TITLE: Proving if A and B are matrices such that A, B, and AB are normal, then BA is also normal.
QUESTION [10 upvotes]: If A and B are matrices such that A, B, and AB are normal, then BA is also normal.
I've seen this statement around, although I've only seen the site/publication/etc... state that it was proven by Wiegmann, but nobody actually gives the proof. Could somebody help me wrap my head around why this is true?
Thank you.

REPLY [7 votes]: There is a simpler proof for this statement (see Problem 42 in Section 2.5 of Matrix Analysis by Horn and Johnson).
You already know that a matrix 
$$A \text{ is normal if and only if } \mathrm{tr}(A^*A)=\sum\limits_{i=1}^n|\lambda_i(A)|^2,$$
where $\lambda_i(A)$ are the eigenvalues of $A$.
Since $A$ and $B$ are normal and $\mathrm{tr}(XY)=\mathrm{tr}(YX)$ for square $X$ and $Y$, we have
$$
\begin{split}
\mathrm{tr}[(AB)^*(AB)]&=\mathrm{tr}(B^*A^*AB)=\mathrm{tr}(B^*AA^*B)\\&=\mathrm{tr}(A^*BB^*A)=\mathrm{tr}(A^*B^*BA)=\mathrm{tr}[(BA)^*(BA)].
\end{split}
$$
So we get
$$
\sum_{i=1}^n|\lambda_i(BA)|^2=\sum_{i=1}^n|\lambda_i(AB)|^2=\mathrm{tr}[(AB)^*(AB)]=\mathrm{tr}[(BA)^*(BA)],
$$
where the first equality follows from the fact that the eigenvalues of $AB$ and $BA$ are equal ($AB$ and $BA$ have the same characteristic polynomials) and the second from the normality of $AB$.
Hence $BA$ is normal.