# 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 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 $$BA=I$$.