# What 's the short proof that for square matrices $AB = I$ implies $BA = I$? [duplicate]

Possible Duplicate:
If $AB = I$ then $BA = I$

I'm trying to remember the one line proof that for square matrices $AB = I$ implies $BA = I$. I think it uses only elementary matrix properties and nothing else. Does anyone know the proof?

I remember it beginning with $BAB = B$ and the result following almost immediately.

## marked as duplicate by user17762, Blue, Jonas Meyer, Andrés E. Caicedo, Dylan MorelandJun 1 '12 at 23:02

The short proof probably goes along the following lines. If a matrix is right-invertible then it is left-invertible (this is the difficult part), or more generally, row rank is equal to column rank. Suppose $AB = I$. Then $A$ has a left inverse $C$, and $C = C(AB) = (CA)B = B$.
For the difficult property, write $A$ in tensor form $A = \sum_{i,j} A_{ij} x_i y_j$, and show that the row rank and the column rank are both equal to the tensor rank, which is the smallest number of rank one tensors that add up to $A$ (a rank one tensor is of the form $\sum_{i,j} a_i b_j x_i y_j$).
Another short proof involves the fact that a linear transformation from a finite dimensional vector space to itself is surjective if and only if it is bijective if and only if it is injective. $AB=I$ amounts to saying $A$ is surjective, hence it is bijective and has a left inverse $C$, so that $CA=I$. Of course, $B=(CA)B=C(AB)=C$, i.e. $BA=I$.