7

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

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

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!

Rodrigo de Azevedo
  • 23,223
is it normal
  • 422

2 Answers2

0

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.

Leo
  • 871
  • 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
0

Suppose $A\in\mathbb{R}^{p\times n}, B\in\mathbb{R}^{n\times q}$. Firstly, note that $$\{Bx~|~ABx=0\}=\mathrm{N}(A)\cap\mathrm{R}(B).$$ The left side can also be regarded as the image of $B$ which is restricted to $\mathrm{N}(AB)$. Denote the restriction of $B$ to $\mathrm{N}(AB)$, that is $B|_{\mathrm{N}(AB)}$, by $\hat{B}$, and then $\mathrm{R}(\hat{B})=\mathrm{N}(A)\cap\mathrm{R}(B)$. According to the rank-nullity theorem, one obtains \begin{align*} \text{dim}(\mathrm{R}(\hat{B})) &= \text{dim}(\mathrm{N}(AB))-\text{dim}(\mathrm{N}(\hat{B}))\\ &=\text{dim}(\mathrm{N}(AB))-\text{dim}(\mathrm{N}(B))\\ &=(q-\text{rank}(AB))-(q-\text{rank}(B))\\ &=\text{rank}(B)-\text{rank}(AB), \end{align*} where the above second equality results from the fact that $$ \mathrm{N}(\hat{B})=\{x\in\mathrm{N}(AB)~|~Bx=0\}=\{x~|~Bx=0\}=\mathrm{N}(B). $$ For a reference, one can refer to Page 133 in Matrix algebra and its applications to statistics and econometrics.

W.J
  • 305