\displaystyle \Leftarrow$: 由 $\displaystyle \left(f\_A(x),f\_B(x)\right)=1$ 知
\begin\{aligned\} &\exists\ u(x),v(x),\mathrm\{ s.t.\} u(x)f\_A(x)+v(x)f\_B(x)=1\\\\ \Rightarrow& E=u(A)f\_A(A)+v(A)f\_B(A)\xlongequal[\tiny\mbox\{Cayley\}]\{\tiny\mbox\{Hamilton-\}\} v(A)f\_B(A)\Rightarrow f\_B(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\}
(1-2)、 $\displaystyle \Rightarrow$: 用反证法. 若 $\displaystyle \left(f\_A(x),f\_B(x)\right)\neq 1$, 则 $\displaystyle A,B$ 有公共特征值 $\displaystyle \lambda$. 由 $\displaystyle |\lambda E-B\mathrm\{T\}|=|\lambda E-B|$ 知 $\displaystyle \lambda$ 也是 $\displaystyle B^\mathrm\{T\}$ 的特征值. 设 $\displaystyle 0\neq \alpha,0\neq \beta$ 分别是 $\displaystyle A,B^\mathrm\{T\}$ 对应于特征值 $\displaystyle \lambda$ 的特征向量, $\displaystyle X=\alpha\beta^\mathrm\{T\}$, 则
\begin\{aligned\} A\alpha=\lambda \alpha, B^\mathrm\{T\} \beta=\lambda \beta, AX=\lambda \alpha\beta^\mathrm\{T\} =\alpha(\lambda \beta)^\mathrm\{T\} =\alpha\beta^\mathrm\{T\} B=XB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &A^\{k-1\}X=XB^\{k-1\}\\\\ \Rightarrow& A^kX=A^\{k-1\}AX=A^\{k-1\}XB=XB^\{k-1\}B=XB^k \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及数学归纳法知 $\displaystyle A^kX=XB^k, \forall\ k\geq 1$. 从而
\begin\{aligned\} f\_B(A)X=Xf\_B(B)=X0=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f\_B(A)$ 可逆知 $\displaystyle X=0$. 这与 $\displaystyle \alpha\neq 0,\beta\neq 0\Rightarrow X=\alpha\beta^\mathrm\{T\}\neq 0$ 矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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1223、 8、 (20 分) $\displaystyle A$ 是 $\displaystyle n$ 阶矩阵, 且有 $\displaystyle n$ 个特征值为偶数. 证明: 关于 $\displaystyle X$ 的矩阵方程 $\displaystyle X+AX-XA^2=0$ 只有零解. (武汉大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle X+AX-XA^2=0\Leftrightarrow (E+A)X=XA^2$. 由 $\displaystyle E+A$ 的特征值都是奇数, $\displaystyle A^2$ 的特征值都是偶数知 $\displaystyle E+A,A^2$ 没有公共特征值, 它们对应的特征多项式互素. 由上一题第 2 问即知 $\displaystyle X=0$.跟锦数学微信公众号. [在线资料](
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1224、 4、 设 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-1&-1\\\\ -1&1&1\\\\ 0&-4&-2\end\{array\}\right), \alpha\_1=\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\-2\end\{array\}\right)$. (1)、 若 $\displaystyle A\alpha\_2=\alpha\_1, A^2\alpha\_3=\alpha\_1$, 求所有的 $\displaystyle \alpha\_2,\alpha\_3$. (2)、 判断 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 是否线性无关, 并证明之. (武汉理工大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle (A,\alpha\_1)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&0&-\frac\{1\}\{2\}&-\frac\{1\}\{2\}\\\\ 0&1&\frac\{1\}\{2\}&\frac\{1\}\{2\}\\\\ 0&0&0&0\end\{array\}\right)$ 知
\begin\{aligned\} \alpha\_2=k\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\-1\\\\2\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{2\}\\\\ \frac\{1\}\{2\}\\\\ 0\end\{array\}\right),\quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由
\begin\{aligned\} (A^2,\alpha\_1)=\left(\begin\{array\}\{cccccccccccccccccccc\}2&2&0&-1\\\\ -2&-2&0&1\\\\ 4&4&0&-2\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&-\frac\{1\}\{2\}\\\\ 0&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
