Given $P \succeq 0$, if $x \neq 0$ and $x^T P x=0$, can we conclude $Px=0$?

Question

For a positive semidefinite matrix $P$ (i.e., for any $x$, $x^TPx\ge0$, it is denoted as $P \succeq 0$). If $x$ is nonzero and $x^TPx=0$, can we conclude $Px=0$? If so, how to prove it?

I do not understand the solution to exercise 10.1 in Boyd & Vandenberghe's Convex Optimization, where $P$ is positive semidefinite, and the solution says if for a nonzero $x$, $x^TPx=0$, since $P\succeq0$, we conclude $Px=0$.

Another related question: if $P\succ0$ and $x^TPx=0$, can we conclude $x=0$? If so, how to prove it?

Thanks.

Answer 1

Using the spectral theorem as indicated by Katie in the comments, there exists a square matrix $Q$ such that $P=Q^TQ$ (take the square roots of the (non-negative) eigenvalues). Now $|Qx|^2=x^TPx=0\Rightarrow Qx=0\Rightarrow Px=0$. Your second question follows directly from the definition as commented by Semiclassical.