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I am considering the order-preserving problem of two square matrices: If $A<B$ and $P$ is positive definite, can we make conclusions about the ordering relation of $PA$ and $PB$? (Likewise, of $AP$ and $BP$.) Does $PA<PB$ hold under some suitable conditions? Or, what is a simple counterexample? (I know that the Hermitian property may not be preserved after multiplying P, so I prefer some counterexamples where all matrices involved are Hermitian.)

Note that the notation $A<B$ here means that $B-A$ is positive.

I roughly know that if $P$ is invertible, then $PAP^{-1}$ and $PBP^{-1}$ will preserve the ordering relation. But I am not sure about the conclusion if we are considering $PA$, $PB$ instead.

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  • $\begingroup$ The product of two hermitian matrices are hermitian iff they commute $\endgroup$
    – Aphyd
    Jan 8 at 6:46

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Let's recap some prelims.

  1. By the spectral theorem, any Hermitian matrix is unitarily diagonalizable with real eigenvalues (see here). Further, a Hermitian matrix is positive definite if and only if its eigenvalues are all positive (see here).

  2. The product of two Hermitian matrices $X$ and $Y$ is Hermitian if and only if they commute i.e. $XY=YX$ (see here).

  3. A set of diagonalizable matrices commutes if and only if the set is simultaneously diagonalizable i.e. there is a single invertible matrix $V$ s.t. $V^{-1}XV$ is diagonal for all $X$ in the set (see here).

Applying these to your setup, if $P$ and $B-A$ are Hermitian and positive definite, their product is Hermitian only if they commute, in which case they must be simultaneously diagonalizable (and hence have the same eigenvectors) with each eigenvalue of $P(B-A)$ given by the product an eigenvalue of $P$ and an eigenvalue of $B-A$, implying $P(B-A)$ is positive definite.

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  • $\begingroup$ To fill in the last step, if $x$ is a common eigenvector of $P$ and $B-A$ corresponding to respective eigenvalues $\lambda_1,\lambda_2,$ then $P(B-A)x=P(\lambda_2 x)=\lambda_1\lambda_2 x$. $\endgroup$ Jan 8 at 14:31

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