Symmetric positive semi definite implies singularity?

Question

So here i have a symmetric matrix $\Sigma$ which is positive semidefinite but not positive definite. How to prove $\Sigma$ is also singular?

From the definition I tried $X^T\Sigma X\geqslant0$.

Given $X\neq0$, it is easy to see that if $X^T\Sigma X>0$, then $\Sigma X>0$.

But when come to the case which $X^T\Sigma X=0$, I don't know if I can suppose $\Sigma X=0$ or not. Because if I can, then the prove is almost done. Any ideas about it? Thanks

Answer 1

Any symmetric matrix is diagonalizable. Similarity transformations preserve positive definiteness and positive semidefiniteness. Any diagonal matrix which is positive semidefinite, yet not positive definite, has a zero entry on the diagonal and is therefore singular.

Therefore any symmetric matrix which is positive semidefinite, yet not positive definite, is singular.

Answer 2

As $\Sigma$ is positive semidefinite but not positive definite, there exists some nonzero vector $x$ such that $x^T\Sigma x=0$. Now, if $\Sigma$ is nonsingular, let $y=\Sigma^{-1}x$. Then $0\le(ax-y)^T\Sigma(ax-y)=y^T\Sigma y-2a\|x\|^2$ for every scalar $a$. Yet this is impossible because $\|x\|>0$. Therefore $\Sigma$ must be singular.