Trace of a matrix to the $n$

Question

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 $t$?

Answer 1

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.

Answer 2

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 $$ \frac{d}{ dt} {\rm tr} \, A^n = {\rm tr} \, [A^n, B]=0 \ . $$

Answer 3

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 $m<0$.