Let $\mathbb{F}$ by a field. Under which conditions a symmetric matrix $S\in \mathbb{F}^{n\times n}$ can be written as $A^TA$, where $A\in\mathbb{F}^{k\times n}$?
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I think that is "Iff every eigenvalue is a square in $\mathbb F$ and $S$ has rank at most $k$" – Exodd Apr 25 '18 at 22:00
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@Exodd I do agree with the rank condition, but where does the other condition come from? – Daniel Apr 25 '18 at 22:03
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over the reals, $A^TA$ is positive semi-definite; the eigenvalues are positive (therefore squares of reals) or zero – Will Jagy Apr 25 '18 at 23:22
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@Will Is it a sufficient condition as well? – Daniel Apr 26 '18 at 14:07
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Hmmm... It seems more complicated than I originally thought. Finite fields usually do not behave well with diagonalization, and even $\mathbb C$ has non-diagonalizable symmetric matrices. My first claim is also false, since there exists a counterexample in $\mathbb F_2$ – Exodd Apr 26 '18 at 18:25