If $AB=I$ show that $A$ has full rank. Let $A,B\in \mathbb R^{n\times n}$. Show that if $AB=I$ then $B$ has full rank.



In fact, I show that $A$ has full rank, which is quite obvious but I really have difficulties to show that $B$ has full rank. I tried by contradiction : there is $u\in \mathbb R^n$ s.t. $Bv\neq u$ for all $v\in \mathbb R^n$. So $v=ABv\neq Au$ which contradict $A$ has full rank. 

It doesn't look correct. What do you think ?
There is a missing thing. How do you know that $Bv\neq u\implies ABv\neq Au$? At this point, you need to use the fact that $A\text{ surjective}\implies A\text{ injective}$. Then, since $AB=\operatorname{Id}$, your argument is correct.
$AB = I$ means $B = A^-1$, so $A$ is invertible. A matrix is only invertible if it is full rank, otherwise when you are in the process of inverting it, you will eventually find a row or colum of zeros.