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It is not clear to me that a positive semidefinite matrix will definitely have zero eigenvalue, or it is just a possibility. When we say $x^TAx \geq 0$, $x$ is not necessarily a eigenvalue, so I don't see how to get any conclusion from there.

Rodrigo de Azevedo
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Sam
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  • @Surb That really depends on the definition. For many authors, a positive semi-definite matrix is positive definite if and only if it is non-singular. For me, a symmetric matrix is positive semi-definite if $x^T A x \ge 0$ and $y^T A y = 0$, for some $y\ne 0$. – PierreCarre Nov 11 '22 at 10:54
  • Yeah just to clarify I am referring to some $A$ for which there exists some $x$ such that $xAx = 0$. – Sam Nov 11 '22 at 10:57

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If $A$ is symmetric, then you can diagonalize $A$ and it is easy to deduce that if there is $x$ such that $x^t Ax=0$ then there is at least one eigenvalue of $A$ which is $0$.

If $A$ is not symmetric(*), it may happen that $x^tAx=0$ for some $x$ even if $\det A\not=0$. E.g., $A=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$.

(*) But in this case it does not make sense to speak of positive-(semi)definiteness.

EDIT: "It does not make sense" might be too strong a statement (good in the complex case, not over the reals).

Anyway, one direction is obvious: if $\det A=0$ there is $x$ (real) s.t. $Ax=0$, whence $x^tAx=0$ as well.

For the other one, we still have counterexamples. Take $$ A=\begin{pmatrix} 1 & 2\\ 0 & 1\end{pmatrix}. $$ Then $$ x^tAx=(x_1+x_2)^2,\qquad x=(x_1,x_2)^t, $$ which is always non-negative, but might be zero (it vanishes when $x_2=-x_1$). Hence, $A$ is only semi-positive definite (in the real sense, without assuming symmetry) but still invertible.

Giulio R
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  • I've seen a answer here claiming that PSD matrix may not be symmetric https://math.stackexchange.com/questions/1954167/do-positive-semidefinite-matrices-have-to-be-symmetric, could you elaborate a bit why 'it does not make sense'? But I can see in the symmetric case it is indeed true – Sam Nov 11 '22 at 11:15
  • OK I might have been too assertive in my statement, let me think and I will edit – Giulio R Nov 11 '22 at 11:25
  • I actually think you are correct as the symmetry is a requirement in the definition found both in wikipedia and Wolfram mathworld. I just asked under that answer but not sure whether i will get a reply since it is from 6 years ago – Sam Nov 11 '22 at 11:29
  • Ok, I have added some more details on the general case. Even if you don't assume symmetry, there are counterexamples. – Giulio R Nov 11 '22 at 13:08
  • thank you so much – Sam Nov 11 '22 at 13:36