rank(a+b) ≤ rank(a) + rank(b) proof

Question

I have searched through the Internet for a proof, but didn't find anything. Can someone give me a full proof of this:

let $a,b$ be matrix $nxn$, prove:

$rank(a+b) ≤ rank(a) + rank(b)$

Already solved at http://math.stackexchange.com/questions/877802/rank-of-a-matrix-sum — Vaneet, Jun 04 '16 at 18:11
Can someone give me a reference to an online book where can I see full proof? — kicklog, Jun 04 '16 at 18:28

score 8 · Answer 1 · answered Jun 04 '16 at 18:20

Let $x_1,\ldots, x_m$ be a basis of the rowspace of $a$ and $y_1,\ldots, y_n$ be the basis of the rowspace of $b$. If every row of $a+b$ can be written as a linear combination of $x_1,\ldots,x_m,y_1,\ldots, y_n$ then $rk(a+b)\leq m+n=rk(a)+rk(b)$. But that's obvious.

rank(a+b) ≤ rank(a) + rank(b) proof

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