I am interested in a proof of the inequality $$\operatorname{rank} (A+B) \leq \operatorname{rank} (A) + \operatorname{rank} (B)$$ that does not use the bases of $A$'s and $B$'s row/column spaces, as described in other answers on this site. My idea is to use rank-$1$ matrices.
So, the proof is: write the matrices $A$ and $B$ as sums of matrices of rank 1. It is known that any matrix $M$ can be written as a sum of at least $rank(M)$ matrices of rank 1. Thus, the decomposition of $A+B$ will have at least $rank(A)+rank(B)$ matrix terms. This decomposition is also a decomposition of $A+B$. Now it's straightforward to see that $rank(A+B) \leq rank(A) + rank(B)$, QED.
Is such a proof valid?
Best Answer
It's valid, it just needs some work to be more strict.
The sentence "any matrix $M$ can be written as a sum of at least $rank(M)$ matrices of rank $1$" is slightly misleading, as in fact, you know (and use) two things about sums of rank $1$ matrices:
In your proof, you first write $A$ and $B$ as sums of $rank(A)$ and $rank(B)$ rank-one matrices, i.e. you use point 1 above. So you write
$$A=A_1 + A_2 +\cdots + A_{rank(A)} \\B=B_1 + B_2 +\cdots + B_{rank(B)}$$
From there, we can write the matrix $A+B$ as $$M=A+B=A_1 + A_2 +\cdots + A_{rank(A)} +B_1 + B_2 +\cdots + B_{rank(B)}$$ and you can now use point 2 above to conclude that, since the sum above is a sum of $rank(A)+rank(B)$ rank-one matrices (i.e. $n=rank(A)+rank(B)$), that therefore, $$rank(A)+rank(B)=n\geq rank(M)=rank(A+B).$$