你知道哪些矩阵运算满足交换律吗？

一、题目

已知 $\boldsymbol{A}$ 是 $n$ 阶矩阵, $\boldsymbol{A}^{\mathrm{T}}$ 是 $\boldsymbol{A}$ 的转置矩阵, $\boldsymbol{A}^{*}$ 是 $\boldsymbol{A}$ 的伴随矩阵, $\boldsymbol{E}$ 是 $n$ 阶单位矩阵, $\boldsymbol{\Lambda}_{1}$, $\Lambda_{2}$ 都是 $n$ 阶对角矩阵, 在下列运算中, 交换律一定成立的是哪个或者哪些？

(1) $A A^{*}=A^{*} A$

(2) $\Lambda_{1} \Lambda_{2}=\Lambda_{2} \Lambda_{1}$

(3) $A^{m} A^{t}=A^{t} A^{m}$

(4) $\boldsymbol{A} \boldsymbol{A}^{\mathrm{\top}}=\boldsymbol{A}^{\mathrm{\top}} \boldsymbol{A}$

(5) $\boldsymbol{A} \boldsymbol{\Lambda}_{1}=\boldsymbol{\Lambda}_{1} \boldsymbol{A}$

(6) $(\boldsymbol{A}+\boldsymbol{E})(\boldsymbol{A}-\boldsymbol{E})=(\boldsymbol{A}-\boldsymbol{E})(\boldsymbol{A}+\boldsymbol{E})$

难度评级：

二、解析

(1)

$$
A A^{*}=A^{*} A=|A| E
$$

(2)

$$
\Lambda_{1} \Lambda_{2}=\Lambda_{2} \Lambda_{1} \Rightarrow
$$

$$
\left[\begin{array}{ccc}a_{1} & 0 & 0 \\ 0 & a_{2} & 0 \\ 0 & 0 & a_{3}\end{array}\right]\left[\begin{array}{ccc}b_{1} & 0 & 0 \\ 0 & b_{2} & 0 \\ 0 & 0 & b_{3}\end{array}\right]=
$$

$$
\left[\begin{array}{ccc}a_{1} b_{1} & 0 & 0 \\ 0 & a_{2} b_{2} & 0 \\ 0 & 0 & a_{3} b_{3}\end{array}\right]=
$$

$$
\left[\begin{array}{ccc}b_{1} & 0 & 0 \\ 0 & b_{2} & 0 \\ 0 & 0 & b_{3}\end{array}\right]\left[\begin{array}{ccc}a_{1} & 0 & 0 \\ 0 & a_{2} & 0 \\ 0 & 0 & a_{3}\end{array}\right]
$$

(3)

$$
A^{m} A^{t}=A^{m+t} \Leftrightarrow A^{t} A^{m}=A^{t+m}
$$

(4)

$$
A A^{\top} \neq A^{\top} A \Rightarrow
$$

$$
(1,2,3)\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)=14
$$

$$
\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)(1,2,3)=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{array}\right]
$$

(5)

$$
A \Lambda_{1} \neq \Lambda_{1} A \Rightarrow
$$

$$
{\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}a & 2 a \\ 3 b & 4 b\end{array}\right]}
$$

$$
{\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]=\left[\begin{array}{ll}a & 2 b \\ 3 a & 4 b\end{array}\right]}
$$

(6)

$$
(A+E)(A-E)=A^{2}-A+A-E=A^{2}-E
$$

$$
(A-E)(A+E)=A^{2}+A-A-Z=A^{2}-E
$$

于是：

$$
(A+E)(A-E)=(A-E)(A+E)
$$

综上可知，一定满足交换律的有：

(1) $A A^{*}=A^{*} A$

(2) $\Lambda_{1} \Lambda_{2}=\Lambda_{2} \Lambda_{1}$

(3) $A^{m} A^{t}=A^{t} A^{m}$

(6) $(\boldsymbol{A}+\boldsymbol{E})(\boldsymbol{A}-\boldsymbol{E})=(\boldsymbol{A}-\boldsymbol{E})(\boldsymbol{A}+\boldsymbol{E})$

可能不满足交换律的有：

(4) $\boldsymbol{A} \boldsymbol{A}^{\mathrm{\top}}=\boldsymbol{A}^{\mathrm{\top}} \boldsymbol{A}$

(5) $\boldsymbol{A} \boldsymbol{\Lambda}_{1}=\boldsymbol{\Lambda}_{1} \boldsymbol{A}$

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