\begin\{aligned\} \alpha\_3=k\left(\begin\{array\}\{cccccccccccccccccccc\}-1\\\\1\\\\0\end\{array\}\right)+\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{2\}\\\\0\\\\0\end\{array\}\right),\quad \forall\ k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 易知 $\displaystyle A\alpha\_1=0$, 而 $\displaystyle A^2\alpha\_2=0, A^3\alpha\_3=0$. 设 $\displaystyle 0=k\_1\alpha\_1+k\_2\alpha\_2+k\_3\alpha\_3$, 则用 $\displaystyle A^2$ 作用得
\begin\{aligned\} 0=k\_3A^2\alpha\_3=k\_3\alpha\_1\Rightarrow k\_3=0\Rightarrow 0=k\_1\alpha\_1+k\_2\alpha\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再用 $\displaystyle A$ 作用得
\begin\{aligned\} 0=k\_2A\alpha\_2=k\_2\alpha\_1\Rightarrow k\_2=0\Rightarrow 0=k\_1\alpha\_1\Rightarrow k\_1=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 线性无关.跟锦数学微信公众号. [在线资料](
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1225、 5、 实对称矩阵 $\displaystyle A$ 的行元素之和为 $\displaystyle 3$, 并且
\begin\{aligned\} \alpha\_1=(-1,2,-1)^\mathrm\{T\}, \alpha\_2=(0,-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 $\displaystyle AX=0$ 的解. (1)、 求 $\displaystyle A$ 的所有特征值及特征向量. (2)、 求正交矩阵 $\displaystyle Q$ 以及对角阵 $\displaystyle \varLambda$ 满足 $\displaystyle Q^\mathrm\{T\} AQ=\varLambda$. (武汉理工大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 取 $\displaystyle \alpha\_3=\left(\begin\{array\}\{cccccccccccccccccccc\}1\\\\1\\\\1\end\{array\}\right)$, 则设 $\displaystyle P=\left(\alpha\_1,\alpha\_2,\alpha\_3\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0&1\\\\ 2&-1&1\\\\ -1&1&1\end\{array\}\right)$ 后,
\begin\{aligned\} AP=P\mathrm\{diag\}(0,0,3)\Rightarrow A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&1\\\\ 1&1&1\\\\ 1&1&1\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle A$ 的特征值为 $\displaystyle 0$ (二重), $\displaystyle 3$ (单重), 对应的特征向量分别为 $\displaystyle \alpha\_1,\alpha\_2; \alpha\_3$. (2)、 将 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3$ 标准正交化为 $\displaystyle \beta\_1,\beta\_2,\beta\_3$, 并设
\begin\{aligned\} Q=(\beta\_1,\beta\_2,\beta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}-\frac\{1\}\{\sqrt\{6\}\}&-\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{2\}\{\sqrt\{6\}\}&0&\frac\{1\}\{\sqrt\{3\}\}\\\\ -\frac\{1\}\{\sqrt\{6\}\}&\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{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 $\displaystyle Q$ 正交, 且 $\displaystyle Q^\mathrm\{T\} AQ=\mathrm\{diag\}(0,0,3)$.跟锦数学微信公众号. [在线资料](
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1226、 9、 已知 $\displaystyle n$ 阶复矩阵 $\displaystyle A,B$ 满足 $\displaystyle A+2B=AB$. 求证: (1)、 $\displaystyle AB=BA$; (2)、 $\displaystyle A$ 与 $\displaystyle B$ 有公共的特征向量. (武汉理工大学2023年高等代数考研试题) [矩阵 ]
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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\} &(A-2E)(B-E)=AB-A-2B+2E=2E\\\\ \Rightarrow&2E=(B-E)(A-2E)=BA-2B-A+2E\\\\ \Rightarrow&BA=2B+A\xlongequal\{\tiny\mbox\{题设\}\} AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle \lambda\_1$ 是 $\displaystyle A$ 的特征值, $\displaystyle V\_\{\lambda\_1\}$ 是对应的特征子空间. 由
\begin\{aligned\} ABX&=BAX=B(\lambda\_1X)=\lambda\_1BX \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle BX\in V\_\{\lambda\_1\}$. $\displaystyle B$ 作为 $\displaystyle V\_\{\lambda\_1\}$ 上的线性变换, 有特征值 $\displaystyle \mu\_1$ 和对应的特征向量 $\displaystyle \eta\_1$. 于是 $\displaystyle \eta\_1$ 就是 $\displaystyle A,B$ 的公共特征向量.跟锦数学微信公众号. [在线资料](
