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?
Let the columns of $A$ and $B$ be $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ respectively. By definition, the rank of $A$ and $B$ are the dimensions of the linear spans $\langle a_1, \ldots, a_n\rangle$ and $\langle b_1, \ldots, b_n\rangle$. Now the rank of $A + B$ is the dimension of the linear span of the columns of $A + B$, i.e. the dimension of the linear span $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$. Since $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$ the result follows.
Edit: Let me elaborate on the last statement. Any vector $v$ in $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$ can be written as some linear combination $v = \lambda_1 (a_1 + b_1) + \ldots + \lambda_n (a_n + b_n)$ for some scalars $\lambda_i$. But then we can also write $v = \lambda_1 (a_1) + \ldots + \lambda_n (a_n) + \lambda_1 (b_1) + \ldots + \lambda_n (b_n)$. This implies that also $v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$. We can do this for any vector $v$, so
$$\forall v \in \langle a_1 + b_1, \ldots, a_n + b_n\rangle: v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$$
This is equivalent to saying $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$.
If $f,g:V\to W$ are linear maps, then we have $$(f+g)(V)\subset f(V)+g(V),$$ which implies $$\mathrm{rk}(f+g)=\dim\ (f+g)(V)\le\dim\ (f(V)+g(V))$$ $$\le\dim f(V)+\dim g(V)=\mathrm{rk}(f)+\mathrm{rk}(g).$$ To justify the first display, note that a vector of $W$ is in $(f+g)(V)$ if and only if it is equal to $f(v)+g(v)$ for some $v$ in $V$, whereas it is in $f(V)+g(V)$ if and only if it is equal to $f(v)+g(v')$ for some $v$ and $v'$ in $V$.
An intuitive picture:
Use the following characterisation of the rank: decompose $A$ into its component column vectors. That is, $A = (a_1, a_2, \ldots, a_n)$, where each $a_i$ is a $m\times 1$ column vector. Then the rank of $A$ is equal to the dimensional of the vector subspace generated by $a_1, \ldots, a_n$.
A vector in the image of $A+B$ is going to be a linear combination of $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$. So we have that the rank of $A+B$ is at most the size of the linear subspace generated by those $2n$ vectors.
But the size of that linear subspace is given by the maximum number of linearly independent vectors one can choose among them. We can choose at most $rank(A)$ many from the $a_i$, and at most $rank(B)$ many from the $b_i$. So this gives an upper bound of $rank(A)+rank(B)$.
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.
$\newcommand{\rank}{{\rm rank}\;{}}$We already have some very good answers to this question, but I would like to add one more using an approach based on partitioned matrices. For a partitioned matrix $$M = \begin{bmatrix}A&B \\ C&D\end{bmatrix},$$ let us define the invertible operations $${\sf rowSwap}(M) = \begin{bmatrix}C&D \\ A&B\end{bmatrix},$$ and $$ {\sf rowSynth}_{X}(M) {}={} \begin{bmatrix} A & B \\ C + XA & D+XB \end{bmatrix}. $$ It is easy to see that ${\sf rowSwap}^{-1} = {\sf rowSwap}$ and ${\sf rowSynth}_{X}^{-1} = {\sf rowSynth}_{-X}$. These operations do not affect the rank of the original matrix.
Now, we start with the observation that $$\rank \begin{bmatrix}A & 0 \\ B & B\end{bmatrix} = \rank A + \rank B.$$ Then, by applying the above elementary operations $$ \begin{bmatrix}A & 0 \\ B & B\end{bmatrix} \overset{{\sf rowSwap}}{\longrightarrow} \begin{bmatrix}B & B \\ A & 0\end{bmatrix} \overset{{\sf rowSynth}_{I}}{\longrightarrow} \begin{bmatrix}B & B \\ A+B & B\end{bmatrix}, $$ and $$ \begin{aligned} \rank A + \rank B {}={}& \rank \begin{bmatrix}A & 0 \\ B & B\end{bmatrix} \\ {}={}& \rank \begin{bmatrix}B & B \\ A+B & B\end{bmatrix} \\ {}\geq{}& \rank \begin{bmatrix}A+B & B\end{bmatrix} \\ {}\geq{}& \rank(A+B). \end{aligned} $$
There's also a proof based on the rank-nullity theorem. Let $n_1$ be the dimension of the nullspace $N(A)$ of $A$ and $n_2$ the dimension of the nullspace $N(B)$ of $B$. Let $n_3$ be the dimension of the nullspace of $A+B$.
Since $N(A) \cap N(B) \subset N(A+B)$, we have $n_1+n_2 - \text{dim}(N(A)+N(B)) \le n_3$. So $-n_3 \le -n_1 - n_2 + \text{dim}(N(A)+N(B)) \le -n_1-n_2+n$. Therefore, $n-n_3 \le n-n_1 +n-n_2$, and so by the Rank-Nullity Theorem, $\text{rank}(A+B) \le \text{rank}(A) + \text{rank}(B)$.
