How to prove $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$? there is a similar thread here Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?, but I'm only looking for a simple linear algebra proof. 

Observe that if $A$ and $B$ are $n\times n$ matrices, $A=(a_{ij})$, and $B=(b_{ij})$, then
$$(AB)_{ii} = \sum_{k=1}^n a_{ik}b_{ki},$$
so
$$
\operatorname{Tr}(AB) = \sum_{j=1}^n\sum_{k=1}^n a_{jk}b_{kj}.
$$
Conclude calculating the term $(BA)_{ii}$ and comparing both traces.