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1227、 3、 (15 分) 设 $\displaystyle A\in\mathbb\{R\}^\{n\times n\}$, 且 $\displaystyle I\_n+A$ 可逆, 其中 $\displaystyle I\_n$ 为 $\displaystyle n$ 阶单位阵, 令 $\displaystyle c(A)=(I\_n-A)(I\_n+A)^\{-1\}$. (1)、 证明: $\displaystyle I\_n+c(A)$ 可逆; (2)、 求 $\displaystyle c\left(c(A)\right)$. (西安电子科技大学2023年高等代数考研试题) [矩阵 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 $\displaystyle c(A)(I+A)=I-A$
\begin\{aligned\} \Rightarrow [I+c(A)] (I+A)=(I+A)+(I-A)=2I\Rightarrow I+c(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)、 由第 1 步知 $\displaystyle \left\[I+C(A)\right\]^\{-1\}=\frac\{I+A\}\{2\}$, 而
\begin\{aligned\} &c\left(c(A)\right)=\left\[I-c(A)\right\]\left\[I+c(A)\right\]^\{-1\} =\left\[I-c(A)\right\]\frac\{I+A\}\{2\}\\\\ =&\frac\{1\}\{2\}\left\\{2I-\left\[I+c(A)\right\]\right\\}(I+A) =(I+A)-I=A. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1228、 8、 (15 分) 设 $\displaystyle n$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}0&1&&\\\\ 1&\ddots&\ddots&\\\\ &\ddots&\ddots&1\\\\ &&1&0\end\{array\}\right)$. (1)、 求 $\displaystyle A$ 的不变因子; (2)、 证明: $\displaystyle A$ 有 $\displaystyle n$ 个两两互异的实特征值. (西安电子科技大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 对 $\displaystyle \forall\ \lambda\in\mathbb\{R\}, \lambda E-A$ 的右上角有一个 $\displaystyle n-1$ 阶的子式 $\displaystyle =(-1)^\{n-1\}\neq 0$, 而 $\displaystyle \mathrm\{rank\}(\lambda E-A)\geq n-1$, 且 $\displaystyle A$ 的行列式因子为 $\displaystyle 1,\cdots,1,|\lambda E-A|$, 不变因子为 $\displaystyle 1,\cdots,1,|\lambda E-A|$. (2)、 由 $\displaystyle A$ 实对称知 $\displaystyle A$ 可对角化, 而 $\displaystyle \mathbb\{R\}^n$ 可表示为 $\displaystyle A$ 的特征子空间的直和:
\begin\{aligned\} \mathbb\{R\}^n=V\_\{\lambda\_1\}\oplus \cdots\oplus V\_\{\lambda\_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 $\displaystyle \lambda\_1,\cdots,\lambda\_s$ 是 $\displaystyle A$ 的互异特征值. 对任意 $\displaystyle 1\leq i\leq s$,
\begin\{aligned\} \mathrm\{rank\} (\lambda\_iE-A)\geq n-1\Rightarrow \dim V\_\{\lambda\_i\}\leq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但 $\displaystyle V\_\{\lambda\_i\}$ 作为特征子空间, 含有非零向量, 而 $\displaystyle \dim V\_\{\lambda\_i\}\geq 1$. 各 $\displaystyle \dim V\_\{\lambda\_i\}=1$,
\begin\{aligned\} n=\dim\mathbb\{R\}^n=\sum\_\{i=1\}^s \dim V\_\{\lambda\_i\}=s. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle A$ 有 $\displaystyle n$ 个互异特征值.跟锦数学微信公众号. [在线资料](
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---
1229、 9、 (15 分) 设矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&-2&6\\\\ -1&0&3\\\\ -1&-1&4\end\{array\}\right)$, 求 $\displaystyle A^k$. (西安电子科技大学2023年高等代数考研试题) [矩阵 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle A$ 的特征值为 $\displaystyle 1$ (三重). 由
\begin\{aligned\} A-E\to\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&-3\\\\ 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
\begin\{aligned\} \mathrm\{rank\}(J-E)=\mathrm\{rank\}(A-E)=1\Rightarrow J=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\\\\ &1&\\\\ &&1\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle A$ 的属于特征值 $\displaystyle 1$ 的特征向量为
\begin\{aligned\} \xi\_1=(-1,1,0)^\mathrm\{T\}, \xi\_2=(3,0,1)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设可逆矩阵 $\displaystyle P=(\eta\_1,\eta\_2,\eta\_3)$ 使得 $\displaystyle P^\{-1\}AP=J$, 则
\begin\{aligned\} (A-E)\eta\_1=0, (A-E)\eta\_2=\eta\_1, (A-E)\eta\_3=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \eta\_1,\eta\_2$ 是 $\displaystyle A$ 的属于特征值 $\displaystyle 1$ 的特征向量. 为求 $\displaystyle \eta\_2$, 由
