# If $AB = I$, the identity matrix prove $\mathrm{rank}(B)$

Let $A$ be an $m \times n$ matrix and $B$ be an $n \times m$ matrix. Show that if $AB = I$, where $I$ is the identity matrix, then $\mathrm{rank}(B) = m$.

I'm not exactly sure how to start this problem. I know that rank is related to the RREF of a matrix but I'm unsure as to how I can relate that to this problem..

• – robjohn
Oct 11 '14 at 1:51

Rank(AB) <= min( Rank(A), Rank(B) ). If rank(B) were not $m$ then Rank(AB) would be less than m, but the rank of $I$ is m. So this would be a contradiction.
Note: here I'm assuming that $m<n$.
• You don't need to assume $m \lt n$. Your argument is $m=$ Rank$(AB) \le$ Rank$(B)$, but also Rank$(B) \le m$ by counting the columns of $B$. Oct 11 '14 at 0:49
Let $L_A, L_B$ the associated linear transformation of each matrix. So $L_AL_B=\text{id}$, from here follows that $L_B$ is injective, i.e., $\ker( L_B)=\ker{B}=\{0\}$. So $m=\text{rank}{B}+\text{dim}(kerB)=\text{rank}{B}$.