# Rank of Products of Matrices

This is somewhat of a reference request.

In several posts on the rank of products of matrices (e.g. Full-rank condition for product of two matrices), it is stated that

$$\mathrm{rank}(AB) = \mathrm{rank}(B) - \dim \big(\mathrm{N}(A) \cap \mathrm{R}(B)\big)$$

It appears that this is a classic result, though I am not familiar with it. If anyone can point me to a textbook that discusses it and other rank inequalities, that would be much appreciated!

• What do you denote $N(A)$ and $R(B)$? Jul 4 '19 at 22:42
• $N(A) = \mbox{nullity of A}$ and $R(B)$ is the rank of $B$, I think. Jul 4 '19 at 22:49
• $\textsf{N}(A)$ is the nullspace of $A$ and $\textsf{R}(B)$ is the range or image of $B$. Jul 4 '19 at 23:01

## 1 Answer

Suppose there exists a $$v$$ with $$B u = v$$ and $$A v = 0$$. What is $$AB u$$? Can you take it from there?

EDIT: typo.

• Thanks! I guess you mean $Av = 0$. It looks like the result comes from counting bases, but I did linear algebra a long time ago and am quite rusty. If you know of a good reference book that would be very helpful! Jul 5 '19 at 13:15
• Every vector that is in both $N(A)$ and $R(B)$ will decrease the rank of $AB$ by one. The rank of $AB$ cannot be larger than $B$ in the first place, so you get the desired result. I'm sorry I don't know a direct reference for this fact. I'd be surprised if you didn't find it in any (vector space-based) linear algebra textbook.
– Leo
Jul 5 '19 at 14:20