Prove/Disprove: $n\cdot tr(AB)=tr(A) \cdot tr(B) $ iff $A$ or $B$ is scalar matrix. A and B are square matrices of size n.
So far I managed to prove that one side is right, left to right, that if A, for example is a scalar matrix so the equality holds but I have difficulties proving the other side of the sentence.
The statement is indeed wrong so let's construct a counter example:
Let $A=B=\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}\in M^{2\times 2}(\mathbb R)$, then we have $$ AB=\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}\begin{pmatrix}0& 1\\ 0 &0\end{pmatrix}=0 $$ and therefore $tr(AB)=0$, we also have $tr(A)=tr(B)=0$ and of course $n=2$ and therefore the equality $$ n\cdot tr(AB)=0=tr(A) \cdot tr(B) $$ but clearly neither $A$ or $B$ are scalar matrices (since they aren't diagonal in the first place), so the statement $$ n\cdot tr(AB)=tr(A) \cdot tr(B)\iff A\text{ or } B \text{ are scalar matrices} $$ is wrong by contradiction.