$\DeclareMathOperator{\Tr}{Tr}$$\Tr(AB)=\Tr(BA)$,所以$\Tr(AB-BA)=0$. 逆命题为真: Only Commutators Have Trace Zero
Lemma 2: 假设 $A$ 不是单位矩阵 $I$ 的非零标量倍，则存在一个相似于 $A$ 的矩阵，其第一个对角元素为0。
Proof 2若 $A$ 为零矩阵, 则 $A$ 的对角元素为0。若 $A$ 不是单位矩阵 $I$ 的标量倍，则存在向量 $v$ 不是 $A$ 的特征向量，则 $v$ 与 $Av$ 线性无关. 将 $v, A v$ 扩充为一组基 $\mathcal{B}$.
因为$Av$就是$\cal B$的第2个元素, 所以 $A$ 关于此基的矩阵的第一列为$\pmatrix{0\\1\\0\\\vdots\\0}$. 其第一个对角元素为0. 证毕.
Proof 3The theorem is obviously valid if $Z$ is 1-by-1 or a bigger zero matrix. Therefore assume that $Z$ is a nonzero square matrix of dimension bigger than 1. Our proof goes by induction; we assume the desired inference valid for all matrices of dimensions smaller than $Z$ 's with Trace zero. Because of that zero Trace, $Z$ cannot be a nonzero scalar multiple of I, so Lemma 2 implies that some invertible $B$ exists making $B^{-1} Z B=\left[\begin{array}{ll}0 & r^T \\ c & K\end{array}\right]$. Observe next that $\operatorname{Trace}(K)=\operatorname{Trace}(Z)=0$. The induction hypothesis implies that $K$ is a commutator; then Lemma 1 implies that $B^{-1} ZB=XY-YX$ is a commutator too for some $X$ and $Y$, whereupon $Z=\left(BXB^{-1}\right)\left(BYB^{-1}\right)-\left(BYB^{-1}\right)\left(BXB^{-1}\right)$ must be a commutator too. End of Proof 3.
又见1957, On matrices of trace zeros. – A. A. Albert, Benjamin Muckenhoupt
Pauli 矩阵
If $\Tr(A)=0$ then $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$ | 
