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Suppose I have an anti-Hermitian matrix $A$: $A=-A^*$, and a positive definite matrix $P$ with Cholesky decomposition $P=LL^*$.

The eigenvalues of $PA$ are purely imaginary: $$PA=LL^* A$$ $$L^{-1}PA L = L^* A L$$ $$(L^* A L)^*=-L^* A L$$ So $PA$ is similar to an anti-Hermitian matrix, so its eigenvalues are purely imaginary.

This argument fails when $P$ is merely positive semi-definite, because $L$ is not invertible in that case. And yet, the eigenvalues of $PA$ remain purely imaginary: consider the positive semi-definite $P$ as a limit of positive definite $P_i$, and the eigenvalues of the limit are the limits of the eigenvalues.

So, question

  1. Is this true?
  2. Is there a way of constructing the similarity transform from $PA$ to an anti-Hermitian matrix, such that it keeps working when $P$ is positive semi-definite?
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Yes.

If $P$ is only positive semidefinite then $P= Q^2$, for some hermitian (positive semidefinite).

Now, like you wrote, $PA = Q^2 A = Q(QA)$. It turns out that for any two matrices $M$, $N$, the matrices $MN$ and $NM$ have the same spectrum ( with multiplicities) ( no need for invertibility). Hence $PA$ has the same spectrum as $Q A Q$, which, as you said, has purely imaginary spectrum, being antihermitian.

Alternatively, you can use the correct fact that the eigenvalues of the limit are the limit of the eigenvalues ( in a properly defined way).

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