$\ker(A + B) = \ker(A) \cap \ker(B)$ where both $A$ and $B$ are positive semidefinite symmetric matrices

Question

Does anyone have a proof for the statement in the title? A link or a book would be fine too. It's not a homework question, I stumbled upon it when reading a paper. I can't really wrap my head around it and my own research didn't lead me anywhere. It looks like a standard theorem, but apparently it's not? Does it have a specific name?

Answer 1

$ \ker A \cap \ker B \subset \ker(A+B)$ is always true.

Let $x\in \ker(A+B)$. Then : $$(x,Ax) + (x,Bx)= 0$$ and therefore, since both terms are non-negative : $$ (x,Ax) = (x,Bx) = 0$$

Since $A,B$ are (symmetric) and positive, this implies $Ax = Bx = 0$.

Answer 2

Given a Hermitian matrix $X$, call $\operatorname{iso}(X)$ its set of isotropic vectors, namely $\operatorname{iso}(X)=\{v\in \Bbb C^n\,:\, v^HXv=0\}$. It is clear that $\ker X\subseteq \operatorname{iso}(X)$. Now, notice that if $A=A^H$, $B=B^H$ are positive semidefinite, then $$v^H(A+B)v=v^HAv+v^HBv\ge 0$$

with equality if and only if both summands are $0$. Therefore $$\ker(A+B)\subseteq\operatorname{iso}(A+B)=\ker A\cap \ker B$$

$\ker A\cap \ker B\subseteq \ker(A+B)$ is obvious.