Rank of Products of Matrices

Question

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!

$N(A) = \mbox{nullity of A}$ and $R(B)$ is the rank of $B$, I think. — Allan Ramos, Jul 04 '19 at 22:49
$\textsf{N}(A)$ is the nullspace of $A$ and $\textsf{R}(B)$ is the range or image of $B$. — azif00, Jul 04 '19 at 23:01

Leo · Answer 1 · 2019-07-05T14:17:38.547

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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.

edited Jul 05 '19 at 14:17

answered Jul 04 '19 at 23:57

Leo

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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! – is it normal Jul 05 '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 05 '19 at 14:20

Rank of Products of Matrices

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