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. 

If $A$ is an $m\times n$ matrix and $B$ an $n\times m$ matrix we have
\begin{eqnarray}
\operatorname{tr}(AB)
&=&\sum_{i=1}^m(AB)_{ii}\cr
&=&\sum_{i=1}^m\sum_{k=1}^n A_{ik}B_{ki}\cr
&=&\sum_{k=1}^n\sum_{i=1}^mB_{ki}A_{ik}\cr
&=&\sum_{k=1}^n(BA)_{kk}\cr
&=&\operatorname{tr}(BA).
\end{eqnarray}
