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 $t$?
 A: For every positive integer $m$ we have
$$
\frac{d}{dt}A^m=A^{m-1}\dot{A}+A^{m-2}\dot{A}A+\ldots+A\dot{A}A^{m-2}+\dot{A}A^{m-1},
$$
where
$$
\dot{A}=\frac{d}{dt}A.
$$
Since 
$$
\text{trace}: M_n(\mathbb{R}) \to \mathbb{R},\ X \mapsto \text{trace}(X)
$$ 
is linear and satisfies
$$
\text{trace}(XY)=\text{trace}(YX) \quad \forall X,Y \in M_n(\mathbb{R}),
$$
it follows that
\begin{eqnarray}
\frac{d}{dt}\text{trace}(A^m)&=&\text{trace}(\frac{d}{dt}A^m)=m\text{trace}(A^{m-1}\dot{A})\\
&=&m\text{trace}[A^{m-1}(AB-BA)]=m[\text{trace}(A^{m-1}AB)-\text{trace}(A^{m-1}BA)]\\
&=&m[\text{trace}(A^mB)-\text{trace}(BA^m)]=0.
\end{eqnarray}
Hence
$$
\text{trace}(A^m(t))=\text{trace}(A^m(0)) \quad \forall t.
$$
Notice that $A^m$ is not necessarily defined for $m<0$.
A: 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. 
A: 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
$$ \frac{d}{ dt} {\rm tr} \, A^n =  {\rm tr} \, [A^n, B]=0  \ . $$
