# 你知道哪些矩阵运算满足交换律吗？

## 一、题目

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

## 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)

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

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

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

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

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

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

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

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

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