Vtyjkyuk

Prove that for square matrices $ A,(mtimes m) $ and $ B,(ntimes n) $, $tr(Atimes B)=tr(A)tr(B).$

$tr(A)=sum_i=1^ma_ii=a_11+a_22+...+a_mm$

$tr(B)=sum_i=1^nb_ii=b_11+b_22+...+b_nn$

Since $Atimes B$ has to be true, the matrices have to be the same size and $m=n$.

$Atimes B=sum_k=1^m a_ikb_kj=a_i1b_1j+...+a_nmb_mn$

I don't know if what I've done is correct so far and I'm not sure how to go on with solving the trace of $Atimes B$.

Suppose that $A$ is $mtimes m$ with typical elements $A_ij$ and $B$ is $ntimes n$ with typical elements $B_hk$. We have
$$
Aotimes B=beginpmatrix
A_11B & A_12B & cdots & A_1mB\
A_21B & A_22B & cdots & A_2mB \
vdots & vdots & ddots & vdots \
A_m1B & A_m2B & cdots & A_mmB
endpmatrix
$$
so the diagonal elements of $Aotimes B$ are
$$
A_11B_11,ldots,A_11B_nn,A_22B_11,ldots,A_22B_nn,ldots,A_mmB_11,ldots,A_mmB_nn.
$$
Summing these yields
beginalign*
&quadoperatornameTr(Aotimes B)\
&=A_11B_11+cdots+A_11B_nn+A_22B_11+cdots+A_22B_nn+cdots+A_mmB_11+cdots+A_mmB_nn\
&=A_11(B_11+cdots+B_nn)+A_22(B_11+cdots+B_nn)+cdots+A_mm(B_11+cdots+B_nn)\
&=(A_11+cdots+A_mm)(B_11+cdots+B_nn)=operatornameTr(A)operatornameTr(B).
endalign*

If you meant $operatornametr(Atimes B)$ as $operatornametr(Aotimes B)$ then as @Sangchul Lee mentioned, we have:

$displaystyle beginbmatrixa&b\c&d\endbmatrixotimes beginbmatrixp&q\r&s\endbmatrix=beginbmatrixacdot beginbmatrixp&q\r&s\endbmatrix&bcdot beginbmatrixp&q\r&s\endbmatrix\ccdot beginbmatrixp&q\r&s\endbmatrix&dcdot beginbmatrixp&q\r&s\endbmatrix\endbmatrix=beginbmatrixacdot p&acdot q&bcdot p&bcdot q\acdot r&acdot s&bcdot r&bcdot s\ccdot p&ccdot q&dcdot p&dcdot q\ccdot r&ccdot s&dcdot r&dcdot s\endbmatrix$

So we have:

$$(operatornametrA)(operatornametrB)=(a+d)(p+s)=acdot p+a cdot s+dcdot p+dcdot s=operatornametr(Aotimes B)$$

Where $otimes$ is the Kronecker Product and $operatornametr(.)$ is the trace of a matrix.

If you meant $operatornametr(Atimes B)$ as $operatornametr(AB)$, which is a simple Matrix multiplication, then :

$displaystyle beginbmatrixa&b\c&d\endbmatrix beginbmatrixp&q\r&s\endbmatrix=beginbmatrixap+br&aq+bs\cp+dr&cq+ds\endbmatrix$

So now we can see that:

$$(operatornametrA)(operatornametrB)=(a+d)(p+s)=acdot p+a cdot s+dcdot p+dcdot s ne acdot p+bcdot r+ccdot q+dcdot s=operatornametr(AB)$$

You can follow the same methodology and it would be true for any $ntimes n$ matrices.