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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.

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    $\begingroup$ If $\mathrm{rank}(B)\lt m$, one can find a vector $v\ne0$ so that $Bv=0$. $\endgroup$
    – robjohn
    Oct 8 '14 at 22:36
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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

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

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

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

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

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  • $\begingroup$ how can I use rank(B)<=m to imply that rank(B)=m ? $\endgroup$
    – Trent
    Oct 8 '14 at 22:43
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    $\begingroup$ @Trent: did you also notice he has $m\le\mathrm{rank}(B)$? I think the $\le$ is a typo; it should be $=$. $\endgroup$
    – robjohn
    Oct 8 '14 at 22:44
  • $\begingroup$ @Trent: The chain of the inequalities gives $m \leq rank(B) \leq m$, so we must have equality. $\endgroup$
    – voldemort
    Oct 8 '14 at 22:44
  • $\begingroup$ How can you say rank(AB)<=rank(B)? I thought the formula was rank(AB)<=min(rankA,rankB) $\endgroup$
    – Trent
    Oct 8 '14 at 22:45
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    $\begingroup$ @Trent: min(rank A, rank B) is definitely $\leq$ rank(B). $\endgroup$
    – voldemort
    Oct 8 '14 at 22:46
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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 $\text{rank}(B)=m$

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