Singular matrix and positive semidefiniteness

Question

Let $C_x$ be a psd matrix. If $b^TC_xb=0$, then why must $C_x$ be a singular matrix and how do you prove it?

Answer 1

First, I assume that when you say 'psd', $C_x$ is assumed to be symmetric. Some people appear to consider the definition of 'psd' to apply to non-symmetric matrices as well, but Wikipedia says it's a term used only for symmetric matrices.
Second, I don't know if there's an elementary way of proving this without resorting to a spectral decomposition of $C_x$, or at least reconstructing some of the proof of a spectral decomposition.
Since $C_x$ is psd, it has a 'square root' - which means a psd matrix $S$ such that $S^2 = C_x$. This is not a bypass of a spectral decomposition but is also proved via spectral decomposition. Now $$0=b^\top C_x b = b^\top S^2 b = (Sb)^\top (Sb) = || Sb ||^2$$ So $Sb= 0 $ since $||Sb|| = 0$. But then $C_x b = S^2 b=S(Sb)=0$. So $b\in \operatorname{Ker} C_x$.