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Let $A $ and $ B $ be symmetric matrices. Can we say something about the eigenvalues of the matrix product $AB$ when $A \preceq B$ in the sense that $ A-B $ is positive semidefinite? Can we conclude maybe that $$ \lambda_{\max}(AB) \leq 1 $$

where $ \lambda_{\max} $ is the largest eigenvalue? If so, how?

We don't assume that $ A $ and $ B $ commute. In this case, they should be simultaneously diagonalizable since they are both symmetric.

Correction of what I wrote: $ A\leqslant B $ means $ B-A $ is semi-definite positive. And I also made an error of what I'm searching: I want to know if there is result on $ \lambda_{\max}(B^{-1}A) $, e.g. $\lambda_{\max}(B^{-1}A) \leq 1 $ (we assume that $ B $ and $ A $ are invertible)

Mickael Alb.
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    Related – Arctic Char Aug 19 '21 at 07:07
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    Remember that real numbers are symmetric matrices in 1D equal to their (only) eigenvalue. And scalar matrices, multiples of $I$, behave the same way in any dimension. So if something isn't true for real numbers it can't be true for symmetric matrices. And why would a product of two real numbers be less than $1$ if all you know about them is that one is less than the other? – Conifold Aug 19 '21 at 08:19
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    I don't understand how $A \preceq B$ is equivalent to $ A-B \succeq 0$. – Rodrigo de Azevedo Aug 19 '21 at 08:24
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    Are you sure that $AB$ has real eigenvalues? – Rodrigo de Azevedo Aug 19 '21 at 08:27
  • If $0\preceq A\preceq B$, then $\lambda_\max(AB)\leq \lambda_\max(A)\lambda_\max(B)$ with equality for example if $A=B$. – MaoWao Aug 19 '21 at 09:20
  • @RodrigodeAzevedo Sorry, that was too quick. It is true if one of the factors is positive semi-definite, and the proof can be found here: https://math.stackexchange.com/questions/4118968/eigenvalues-of-the-product-of-symmetric-and-positive-definite-matrices. – MaoWao Aug 19 '21 at 09:48
  • Correction of my question: $ A\leq B \Leftrightarrow B-A $ is semi-definite positive – Mickael Alb. Aug 19 '21 at 14:07
  • there are a lot of problems here. In addition to twice mentioned issue that question doesn't make sense for non PSD $B$ (i.e. PD since invertible), the title uses $B^{-1}$ and the text seems to not. If $B$ is PD and the title is the actual question then this can be done via congruence. – user8675309 Aug 19 '21 at 17:53
  • @user8675309 Can you please elaborate more on how to use congruence to prove maximum eigenvalue of $AB^{-1}$ is bounded by 1? – Chuanhao Li Feb 16 '23 at 18:22

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