Trace of a matrix to the $n$ Why is it that if $A(t), B(t)$ are two $n\times n$ complex matrices and $${d\over dt}A=AB-BA$$ then the trace of the matrix $A^n$ where $n\in \mathbb Z$ is a constant for all $t$?

Note that Trace(FE)=Trace(EF) in general.

$n>0$ : Trace$(A^n)' = n [$Trace$ (A'(t) A^{n-1})] = n[ $Trace$ ((AB - BA)A^{n-1})] = 0$
$ n=0$ : $A^0 = I$ So we are done 
$n <0$ : Check $(A^{-1})' = A^{-1} B - BA^{-1}$ So this case is reduced to the first case. 
