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.
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.
Here is a proof as taken from this paper: Kaplansky, Irving Products of normal operators. Duke Math. J. 20, (1953). 257-260. This paper is also contained in 'Selected Papers and Other Writings of Irving Kaplansky'. It is supposed to be proven originally by Wiegmann, but I do not have access to this paper.
Theorem 1: Assume $A$ and $AB$ are normal. If $B$ commutes with $A^*A$ then $BA$ is normal.
Proof: Take the polar decomposition of $A$, $A=UP$, $U$ unitary, $P=P^*$. Since $A$ is normal it follows $$ AA^* = UPP^*U^* = A^*A = P^*P, $$ hence $UP^2 = P^2U$, which implies that $U$ and $P$ commute.
As $B$ commutes with $A^*A$ it also commutes with $P=(A^*A)^{1/2}$.
This implies $$ U^*ABU = U^*UPBU = PBU = BPU = BA, $$ hence $BA$ is unitary equivalent to a normal matrix, and thus is normal itself.
Theorem 2: Assume $A$, $B$, $AB$ normal. Then $B$ commutes with $A^*A$.
Proof: Set $C:=BA^*A-A^*AB= [B,A^*A]$ where $[C,D]=CD-DC$ denotes the commutator of two matrices $C$ and $D$.
Then $$ C^*C=(A^*AB^*-B^*A^*A)(BA^*A-A^*AB) = A^*AB^*BA^*A - A^*AB^*A^*AB -B^*A^*ABA^*A +B^*A^*AA^*AB. $$ Now, by normality of $AB$ and $A$ we obtain $$ B^*A^*AA^*AB- A^*AB^*A^*AB = (B^*A^*)(AA^*AB)- (A^*AAB)(B^*A^*)= [B^*A^*,AA^*AB]. $$ Since $$ A^*AB^*BA^*A-B^*A^*ABA^*A = A^*AB^*BA^*A-ABB^*A^*A^*A= (A^*)(AB^*BA^*A)-(AB^*BA^*A)(A^*) = [A^*,AB^*BA^*A]. $$ This proves $$ C^*C = [B^*A^*,AA^*AB] + [A^*,AB^*BA^*A]. $$ Since the trace of commutators is zero due to $tr(CD)=tr(DC)$, it follows $tr(C^*C)=0$, hence $C=0$, and $B$ commutes with $A^*A$.
Theorem 1 + 2 now prove the claim.
