range of sum of positive semidefinite matrices

Question

Let $A$ and $B$ be positive semidefinite matrices. Does it follow that $\text{range} \ A \subseteq \text{range} \ (A + B)$? I think the answer is yes, as I have not been able to think of a counter example, but I have not been able to prove it either.

Answer 1

Note first that if $A$ is positive semidefinite and $\left< Au, u \right> = 0$ then $Au = 0$. To see why, let $v_1,\dots,v_n$ be an orthonormal basis of eigenvectors of $A$ (so $Av_i = \lambda_i v_i$) and write $u = \sum_{i=1}^n \left< u, v_i \right> v_i$. Then

$$ \left< Au, u \right> = \sum_{i=1}^n \left<u, v_i \right>^2 \lambda_i = 0 $$ together with $\lambda_i \geq 0$ implies that $\left< u, v_i \right> = 0$ if $\lambda_i > 0$ so $u \in \ker(A)$.

To prove that $\operatorname{range}(A) \subseteq \operatorname{range}(A + B)$ it is enough to prove that

$$ \ker(A + B) = \operatorname{range}(A + B)^{\perp} \subseteq \operatorname{range}(A)^{\perp} = \ker(A)$$

so let $u \in \ker(A + B)$. Then

$$ 0 = \left< (A + B)u, u \right> = \left< Au, u \right> + \left< Bu, u \right> $$

which implies that $\left< Au, u \right> = 0$ so $u \in \ker(A)$.