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Let $E/F$ be a quadratic extension of $p$-adic fields. Consider $M_n(E)$ with the unitary (aka 2nd kind) involution $X \mapsto \sigma(X)^{tr}$, where $\sigma(X)$ denotes the entry-wise application of $ \text{id} \neq \sigma \in \text{Gal}(E/F)$.

With respect to this involution, a matrix $X$ can be "hermitian". A way to construct such matrices is by symmetrization: $X=Y \cdot \sigma(Y)^{tr}$.

Given a hermitian $X$, what do we know about the existence of such a $Y$?

If we were talking about $\textbf{C}/\textbf{R}$, the existence of such a $Y$ would be equivalent to $X$ being "positive semidefinite" and the expression $X=Y \cdot \sigma(Y)^{tr}$ would be called the "Cholesky decomposition". But we're not.

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    $\begingroup$ I've usually seen Cholesky defined with the requirement that Y be lower triangular with a positivity condition on the diagonal (that I am not sure I should translate into the p-adic world). If triangularity is a condition you want, then my first guess would be that it is necessary and sufficient for the eigenvalues of X to be norms of elements of E. $\endgroup$
    – S. Carnahan ♦
    Apr 22, 2010 at 6:01
  • $\begingroup$ Yeah, the usual form of it uses the stronger "positive definite" of X and the payoff in the theorem is the triangularity of Y. I don't know that much about it, but I heard that if you place no restrictions on the Y at all, it is equivalent to "positive semi-definite". For my application, no extra assumption on Y is necessary. Thanks. I'll think about your suggestion. $\endgroup$
    – Sean Rostami
    Apr 22, 2010 at 13:05
  • $\begingroup$ Yup, that's pretty useful. I guess I forgot the circumstances under which the spectral theorem works =) Thanks again! $\endgroup$
    – Sean Rostami
    Apr 22, 2010 at 16:18

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