TITLE: When is spectral norm of AB equal to that of BA?
QUESTION [8 upvotes]: I have $A^{1/2} B A^{1/2} \preceq I$ for two PSD matrices $A$ and $B$, and I'd like to know if that implies $\|AB\|_2 \leq 1.$ 
The argument I was using to show this is that for any two square matrices $A$ and $B,$ it is always the case that $\|AB\|_2 = \|BA\|_2.$ I thought I read that this equality does hold in a reputable source, but I don't have access to it right now and I was unsuccessful in reproducing a proof. 
I know the eigenvalues of $AB$ and $BA$ are the same modulo possibly having different numbers of zeros, so I'm worried that I might have ``remembered'' something that isn't true!
Any references/counterexamples for either question?

REPLY [3 votes]: Going back to the original motivation, the following observations might still be useful. 

Denis already mentioned this: $A^{1/2}BA^{1/2} \sim A^{1/2}A^{1/2}BA^{1/2}A^{-1/2}$ for strictly positive $A$. This implies that $\sigma(A^{1/2}BA^{1/2})=\sigma(AB)$.
If $A$ and $B$ are arbitrary (square) matrices such that their product $AB$ is normal, then one can show that $\sigma(AB) \prec_w \sigma(BA)$ (where $\prec_w$ denotes weak majorization); equivalently, for any unitarily invariant norm, we have $\|AB\| \le \|BA\|$.
If $\|AB\| \le 1$ for positive definite $A$ and $B$, then $\|A^{1/2}B^{1/2}\| \le 1$.