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I am trying to prove this just to be clear about this but I don't have enough conditions to force this idea to be true, so I doubt it is.

Are symmetric matrices always at least positive semi-definite?

I know the other way around, by convention, positive-definite / positive semi-definite matrices are always symmetric.

I currently have on my piece of paper a 1x1 block, after using the definition of positive-definiteness, that consists of a polynomial that is quadratic in the variables x, y, z, which are the components of my chosen vector. But the coefficients attached are coming from the symmetric matrix, and there's no reason why a symmetric matrix needs to have all positive entries.

Thanks,

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    $\begingroup$ No. Just take the matrix with $-1$ all along the diagonal, this is negative definite. Or a mix of $-1$'s and $1$'s $\endgroup$
    – jxnh
    May 6, 2015 at 4:40
  • $\begingroup$ Awesome. Thanks so much, @Jhance. $\endgroup$
    – user238220
    May 6, 2015 at 4:40
  • $\begingroup$ All you can say is that real symmetric matrices have only real eigenvalues. $\endgroup$
    – Lelouch
    Oct 29, 2022 at 10:58

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Positive semi-definite by definition are symmetric matrices that has positive eigen-values. Pick for example $2 \times 2$ identity matrix and multiply first $a_11$ component by negative one you'll see that this matrix is symmetric however have negative eigen-values. In particular it has $-1$ as an eigen-value.

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