Prove if matrix has right inverse then also has left inverse. I tried to prove that if $A$ and $B$ are both $n\times n$ matrices and if $AB = I_n$ then $BA = I_n$ (i.e. the matrix $A$ is invertible). So first I managed to conclude that if exists both $B$ and $C$ such that $AB = I_n$ and $CA = I_n$, then trivially $B=C$ . However to conclude the proof we need to show that if such a right inverse exists, then a left inverse must exist too. 
No idea how to proceed. All I can use is definition of matrices, and matrix multiplication, sum , transpose and rank.
(I saw proof of this in other questions, but they used things like determinants or vectorial spaces, but I need a proof without that).
 A: A matrix $A\in M_n(\mathbb{F})$ has a right inverse $B$ (which means $AB=I$) if and only if it has rank $n$. I assume you know that. So now you need to prove that $BA=I$. Well, let's multiply the equation $AB=I$ by $A$ from the right side. We get $A(BA)=A$ and hence $A(BA-I)=0$. Well, now we can split the matrix $BA-I$ into columns. Let's call its columns $v_1,v_2,...,v_n$ and so this way we get $Av_1=0,Av_2=0,...,Av_n=0$. But because the rank of $A$ is $n$ we know that the system $Ax=0$ can have only the trivial solution. Hence $v_1=v_2=...=v_n=0$, so $BA-I$ is the zero matrix and hence $BA=I$. 
A: Let B be a right inverse of A, then AB=I. The columns of I are in the column space of A (ColA). Since Ix=0 only if x=0, the column vectors of I are linearly independent and form a basis of R^n as there are n of them.
This means that R^n = colA. Since there are n column vectors of A that span R^n, they must be independent hence
Av=0 implies v=0.
For all vectors x in R^n, BAx=y, a vector in R^n. Therefore, A(BA) x=
(AB) Ax=Ax=Ay, so A(x-y) =0 implies x=y.
Hence BA=I.
B is also unique for if AB=AC=I, then BAB=BAC implies B=C
