In this problem I'm given that A and B are both positive semi-definite matrices. I have to show that A+B is also a positive semi-definite matrix, but that part is simple. I'm also responsible for showing that the column space of A must be contained in the column space of the matrix A+B, and this is the part that evades me…
This looks as if it may be possible to solve by contradiction (if an element from the column space of A is not in the column space of A+B then perhaps somehow this violates the positive semi-definiteness of A, B, or A+B?), but all of my attempts to show this (by contradiction or otherwise) have gone nowhere.
Any hints would be greatly appreciated!
Best Answer
First, notice that for a symmetric matrix $P$ holds $range(P)\perp null(P) \Leftrightarrow range^\perp(P) = null(P)$, because every symmetric matrix can be diagonalized with an orthonormal basis.
Now notice that since $A\succeq 0$, $B\succeq 0$ then from $(A+B)v =0\Rightarrow v^T(A+B)v =0 \Leftrightarrow v^TAv = -v^TBv$ follows $v^TAv \geq 0$, $-v^TBv \leq 0$ and therefore $v^TAv=v^TBv=0 \Leftrightarrow v\in null(A), v\in null(B)$.
Now we proved $null(A+B) \subseteq null(A)$ and $null(A+B) \subseteq null(B)$ and by taking the orthogonal complement of both relations we get
$range(A) = null^\perp(A) \subseteq null^\perp(A+B) = range(A+B)$
and
$range(B) = null^\perp(B) \subseteq null^\perp(A+B) = range(A+B)$