问题及解答

设 $A$, $B$ 分别为 $m\times n$, $n\times m$ 型矩阵, 证明: $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$.

(即 $AB$ 和 $BA$ 的迹相等, $tr(AB)=tr(BA)$.)

设 

\[
A_{m\times n}=
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots &\ddots &\vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}\\
\end{pmatrix},\qquad
B_{n\times m}=
\begin{pmatrix}
b_{11} & b_{12} & \cdots & b_{1m}\\
b_{21} & b_{22} & \cdots & b_{2m}\\
\vdots & \vdots &\ddots &\vdots\\
b_{n1} & b_{n2} & \cdots & b_{nm}\\
\end{pmatrix}
\]

令 $C=AB$, $D=BA$, 则 $C$, $D$ 分别为 $m$-阶 和 $n$-阶方阵. 记 $C=(c_{ij})_{m\times m}$, $D=(d_{ij})_{n\times n}$, 由矩阵乘法的定义, 

\[
\begin{aligned}
c_{ij}&=\sum_{k=1}^{n}a_{ik}b_{kj},\\
d_{ij}&=\sum_{\ell=1}^{m}b_{i\ell}a_{\ell j}.
\end{aligned}
\]

于是, 

\[
\mathrm{Tr}(AB)=\sum_{i=1}^{m}c_{ii}=\sum_{i=1}^{m}\sum_{k=1}^{n}a_{ik}b_{ki}
\]

\[
\begin{split}
\mathrm{Tr}(BA)&=\sum_{i=1}^{n}d_{ii}=\sum_{i=1}^{n}\sum_{\ell=1}^{m}b_{i\ell}a_{\ell i}\\
&=\sum_{\ell=1}^{m}\sum_{i=1}^{n}a_{\ell i}b_{i\ell}\\
&=\sum_{i=1}^{m}\sum_{k=1}^{n}a_{ik}b_{ki}\\
&=\mathrm{Tr}(AB).
\end{split}
\]