Under what conditions does $\text{Rank}(A+B) = \text{Rank}(A)+\text{Rank}(B)$?

Question

This is part of a larger problem. I have the expression

$$E = A^{-1}B(C^{-1} + D)^{-1}A,$$

where all matrices are of appropriate dimensions, in my notes. I am told that E is of rank n. Given that all matrices are of full rank. The only remaining part of the puzzle to to see why the summation of these matrices also results in a rank $n$ matrix.

one necessary contition is that the images have trivial intersection — Asinomás, May 03 '18 at 18:34
Basically I am looking for "If (something) then rank(A+B) = ...". — , May 03 '18 at 18:39
it might be easier if you post the actual problem since people know what stuff can be assumed — Asinomás, May 03 '18 at 18:40

score 2 · Accepted Answer · answered May 03 '18 at 18:42

2

Let $\mathcal{R}_A,\mathcal{R}_B,\mathcal{C}_A,\mathcal{C}_B$ be the row spaces for $A,B$ and the column spaces for $A,B$ respectively. Then $\text{rank}(A+B) = \text{rank}(A)+\text{rank}(B)$ if and only if $\text{dim}(\mathcal{R}_A\cap \mathcal{R}_B) = \text{dim}(\mathcal{C}_A\cap \mathcal{C}_B) = 0$.

The proof for this is given here: http://www.dtic.mil/dtic/tr/fulltext/u2/600471.pdf

answered May 03 '18 at 18:42

SlipEternal

10,555
1
17
38

2

One can also look at When is "Rank" Additive? by Callan (found at https://www.maa.org/sites/default/files/0746834249131.di020781.02p03912.pdf). – Clarinetist May 03 '18 at 18:46
oh wow, nice.${}{}$ – Asinomás May 03 '18 at 18:55

Under what conditions does $\text{Rank}(A+B) = \text{Rank}(A)+\text{Rank}(B)$?

1 Answers1