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I know that the product $AB$ of a Hermitian and positive definite matrix $A$ and a Hermitian matrix $B$ is itself Hermitian.

From simulations (where $A$ is diagonal, but I don't think this matters), I have a suspicion that the signature (number of positive/negative eigenvalues) of $AB$ is the same as $B$. I think Sylvester's law of inertia is applicable, but I don't know how to view $A$ as a transformation under which the signature is invariant. Also, I don't know anything about quadratic forms, so please bear with me.

I'd like to use this result in my (physics) bachelor's thesis, so any help is greatly appreciated!

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Bobson Dugnutt
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    If $A$ is a positive definite Hermitian matrix then there is a positive definite Hermitian $C$ such that $C^2=A$. Now $AB=C^2B$, hence $C^{-1}ABC=CBC=CBC^$. By Sylvester's law, the signature of $CBC^$ is the same of $B$. Now, $CBC^*$ and $AB$ are similar, hence they have the same eigenvalues. Therefore the signature of $AB$ is the same of $B$. Notice that if $A$ is not positive definite then the signature may change. Example: $A$ is any diagonal matrix with negative eigenvalues and $B=Id$. – Daniel Dec 18 '16 at 01:56
  • @Daniel Thank you! I actually just realized this (when I was about to fall asleep), but it is very nice to have it written out so clearly. If you post your comment as an answer, I'd gladly accept it! – Bobson Dugnutt Dec 18 '16 at 02:04
  • You are welcome. – Daniel Dec 18 '16 at 02:05

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I know that the product $AB$ of a Hermitian and positive definite matrix $A$ and a Hermitian matrix $B$ is itself Hermitian.

No, that isn't true. Counterexample: $\pmatrix{1&0\\ 0&2}\pmatrix{1&1\\ 1&1}=\pmatrix{1&1\\ 2&2}$ is not Hermitian.

However, in case $AB$ is Hermitian, we get $AB=(AB)^\ast=B^\ast A^\ast=BA$. Hence $B$ also commutes with $A^{1/2}$ and $AB=A^{1/2}BA^{1/2}$. Therefore, by Sylvester's law of inertia, $AB$ and $B$ have identical signature.

user1551
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  • Thank you! In this answer is the point then that $AB=FFB$ is not necessarily* Hermitian, but that it's eigenvalues are real (because it shares spectrum with $FBF$ which is* Hermitian)? In the first case (where $AB$ is not Hermitian but has real eigenvalues), can we then still say that $AB$ and $B$ share signature? – Bobson Dugnutt Dec 18 '16 at 08:40
  • @Lovsovs We speak of the signature of a matrix only when it is Hermitian. So, in your case, "signature" is not the correct term to use. But surely, $AB$ and $B$ have the same number of positive/zero/negative eigenvalues, because $AB$ is similar to $A^{1/2}BA^{1/2}$, and the latter is congruent to $B$. – user1551 Dec 18 '16 at 08:52
  • I see. So invoking Sylvester's law is, in that case, not really necessary at all? Also, why/how do you differentiate between the terms "similar" and "congruent"? I'm a bit confused as to what their actual difference in this context is. – Bobson Dugnutt Dec 18 '16 at 09:06
  • @Lovsovs (a) $X$ is simiiar to $Y$ if $X=P^{-1}YP$ for some invertible matrix $P$; $X$ is congruent to $Y$ if $X=P^\ast YP$ for some invertible matrix $P$. Similarity is about change of basis for a linear transformation, while congruence is about change of basis for a bilinear or sesquilinear form. (b) You still need Sylvester's law of inertia to justify that $A^{1/2}BA^{1/2}$ and $B$ have identical signature. – user1551 Dec 18 '16 at 09:18
  • Ah, I get it now, thank you so much for your time! – Bobson Dugnutt Dec 18 '16 at 09:41