TITLE: Is the rank of the sum of two positive semi-definite matrices larger than their individual ranks?
QUESTION [5 upvotes]: How to prove $\operatorname{rank}(X+Y) \geq \min(\operatorname{rank}(X),\operatorname{rank}(Y))$, where $X$ and $Y$ are both positive semi-definite matrices?

REPLY [12 votes]: This answer does not involve any decompositions. 
Let $N(X)$ be the null space of $X$. For any $v\in N(X+Y)$, we have $$v^T(X+Y)v=0\Leftrightarrow v^TXv=0,v^TYv=0$$
Consequently, $v\in N(X)$ and $v\in N(Y)$ and hence
$$N(X+Y)=N(X)\cap N(Y)$$
Then,
\begin{align}
rank(X+Y)
&=n-dim(N(X+Y))\\
&=n-dim(N(X)\cap N(Y))\\
&\ge n-dim(N(X))\\
&=rank(X)
\end{align}
Similarly, it can be shown $rank(X+Y)\ge rank(Y)$