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 $t$?

For $n=1$.
$$
\dot{ {\rm tr} \, A}={\rm tr}\, \dot{A} ={\rm tr}\, [A, B] =0 \ . $$
For $n=2$ 
$$ \dot{ {\rm tr} \, A^2}={\rm tr}\, \dot{A^2} ={\rm tr}\, (A \dot{A}+\dot{A} A) = $$
$$tr(A(AB-BA)+(AB-BA)A)=tr(A^2B-BA^2)=tr[A^2, B]=0 $$
More generally by an easy induction
$$ \frac{d}{ dt} {\rm tr} \, A^n =  {\rm tr} \, [A^n, B]=0  \ . $$