\begin\{aligned\} (A-E,k\xi\_1+l\xi\_2)&=\left(\begin\{array\}\{cccccccccccccccccccc\} -2&-2&6&-k+3l\\\\ -1&-1&3&k\\\\ -1&-1&3&k \end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&0&-k+l\\\\ 0&0&0&k-l\\\\ -1&-1&3&l\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle k=l$ 时, $\displaystyle (A-E)x=k\xi\_1+l\xi\_2$ 有界. 取 $\displaystyle k=l=1$ 后知可取
\begin\{aligned\} \eta\_1=\xi\_1+\xi\_2=(2,1,1)^\mathrm\{T\}, \eta\_2=(-1,0,0)^\mathrm\{T\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再取 $\displaystyle \eta\_3=\xi\_1=(-1,1,0)^\mathrm\{T\}$, 并令
\begin\{aligned\} T=(\eta\_1,\eta\_2,\eta\_3)=\left(\begin\{array\}\{cccccccccccccccccccc\}2&-1&-1\\\\ 1&0&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\}
则
\begin\{aligned\} &T^\{-1\}AT=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&\\\\ &1&\\\\ &&1\end\{array\}\right)\\\\ \Rightarrow& A^k=T\left(\begin\{array\}\{cccccccccccccccccccc\}1&k&\\\\ &1&\\\\ &&1\end\{array\}\right)T^\{-1\} =\left(\begin\{array\}\{cccccccccccccccccccc\} 1-2k&-2k&6k\\\\ -k&1-k&3k\\\\ -k&-k&1+3k \end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1230、 (2)、 求 $\displaystyle n$ 阶矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&1&\cdots&1\\\\ 1&2&\cdots&1\\\\ \vdots&\vdots&&\vdots\\\\ 1&1&\cdots&2\end\{array\}\right)$ 的逆. (西安交通大学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)、 我们有如下求一类矩阵的逆的好方法: Sherman-Morrison 公式. 设 $\displaystyle A$ 是 $\displaystyle n$ 阶可逆阵, $\displaystyle \alpha,\beta$ 是 $\displaystyle n$ 维列向量, 且 $\displaystyle 1+\beta^\mathrm\{T\} A^\{-1\}\alpha\neq 0$. 求证:
\begin\{aligned\} (A+\alpha\beta^\mathrm\{T\})^\{-1\}=A^\{-1\}-\frac\{1\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\} A^\{-1\}\alpha \beta^\mathrm\{T\} A^\{-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最笨的办法是直接验算:
\begin\{aligned\} &(A+\alpha \beta^\mathrm\{T\})\left(A^\{-1\}-\frac\{1\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\} A^\{-1\}\alpha \beta^\mathrm\{T\} A^\{-1\}\right)\\\\ =&E-\frac\{\alpha\beta^\mathrm\{T\} A^\{-1\}\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\} +\alpha \beta^\mathrm\{T\} A^\{-1\} -\frac\{\alpha (\beta^\mathrm\{T\} A^\{-1\}\alpha)\beta^\mathrm\{T\} A^\{-1\}\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\}\\\\ =&E+\alpha\beta^\mathrm\{T\} A^\{-1\} -\frac\{\alpha\beta^\mathrm\{T\} A^\{-1\}\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\} -\frac\{(\beta^\mathrm\{T\} A^\{-1\}\alpha) \alpha\beta^\mathrm\{T\} A^\{-1\}\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\}\\\\ =&E+\alpha\beta^\mathrm\{T\} A^\{-1\} -\alpha\beta^\mathrm\{T\} A^\{-1\} =E. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但你知道这公式是如何得到的么? 事实上, 由
\begin\{aligned\} (1-x)^\{-1\}=\sum\_\{n=0\}^\infty x^n, |x| < 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知我们可形式计算如下:
\begin\{aligned\} &\quad (A+\alpha \beta^\mathrm\{T\})^\{-1\}=\left\[A(E+A^\{-1\}\alpha\beta^\mathrm\{T\})\right\]^\{-1\} =\left(E+A^\{-1\}\alpha \beta^\mathrm\{T\}\right)^\{-1\}A^\{-1\}\\\\ &=\left\[E+\sum\_\{k=1\}^\infty (-1)^k (A^\{-1\}\alpha \beta^\mathrm\{T\})^k\right\] A^\{-1\}\\\\ &=\left\[E+A^\{-1\}\alpha \cdot\sum\_\{k=1\}^\infty (-1)^k (\beta^\mathrm\{T\} A^\{-1\}\alpha)^\{k-1\}\cdot \beta\right\] A^\{-1\}\\\\ &=\left\[E-A^\{-1\}\alpha \frac\{1\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\}\beta^\mathrm\{T\}\right\]A^\{-1\} =A^\{-1\}-\frac\{A^\{-1\}\alpha\beta^\mathrm\{T\} A^\{-1\}\}\{1+\beta^\mathrm\{T\} A^\{-1\}\alpha\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 设 $\displaystyle e=(1,\cdots,1)^\mathrm\{T\}$, 则 $\displaystyle A=E+ee^\mathrm\{T\}$, 而由第 i 步知
