## 每週問題 March 10, 2014

Let $A=[a_{ij}]$ be an $n\times n$ matrix with $a_{ii}=0$ for all $i$. Show that there exist $n\times n$ matrices $B$ and $C$ such that $A=BC-CB$.

$B=[b_{ij}]$$C=[c_{ij}]$$n\times n$ 階矩陣且 $B$ 為對角矩陣，則

\displaystyle\begin{aligned} BC-CB&=\begin{bmatrix} b_{11}&&&\\ &b_{22}&&\\ &&\ddots&\\ &&&b_{nn} \end{bmatrix}\begin{bmatrix} c_{11}&c_{12}&\cdots&c_{1n}\\ c_{21}&c_{22}&\cdots&c_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ c_{n1}&c_{n2}&\cdots&c_{nn} \end{bmatrix}\\ &~~~~~-\begin{bmatrix} c_{11}&c_{12}&\cdots&c_{1n}\\ c_{21}&c_{22}&\cdots&c_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ c_{n1}&c_{n2}&\cdots&c_{nn} \end{bmatrix}\begin{bmatrix} b_{11}&&&\\ &b_{22}&&\\ &&\ddots&\\ &&&b_{nn} \end{bmatrix}\\ &=\begin{bmatrix} 0&(b_{11}-b_{22})c_{12}&\cdots&(b_{11}-b_{nn})c_{1n}\\ (b_{22}-b_{11})c_{21}&0&\cdots&(b_{22}-b_{nn})c_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ (b_{nn}-b_{11})c_{n1}&(b_{nn}-b_{22})c_{n2}&\cdots&0 \end{bmatrix}. \end{aligned}

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