Understanding Property of PSD Matrix

Question

For some reason I can't wrap my mind around several concepts.

I'm reading the proofs:

The column space of a positive semi-definite matrix A is contained in the column space of A+B?

Rank of the sum of two positive semi-definite matrices

I can't seem to understand one line. Suppose $X \succeq 0$ is positive semidefinite. Let $v \neq 0$ and suppose $v^{T}Xv=0$. The second and first link say then that $v$ is then in the null space of $X$, meaning $Xv=0$

Why must this be the case?

Thanks.

Answer 1

Any positive semidefinite matrix $X$ has a positive semidefinite square root $Y$, that is, $Y^2=X$.

$X$ is symmetric, hence diagonalizable in an orthonormal basis, with nonnegative eigenvalues. Take the diagonal matrix with the square roots of the eigenvalues, in the same basis.

Consequently, $0=v^TXv=v^TY^2v=v^TY^TYv=\|Yv\|^2$ implies $Yv=0$ whence $Xv=Y(Yv)=0$ as well.

Answer 2

On a more fundamental level, if $X$ is positive semidefinite, $(u,v):=u^TXv$ defines a semi-inner product. Hence it satisfies Cauchy-Schwarz inequality. So, if $(v,v)=0$, then $|u^TXv|=|(u,v)|\le\sqrt{(u,u)(v,v)}=0$ for every vector $u$. Hence $u^TXv$ is identically zero. Since $u$ is arbitrary, $Xv$ must be the zero vector.