\begin\{aligned\} A^\{-1\}=&E-\frac\{1\}\{1+e^\mathrm\{T\} e\}ee^\mathrm\{T\} =E-\frac\{1\}\{n+1\}ee^\mathrm\{T\}\\\\ =&\left(\begin\{array\}\{cccccccccccccccccccc\}\frac\{n\}\{n+1\}&-\frac\{1\}\{n+1\}&\cdots&-\frac\{1\}\{n+1\}\\\\ -\frac\{1\}\{n+1\}&\frac\{n\}\{n+1\}&\cdots&-\frac\{1\}\{n+1\}\\\\ \vdots&\vdots&&\vdots\\\\ -\frac\{1\}\{n+1\}&-\frac\{1\}\{n+1\}&\cdots&\frac\{n\}\{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\}
跟锦数学微信公众号. [在线资料](
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---
1231、 2、 已知 $\displaystyle A,B$ 均为 $\displaystyle n$ 阶实对称矩阵, 证明: (1)、 $\displaystyle AB-BA$ 的特征值为 $\displaystyle 0$ 或纯虚数; (2)、 $\displaystyle \mathrm\{tr\}\left\[(AB)^2\right\]\leq \mathrm\{tr\}(A^2B^2)$, 其中 $\displaystyle \mathrm\{tr\} M$ 表示矩阵 $\displaystyle M$ 的迹. (西安交通大学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)、 先证一个结论: 反对称实矩阵的特征值是零或纯虚数. 事实上, 设 $\displaystyle \lambda$ 是反对称实矩阵 $\displaystyle A$ 的特征值, 则存在 $\displaystyle \alpha\neq 0$ 使
\begin\{aligned\} &A \alpha =\lambda\alpha \\\\ \Rightarrow& \lambda \overline\{\alpha\} ^\mathrm\{T\}\alpha =\overline\{\alpha\} ^\mathrm\{T\} A \alpha =-\overline\{\alpha\} ^\mathrm\{T\} A ^\mathrm\{T\}\alpha =-\left(A \overline\{\alpha\} \right)^\mathrm\{T\}\alpha =-\left(\overline\{A \alpha \}\right)^\mathrm\{T\}\alpha =-\overline\{\lambda\} \overline\{\alpha\}^\mathrm\{T\}\alpha \\\\ \Rightarrow& \lambda=-\overline\{\lambda\} \Rightarrow\lambda=0\mbox\{或纯虚数\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle C=AB-BA$, 则 $\displaystyle C^\mathrm\{T\}=B^\mathrm\{T\} A^\mathrm\{T\}-A^\mathrm\{T\} B^\mathrm\{T\}=BA-AB=-C$. 由第 1 步即知 $\displaystyle C$ 的特征值要么为 $\displaystyle 0$ 要么为纯虚数. (3)、 由
\begin\{aligned\} C^\mathrm\{T\} C&=-C^2=-(AB-BA)(AB-BA)\\\\ &=-(AB)^2+AB^2A+BA^2B-(BA)^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle C^\mathrm\{T\} C$ 半正定知
\begin\{aligned\} 0\leq \mathrm\{tr\}(C^\mathrm\{T\} C)=&-\mathrm\{tr\}\left\[(AB)^2\right\]+\mathrm\{tr\}(AB^2\cdot A)\\\\ &+\mathrm\{tr\}(B\cdot A^2B)-\mathrm\{tr\}(B\cdot ABA). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由公式 $\displaystyle \mathrm\{tr\}(AB)=\mathrm\{tr\}(BA)$ 知
\begin\{aligned\} &\mathrm\{tr\}(AB^2\cdot A)=\mathrm\{tr\}(A^2B^2), \mathrm\{tr\}(B\cdot A^2B)=\mathrm\{tr\}(A^2B^2),\\\\ &\mathrm\{tr\}(B\cdot ABA)=\mathrm\{tr\}(ABAB). \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\} 0\leq 2\left\\{\mathrm\{tr\}(A^2B^2)-\mathrm\{tr\}\left\[(AB)^2\right\]\right\\}\Rightarrow \mathrm\{tr\}\left\[(AB)^2\right\]\leq \mathrm\{tr\}(A^2B^2). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1232、 4、 解答如下问题: (1)、 已知 $\displaystyle n$ 阶方阵 $\displaystyle A,B$ 满足 $\displaystyle A-B+AB=0$. 证明: $\displaystyle AB=BA$. (西安交通大学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) /
