$\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B)$

Question

Let $A, B ∈ M_{m×n}(F)$. Could someone give a hint as to how to prove that $$\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B).$$

possible duplicate of Rank of the difference of matrices. Actually Prove that $\def\rk{\operatorname {rank}}\rk (A)+\rk(B)\geq\rk(A+B)$ is better as duplicate — Marc van Leeuwen, May 09 '13 at 17:24

Federica Maggioni · Answer 1 · 2013-05-09T07:22:47.977

5

You can view it in terms of linear maps associated to matrices. Let $f_A, f_B$ the linear maps associated to $A$ and $B$ in some fixed basis. Then, it suffices to prove that $$\operatorname{Im}(f_A+f_B)\subseteq \operatorname{Im}(f_A)+\operatorname{Im}(f_B)$$ but this is clearly true, since

$$(f_A+f_B)(x)=f_A(x)+f_B(x)\in\operatorname{Im}(f_A)+\operatorname{Im}(f_B)$$ for every vector $x$.

edited May 09 '13 at 07:22

answered May 09 '13 at 07:08

Federica Maggioni

8,632

What is $x$? How are the linear maps associated? – Yosef Qian May 09 '13 at 17:52
@YosefQian $x\in F^n$ and $f_A(x)=A\cdot x$ (matrix product) – Federica Maggioni May 09 '13 at 17:54

score 4 · Answer 2 · answered May 09 '13 at 07:03

4

Hints:

$rank(A)=\dim\,im(A)=\dim\{Ax\,\mid\,x\in F^n\}$
$im(A+B)\subseteq im(A)+im(B)$
$\dim(U+V)\le\dim U+\dim V$.

answered May 09 '13 at 07:03

Berci

92,013

1

How do you come to action 3.)? – Yosef Qian May 09 '13 at 17:38

$\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B)$

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