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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?

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  • $\begingroup$ This is false in general. You can find $A, B$ such that $AB = 0$ but $BA \neq 0$ (although the result you cite shows that $BA$ is nilpotent). The spectral norm is defined in terms of singular values, not eigenvalues. $\endgroup$
    – Qiaochu Yuan
    Commented Feb 22, 2013 at 7:27
  • $\begingroup$ Did I miss something or your assumptions imply that $BA$ is the transpose of $AB$ ? $\endgroup$
    – Fabien Besnard
    Commented Jan 11, 2021 at 16:19

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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 $BA\neq 0_n$.

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Going back to the original motivation, the following observations might still be useful.

  1. 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 $\sigma(A^{1/2}BA^{1/2})=\sigma(AB)$.

  2. 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 $\|AB\| \le \|BA\|$.

  3. If $\|AB\| \le 1$ for positive definite $A$ and $B$, then $\|A^{1/2}B^{1/2}\| \le 1$.

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  • $\begingroup$ In (1), the assumption that $A$ is strictly positive is unnecessary since strictly positive matrices are dense in all positive matrices. $\endgroup$
    – Eric Wofsey
    Commented Feb 22, 2013 at 19:28
  • $\begingroup$ @Eric: yes indeed; I just wrote it so that it was ok to write $A^{-1/2}$ instead of $(A+\epsilon I)^{-1/2}$, and then let $\epsilon \to 0$ $\endgroup$
    – Suvrit
    Commented Feb 22, 2013 at 19:32
  • $\begingroup$ @Suvrit: I was wondering how much can be said if $A$ is positive definite but $B$ is only symmetric. If $B$ is any symmetric matrix, using 1, I have that the eigenvalues of $A^{1/2}BA^{1/2}$ and $AB$ are equal. Now since $A^{1/2}BA^{1/2}$ is symmetric, its max absolute eigenvalue is its spectral norm. Can I then say that $\Vert AB\Vert \le \Vert A^{1/2}BA^{1/2}\Vert$? $\endgroup$
    – Arnab Auddy
    Commented Feb 25, 2020 at 4:08

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