\begin\{aligned\} &-I=A-B+AB-I=(A-I)(B+I)\\\\ \Rightarrow& -I=(B+I)(A-I)=BA-B+A-I\\\\ \Rightarrow&BA=B-A\xlongequal\{\tiny\mbox\{题设\}\} AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1233、 (2)、 是否存在 $\displaystyle 2$ 阶实矩阵 $\displaystyle A$, 使得 $\displaystyle A^2=\left(\begin\{array\}\{cccccccccccccccccccc\}-1&0\\\\ 0&-2\end\{array\}\right)$. 若存在, 请举例; 若不存在, 请加以证明. (西安交通大学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) / 不存在. 用反证法. 若存在满足题设的二阶实矩阵 $\displaystyle A$, 则 $\displaystyle A$ 的特征值 $\displaystyle \lambda$ 满足
\begin\{aligned\} \lambda^2=-1\mbox\{或\} -2\Rightarrow \lambda^2\in\mathbb\{R\}, \lambda^2 < 0\Rightarrow \lambda=b\mathrm\{ i\}, b\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle A\in\mathbb\{R\}^\{2\times 2\}$ 的虚特征值成对出现蕴含 $\displaystyle A$ 的特征值为 $\displaystyle \pm b\mathrm\{ i\}$. 由 Jordan 标准形理论知存在复可逆矩阵 $\displaystyle P$ 使得
\begin\{aligned\} P^\{-1\}AP=\mathrm\{diag\}(b\mathrm\{ i\},-b\mathrm\{ i\})\Rightarrow P^\{-1\}A^2P=\mathrm\{diag\}(-b^2,-b^2)=-b^2I\_2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与题设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
1234、 5、 已知 $\displaystyle A=(a\_\{ij\})$ 是 $\displaystyle n$ 阶矩阵, $\displaystyle a\_\{ij\}=\left\\{\begin\{array\}\{llllllllllll\}1,&1\leq i < j\leq n,\\\\ 0,&\mbox\{其它\}.\end\{array\}\right.$ 证明: (1)、 $\displaystyle A^n=0$; (2)、 若 $\displaystyle n$ 阶矩阵 $\displaystyle B$ 满足 $\displaystyle AB+BA=B$, 证明: $\displaystyle B=0$. (西安交通大学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)、 $\displaystyle A$ 的特征值全为 $\displaystyle 0$, 且 $\displaystyle \mathrm\{rank\} A=n-1$. 而 $\displaystyle A$ 的 Jordan 标准形 $\displaystyle J$ 满足 $\displaystyle \mathrm\{rank\} J=n-1$. 这表明
\begin\{aligned\} J=J\_n(0)\Rightarrow J^n=0\Rightarrow A^n=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 为证第 2 问. 我们给出一个结果. 设 $\displaystyle A,B$ 为数域 $\displaystyle \mathbb\{P\}$ 上的 $\displaystyle n$ 阶矩阵, 若 $\displaystyle A,B$ 无公共特征值, 则矩阵方程 $\displaystyle AX=XB$ 只有零解. 设 $\displaystyle A$ 的特征值为 $\displaystyle \lambda\_1,\cdots,\lambda\_n$, 则 $\displaystyle A$ 的特征多项式
\begin\{aligned\} f(\lambda)=\prod\_\{i=1\}^n(\lambda-\lambda\_i). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再设 $\displaystyle B$ 的特征值为 $\displaystyle \mu\_1,\cdots,\mu\_n$, 则 $\displaystyle f(B)$ 的特征值为 $\displaystyle f(\mu\_1),\cdots,f(\mu\_n)$. 由于 $\displaystyle \lambda\_i\neq \mu\_j\ (i\neq j)$, 我们知 $\displaystyle f(\mu\_j)\neq 0\ (\forall\ j)$. 据此,
\begin\{aligned\} |f(B)|=\prod\_\{j=1\}^n f(\mu\_j)\neq 0\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\}
又由
\begin\{aligned\} AX=XB\Rightarrow& A^2X=AAX=AXB=XBB=XB^2\Rightarrow \cdots\\\\ \Rightarrow& A^kX=XB^k\ (\forall\ k) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f(A)X=Xf(B)$. 由 Hamilton-Caylay 定理, $\displaystyle 0=f(A)\Rightarrow 0=Xf(B)$. 因 $\displaystyle f(B)$ 可逆, 我们最终得到 $\displaystyle X=0$. (3)、 回到题目. 由题设, $\displaystyle (A-E)B=B(-A)$. 注意到 $\displaystyle A-E$ 的特征值全为 $\displaystyle -1$, $\displaystyle -A$ 的特征值全为 $\displaystyle 0$, 我们可由第 2 步知 $\displaystyle B=0$.跟锦数学微信公众号. [在线资料](
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https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
1235、 6、 已知 $\displaystyle n$ 阶的三对角矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}3&-1&&\\\\ -2&\ddots&\ddots&\\\\ &\ddots&\ddots&-1\\\\ &&-2&3\end\{array\}\right)$. 证明 $\displaystyle A$ 可逆. (西安交通大学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 $\displaystyle |A|=D\_n$, 则按第一行展开知 $\displaystyle D\_n=3D\_\{n-1\}-2D\_\{n-2\}$. 特征方程为 $\displaystyle \lambda^2-3\lambda+2=0$. 故可设
