Vrftjkry

The commutator of square matrices A,B∈Mn×n(F) is defined as

$[A,B]=AB−BA$
The trace of a square matrix $C∈Mn×n(F)$ is defined as the sum of diagonal entries:

$tr(C)=C_1$$_ 1$+…+$C_n$$_n$.

a) Show that for all square matrices, tr([A,B])=0.

b) Show that if S is an invertible matrix, then $tr(SCS^−$$^1)=tr(C)$.

in part A, i prove tr(AB)=tr(BA),so tr(AB)-tr(BA)=0, is it right begin{align*}
mathrm{tr}(AB) &= sum_{i=1}^n (AB)_{ii}\
&=sum_{i=1}^n sum_{j=1}^m A_{ij}B_{ji}\
&= sum_{j=1}^m sum_{i=1}^n B_{ji}A_{ij}\
&= sum_{j=1}^m (BA)_{jj}\
&= mathrm{tr}(BA)
end{align*}
part b I did : if S is an invertible then tr(SC$S^-$$^1$)=tr(C) * tr(S) * tr($S^-$$^1$) = tr(C)*I=tr(C) , am I right for this one

Your answer for part (a) looks good. I would maybe add a quick sentence saying that $operatorname{tr}(AB)-operatorname{tr}(BA)=operatorname{tr}(AB-BA)$.

For part (b) this does not work. The trace function is not multiplicative. What you can do is use $operatorname{tr}(AB)=operatorname{tr}(BA)$ that you proved in part (a). Try setting $A=SC$ and $B=S^{-1}$.

part a) is correct.

part b) is not correct: the trace is not multiplicative, i.e. we don't have $tr(AB)=tr(A)cdot tr(B)$.

But, we only need part a) for this: observe that $C=S^{-1}(SC)$.