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.
Must a positive semidefinite matrix have zero eigenvalue
linear algebramatricespositive-semidefinitesymmetric matrices
Best Answer
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.