TITLE: Show that $\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$
QUESTION [17 upvotes]: I know about the fact that $\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$, where $A$ and $B$ are $m \times n$ matrices.
But somehow, I don't find this as intuitive as the multiplication version of this fact. The rank of $A$ plus the rank of $B$ could have well more than the columns of $(A+B)$! How can I show to prove that this really is true?

REPLY [5 votes]: It suffices to show that, Row rank $(A + B)$ ≤ Row rank $A + $Row rank $B$ $(see~here)$
i.e. to show $\dim \langle a_1 + b_1, a_2 + b_2, …, a_n + b_n\rangle \leq \dim \langle a_1, a_2, … , a_n\rangle+\dim \langle b_1, b_2,..., b_n\rangle$ [Letting the row vectors of A and B as $a_1, a_2, … , a_n$ and $b_1, b_2,…, b_n$ respectively] 
Let $\{A_1, A_2, …, A_p\}$ & $\{B_1, B_2, … , B_q\}$ be the bases of $\langle a_1, a_2, … , a_n\rangle$ and $\langle b_1, b_2,…, b_n\rangle$ respectively.
Case I: $p, q ≥ 1$
Choose $v\in\langle a_1 + b_1, a_2 + b_2, …, a_n + b_n\rangle$
Then $v = c_1(a_1 + b_1) + … + c_n(a_n + b_n) [$for some scalars $c_i] = ∑c_i (∑g_jA_j) + ∑c_i(∑h_kB_k)$ [for some scalars $g_j, h_k$]
i.e. $\dim \langle a_1 + b_1, a_2 + b_2, …, a_n + b_n \rangle \le p + q$. Hence etc.
Case II: $p = 0$ or $q = 0$: One of the bases is empty & the corresponding matrix becomes zero. Rest follows immediately.