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?



When is spectral norm of AB equal to that of BA?
I suppose that your norm is either the Schur-Frobenius $\|F\|=({\rm Tr}FF^T)^{1/2}$ or that subordinated to the Euclidan norm $\|F\|=\rho(FF^T)$. In either case, the equality $\|AB\|=\|BA\|$ follows from the equality of the spectra of $ABBA$ and $BAAB$ (mind that $A$ and $B$ are symmetric). This is true because both matrices are similar to $A^2B^2$ (mind that $A$ and $B$ are non-singular).


If you drop your assumption that $A,B$ are PSD, then the equality does not hold in general. A simple counter-example is given by a pair $(A,B)$ such that $AB=0\_n$ but $BA\neq 0\_n$.