\begin\{aligned\} D\_n=c\_1+c\_2 2^n\stackrel\{D\_1=3, D\_2=7\}\{\Rightarrow\}c\_1=-1, c\_2=2 \Rightarrow D\_n=2^\{n+1\}-1 > 0\Rightarrow 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\}
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---
1236、 7、 已知 $\displaystyle n$ 阶矩阵 $\displaystyle A,B$ 满足
\begin\{aligned\} A^2=A, B^2=B, M=AB. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明:
\begin\{aligned\} \mathrm\{rank\}(I\_n-M)\leq 2\left(n-\mathrm\{rank\} M\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle I\_n$ 为 $\displaystyle n$ 阶单位矩阵. (西安交通大学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)、 由
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A&\\\\ &A-I\_n\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ &A-I\_n\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A&A\\\\ -A&-I\_n\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}A-A^2&\\\\ -A&-I\_n\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A-A^2&\\\\ &I\_n\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}0&\\\\ &I\_n\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathrm\{rank\} A+\mathrm\{rank\}(A-I\_n)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理, $\displaystyle \mathrm\{rank\} B+\mathrm\{rank\}(B-I\_n)=n$. (2)、
\begin\{aligned\} &\mathrm\{rank\}(I\_n-M)=\mathrm\{rank\}(I\_n-A+A-AB)\\\\ \leq& \mathrm\{rank\}(I\_n-A)+\mathrm\{rank\}\left\[A(I\_n-B)\right\] \leq \mathrm\{rank\}(I\_n-A)+\mathrm\{rank\}(I\_n-B)\\\\ =&n-\mathrm\{rank\} A+n-\mathrm\{rank\} B \leq n-\mathrm\{rank\} M+n-\mathrm\{rank\} M\\\\ =&2(n-\mathrm\{rank\} M). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1237、 10、 设 $\displaystyle n$ 阶矩阵 $\displaystyle A$ 的元素均为 $\displaystyle 0$ 或 $\displaystyle 1$, 且 $\displaystyle A$ 的特征值均大于零. 证明: $\displaystyle \det A=1$. (西安交通大学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 $\displaystyle a\_\{ij\}=1\mbox\{或\} 0$ 及行列式的定义知 $\displaystyle \det A\in\mathbb\{Z\}$. 设 $\displaystyle A$ 的特征值为 $\displaystyle \lambda\_1,\cdots,\lambda\_n$, 则由题设, $\displaystyle \lambda\_i > 0$,
\begin\{aligned\} \det A=\lambda\_1\cdots\lambda\_n\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再者,
\begin\{aligned\} \det A=\lambda\_1\cdots\lambda\_n\stackrel\{\tiny\mbox\{均值\}\}\{\leq\}\left(\frac\{\lambda\_1+\cdots+\lambda\_n\}\{n\}\right)^n =\left(\frac\{\mathrm\{tr\} A\}\{n\}\right)^n \stackrel\{a\_\{ii\}=0\mbox\{或\} 1\}\{\leq\}\left(\frac\{n\}\{n\}\right)^n=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \det A=1$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
1238、 6、 已知 $\displaystyle A,B$ 为 $\displaystyle 3$ 阶方阵且满足 $\displaystyle 2A^\{-1\}B=B-4E$. (1)、 证明: $\displaystyle A-2E$ 可逆, 并求 $\displaystyle A-2E$ 的逆; (2)、 若 $\displaystyle B=\left(\begin\{array\}\{cccccccccccccccccccc\}1&-2&0\\\\ 1&2&0\\\\ 0&0&2\end\{array\}\right)$, 求 $\displaystyle A$. (西北大学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\} &2B=AB-4A\Rightarrow AB-4A-2B=0\\\\ \Rightarrow& (A-2E)(B-4E)=8E \Rightarrow (A-2E)^\{-1\}=\frac\{B-4E\}\{8\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由第 1 步知
\begin\{aligned\} A=2E+8(B-4E)^\{-1\}=\left(\begin\{array\}\{cccccccccccccccccccc\}0&2&0\\\\ -1&-1&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](
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---
