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!

  • $\begingroup$ What do you denote $N(A)$ and $R(B)$? $\endgroup$
    – Bernard
    Jul 4 '19 at 22:42
  • $\begingroup$ $N(A) = \mbox{nullity of A}$ and $R(B)$ is the rank of $B$, I think. $\endgroup$ Jul 4 '19 at 22:49
  • 1
    $\begingroup$ $\textsf{N}(A)$ is the nullspace of $A$ and $\textsf{R}(B)$ is the range or image of $B$. $\endgroup$
    – azif00
    Jul 4 '19 at 23:01

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.

  • $\begingroup$ 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! $\endgroup$ Jul 5 '19 at 13:15
  • $\begingroup$ 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. $\endgroup$
    – Leo
    Jul 5 '19 at 14:20

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