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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 $\endgroup$ – JHance May 6 '15 at 4:40
  • $\begingroup$ Awesome. Thanks so much, @Jhance. $\endgroup$ – user238220 May 6 '15 at 4:40
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Positive semi-definite by definition are symmetric matrices that has positive eigen-values pick for example 2x2 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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