If $A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix such that $AB=I$, prove that rank$(B)=m$

Question

If $A$ is an $m\times n$ matrix and $B$ is an $n\times m$ matrix such that $AB=I$, prove that rank$(B)=m$.

I am not sure where to begin with this proof. I have that rank$(AB) = m$, but I can't find anything to help me get further. Can anyone help me out? Thanks.

Answer 1

We know that rank($B$) $\leq m$.

Also, $m=$ rank($AB$) $\leq$ rank($B$) $\leq m$. (See here)

So, rank($B$) $= m$

Answer 2

We can see the matrix $B$ as a linear transformation

$$B:\Bbb R^m\to \Bbb R^n,\quad x\mapsto Bx$$ and recall a famous and simple result from the basic set theory:

If $f\circ g$ is injective then $g$ is injective

so from the hypothesis $AB=I$ we get that $B$ is injective and then by the rank-nullity theorem

$$\operatorname{rank}(B)=\dim\Bbb R^m=m$$

Answer 3

An other way using the robjohn hint in comments above:

By rank theorem, we have: $\text{rank}(B)+\dim (\ker B)=m$

$X \in \ker(B) \Rightarrow BX=0 $ and $BX=0 \Rightarrow (AB) X=0$ . Since $AB=I$ we have $\ker B=\{0\}$ then $\text{rank}(B)=m$