1239、 7、 设 $\displaystyle A$ 是 $\displaystyle n$ 阶方阵, $\displaystyle A^\star$ 是 $\displaystyle A$ 的伴随矩阵. 证明: 若 $\displaystyle A^2=E$, 则 (1)、 $\displaystyle \mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)=n$; (2)、 $\displaystyle \mathrm\{rank\}(A^\star+E)+\mathrm\{rank\}(A^\star-E)=n$. (西北大学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
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&0\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}A+E&A-E\\\\ 0&A-E\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}2E&A-E\\\\ E-A&A-E\end\{array\}\right)\\\\ \to&\left(\begin\{array\}\{cccccccccccccccccccc\}2E&0\\\\ E-A&(A-E)+\frac\{1\}\{2\}(E-A)(E-A)\end\{array\}\right)\to\left(\begin\{array\}\{cccccccccccccccccccc\}E&0\\\\ 0&A^2-E\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mathrm\{rank\}(A+E)+\mathrm\{rank\}(A-E)=\mathrm\{rank\}(E\_n)+\mathrm\{rank\}(A^2-E)=n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 $\displaystyle (A^\star)^2=A^\star A^\star=(AA)^\star=E^\star=E$. 在第 1 步的结果中 $\displaystyle A\to A^\star$ 即得结论.跟锦数学微信公众号. [在线资料](
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---
1240、 2、 (20 分) 已知 $\displaystyle A$ 为 $\displaystyle 4$ 阶方阵, 且 $\displaystyle A^\{-1\}XA-A^\star=A^\{-1\}X$, 其中 $\displaystyle A^\star$ 为 $\displaystyle A$ 的伴随矩阵. 若 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}1&1&0&0\\\\ 0&1&2&0\\\\ 0&0&1&3\\\\ 4&0&0&1\end\{array\}\right)$, 求矩阵 $\displaystyle X$. (西南财经大学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) / 由 $\displaystyle |A|=-23$ 知 $\displaystyle A^\star=|A|A^\{-1\}=-23A^\{-1\}$, 而
\begin\{aligned\} &A^\{-1\}XA+23A^\{-1\}=A^\{-1\}X \stackrel\{\mbox\{左乘\}A\}\{\Rightarrow\}XA+23E=X\\\\ \Rightarrow&X(E-A)=23E\Rightarrow X=23(E-A)^\{-1\}=\frac\{23\}\{12\}\left(\begin\{array\}\{cccccccccccccccccccc\}&&&-3\\\\ -12&&&\\\\ &-6&&\\\\ &&-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\}
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---
1241、 3、 矩阵 $\displaystyle A=\left(\begin\{array\}\{cccccccccccccccccccc\}2&0&0\\\\ a&2&0\\\\ b&c&-1\end\{array\}\right)$. 证明: $\displaystyle A$ 可对角化当且仅当 $\displaystyle a=0$. (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) / 易知 $\displaystyle A$ 的特征值为 $\displaystyle 2,2,-1$. 于是 $\displaystyle A$ 可对角化 $\displaystyle \Leftrightarrow A\sim \mathrm\{diag\}(2,2,-1)$
\begin\{aligned\} \Leftrightarrow 1=\mathrm\{rank\}(A-2E)=\mathrm\{rank\}\left(\begin\{array\}\{cccccccccccccccccccc\}0&0&0\\\\ a&0&0\\\\ b&c&1\end\{array\}\right)\Leftrightarrow a=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
1242、 4、 设 $\displaystyle A\_i, i=1,\cdots,4$ 是三阶方阵. 证明: 存在不全为零的 $\displaystyle x\_i, i=1,\cdots,4$, 使得
\begin\{aligned\} \det(x\_1A\_1+x\_2A\_2+x\_3A\_3+x\_4A\_4)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(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) / 设 $\displaystyle A\_i=(\alpha\_i,\beta\_i,\gamma\_i)$, 则 $\displaystyle 4$ 个三维向量组 $\displaystyle \alpha\_1,\alpha\_2,\alpha\_3,\alpha\_4$ 是线性相关的, 即存在不全为 $\displaystyle 0$ 的 $\displaystyle x\_i$, 使得
\begin\{aligned\} \sum\_\{i=1\}^n x\_i\alpha\_i=0\Rightarrow \sum\_\{i=1\}^4 x\_iA\_i\mbox\{第一列为零向量\}\Rightarrow \det\left(\sum\_\{i=1\}^4 x\_iA\_i\right)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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发表于 2023-3-5 13:12:49
