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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?

Aaron Hendrickson
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eet
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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 $b\in \operatorname{Ker} C_x$.

  • How does $b\in \operatorname{Ker} C_x$ prove singularity? Any reference? – eet Jun 15 '21 at 15:32
  • First, the problem must state that $b\ne 0$. For if $b=0$ then obviously $b^\top C_x b=0$ but $C_x$ doesn't have to be singular. Now, if $b\ne 0$ and $C_x b=0$ then $C_x$ cannot be invertible, because the inverse would have to map $0$ to $b\ne 0$. What's your definition of singularity? –  Jun 15 '21 at 16:05
  • Now I understand your point. thanks. – eet Jun 15 '21 at 17:02