How to prove that a matrix is invertible $\iff$ invertible at right $\iff$ invertible at left? Let $A$ a $n\times n$ invertible at left. In fact, I just want to prove that it's invertible at right (the rest is obvious). All what I can say is that there is a $B$ s.t. $BA=I.$
To prova $AB=I$, I have problem. I have that $$AB=AB^2A=BA^2B,$$
but I can't conclude that it's $I$.
 A: Assuming $A\in \mathbb{R}^{n\times n}$, let $f_A : \mathbb{R}^n \to \mathbb{R}^n, x\mapsto Ax$ be the associated linear map. Since $BA = I$, we have $f_B\circ f_A = \operatorname{id}$, so $f_A$ is injective. But then $f_A$, as a linear map between finite-dimensional vector spaces of same dimension, is also surjective, hence bijective and so $A$ is invertible.
This is because $n = \operatorname{dim}(\ker f_A) + \operatorname{dim}(\operatorname{im} f_A)$. Since $f_A$ is injective, $\operatorname{dim}(\ker f_A) = 0$, so $\operatorname{dim}(\operatorname{im} f_A) = n$, hence $f_A$ is surjective.
A: If you can try $B = \text{adj}(A)/\det(A)$ and show by brute force computation that $BA = I = AB$.
You would only need $B^jA_i = \delta_{ij}$ and $A^iB_j = \delta_{ij}$. (Superscript means row vector and subscript meaning column vector index, and $\delta_{ij}$ means indicator of $i = j$).  
You only need to think about the relationship between the definitions of $\text{adj}$ and $\det$. See https://en.wikipedia.org/wiki/Adjugate_matrix
If additionally you know how transpose relates to inverse, I think you only need $B^jA_i = \delta_{ij}$.
A: I'll assume that you have the result that $\det AB=\det A \det B$ and that the determinant of a matrix is the product of its eigenvalues.
Suppose $BA=I$ and that $AB\neq I$. Then there exists $u$ such that $ABu=v$ for some $v\neq u$ and so $BABu=Bv$ hence $Bu=Bv$. This gives us $B(u-v)=0$ and so $u-v$ is an eigenvector associated to the eigenvalue $0$.
It follows that $\det AB =(\det A) (\det B) = (\det A) 0 =0$. But $\det AB = \det I =1$ which is a contradiction.
It follows that if $BA=I$ then $AB=I$.
