Condition for right handed invertibility Suppose that $A$ is an m by n matrix and is right invertible, such that there exists and an n by m matrix $B$ such that $AB = I_m.$ Prove that $m\leq n.$ 
I'm not really sure how go about this problem; any help would be appreciated.
 A: Here's a hint: Show that for every $\mathbf y\in \Bbb R^m$, you can solve the equation $A\mathbf x=\mathbf y$. This means there are no inconsistent systems, and so the rank of $A$ must be $m$.
A: Suppose $m>n$. Complete $A$ with rows of zeros to get an $m\times m$ matrix called $A_0$. Complete $B$ with columns of zeros to get another $m\times m$ matrix $B_0$.
Then $A_0B_0=I_m$, but this is a contradiction because $\det A_0=0$, that is, $A_0$ is not invertible.
A: View $A$ as a map from an $n$-dimensional vector space to an $m$-dimensional vector space. The first thing to notice is that $A$ must be surjective, for $I_{m}$ is clearly surjective onto its image, and had $A$ not been, then $I_{m}$ couldn't have been either.
But if $A$ is surjective, then the dimension of the space of the image must be less than or equal to the dimension of the space of the source. Thus $m\leq n$. To further clarify this point: $n=rk(A)+nl(A)=m+nl(A)\geq m$. The rank of $A$ is $m$ since it is surjective, as mentioned above.
