Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$

Question

Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$, where $A,B \in M_{m\times n}(\mathbb{F})$.

I'm trying to think in terms of linear transformations.
We can define $T_a, T_b:\mathbb{F}^n\rightarrow \mathbb{F}^m$.
I know that $\dim_{\mathbb F}\operatorname{Im} T_a, \dim_{\mathbb F}\operatorname{Im} T_b \le m$.

What should I do next?

See also: Show that $\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$ — Martin Sleziak, Apr 22 '17 at 09:36
See also accepted answer to: https://math.stackexchange.com/questions/877802/rank-of-a-matrix-sum/877806#877806 — user126154, Jun 20 '22 at 08:51

Git Gud · Accepted Answer · 2014-06-29T20:38:17.073

25

Hint: It suffices to prove that $C(A+B)\subseteq C(A)+C(B)$, because $$\begin{align} C(A+B)\subseteq C(A)+C(B) &\implies \dim \left(C(A+B)\right)\leq \dim \left(C(A)+C(B)\right)\\ &\implies \text{rank}(A+B) \leq \dim(C(A))+\dim(C(B))-\dim(C(A)\cap C(B)).\end{align}$$

edited Jun 29 '14 at 20:38

answered Jun 29 '14 at 20:16

Git Gud

31,356

Hmm.. the columns of $C(A+B)$ are linear combination of the columns of $C$. Is that what the hint leading to? – AnnieOK Jun 29 '14 at 20:32
@AnnieOK $C(X)$ denotes the column space (or range). – Git Gud Jun 29 '14 at 20:33
Maybe: Since every $C(A+B)_i$ is a linear combination of $C(A)_i, C(B)_i$ then we can conclude $C(A+B) \subseteq C(A) + C(B)$? Is that right? – AnnieOK Jun 29 '14 at 20:39
7

@AnnieOK That proves the inclusion, yes. But I'd rather do it like this: let $X\in C(A+B)$, then $X=(A+B)x$, for some $x$, thus $X=\underbrace{Ax}{\in C(A)}+\underbrace{Bx}{\in C(B)}$. – Git Gud Jun 29 '14 at 20:41
Great, I get it now :) – AnnieOK Jun 29 '14 at 20:44
Comment, down voter? – Git Gud Jul 04 '14 at 20:28
@GitGud in your comment why X=(A+B)x for some x? – gbox Aug 20 '14 at 20:35
4

@gbox It's the definition of $C(M)$ for a $p\times q$ matrix $M$, more precisely $C(M):={Mx\colon x\in \mathbb R^{q\times 1}}$. – Git Gud Aug 20 '14 at 20:55
@GitGud shouldn't it be xM? thanks – gbox Aug 20 '14 at 21:06
@gbox No. – Git Gud Aug 20 '14 at 21:08
@GitGud sorry you are right, my bad – gbox Aug 20 '14 at 21:09
1

Nice solution +1! – RFZ Nov 25 '19 at 03:07

score 7 · Answer 2 · answered Jun 29 '14 at 20:21

7

How about this: Let $rk A: =a$ , $rk B:=b$. Now we can do Gaussian elimination on both, to end with two matrices $A'$, $B'$ with , respectively, $a$ and $b$ non-zero rows. But the sum $A'+B'$ will have at most max{a,b} nonzero rows. Then $a+b \geq$ Max{$a,b$}

answered Jun 29 '14 at 20:21

user99680

6,708

I like this solution, though it somewhat technical/arithmetical. – AnnieOK Jun 29 '14 at 20:33
Thanks; do you want to suggest changes? – user99680 Jun 29 '14 at 20:37
Oops.. auto-correct made it arithmetical. I meant "algorithmical" (If that's a word). The proof is $100$ percent fine, I just want to understand it through another perspective. – AnnieOK Jun 29 '14 at 20:42
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But if you are doing Gaussian elimination on both of them independently, you can't really just add them like that in the end. You have $A'+B' = E_1 A E_1^{-1} + E_2 B E_2^{-1}$, which does not really say much about $A+B$. – Ondrej Draganov Mar 25 '19 at 18:53
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$rank(A+B)$ is equal to $rank((A+B)')$, but not necessarily equal to $rank(A'+B')$ – SOFe May 17 '19 at 04:35
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I was thinking exactly along your lines (because it applies to the product AB), but then I wondered why is this constraint so loose? Why not say $rank(A+B) \le max(rank(A), rank(B))$, instead? Then I thought of a counterexample. Let A = [1 0; 2 0] and B = [0 4; 0 3]. A + B = [1 4; 2 3] (Julia notation). Then the ranks of A and B are 1, but the rank(A+B) = 2. So there is some faulty logic in this proof. – kfmfe04 Nov 28 '21 at 19:12

Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$

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