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For a matrix $A \in \mathbb R^{m\times n}$ with columns $a_i \in \mathbb R^m$, we know that $A^\top A$ is Positive Semidefinite (PSD).

I am interested in checking if B is positive semidefinite where $B_{ij} = (a_i^\top a_j)^2$. Each component of $B$ is the square of corresponding component of $A^\top A$.

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  • $\begingroup$ if $A^\top A$ is diagonal it's obviiously trivial, as a feel I think that you can't conclude nothing in general, but I'm curious to see tha aswers to this question $\endgroup$ – gimusi Dec 5 '17 at 0:33
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Yes. This is a direct consequence of the Schur product theorem: the elementwise product of two positive semidefinite matrices is also positive semidefinite.

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  • $\begingroup$ The proof based on eigen decomposition is very illuminating. Thank you!. Was this very obvious to you or did you need to search for this ? $\endgroup$ – Hestenes Dec 5 '17 at 4:25
  • $\begingroup$ I had met this fact before, but I had forgotten its name, so it took a small bit of searching to find again. $\endgroup$ – kimchi lover Dec 5 '17 at 7:04

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