Prove $\operatorname{rank}(AB) = \operatorname{rank}(B)$

Question

I am given the info that A is an invertible m by m matrix and B is a m by n matrix. By solving for the null space $ABx=0$, I get $A^{-1}ABx=A^{-1}0$, therefore $Bx=0$. I concluded that the null spaces of A and AB must be equivalent, and therefore the nullity(AB) = nullity(B) = some number p. Therefore rank(AB) = n-p = rank(B) (because both AB and B are m by n matrices). Is this proof valid, specifically the null space step?

Answer 1

The proof is incomplete. You've essentially justified $\text{Null}(AB) \subseteq \text{Null}(B)$, but you haven't proven the reverse containment $\text{Null}(B) \subseteq \text{Null}(AB)$. If $Bx = 0$, then $ABx = A0 = 0$. Hence $x \in \text{Null}(B)$ $\implies$ $x \in \text{Null}(AB)$. Thus $\text{Null}(B) \subseteq \text{Null}(AB)$, and given that you justified $\text{Null}(AB) \subseteq \text{Null}(B)$, consequently we have $\text{Null}(AB) = \text{Null}(B)$. So $\text{dim}\, \text{Null}(AB) = \text{dim}\, \text{Null}(B)$, i.e., $\text{rank}(AB) = \text{rank}(B)$.