I am aware that very similar questions have been answered in various approaches previously. But I'm looking for a proof through a very specific approach.
$\exists$ left inverse of a square matrix $\Rightarrow$ $\exists$ right inverse? – is very relavent to mine, but has an approach involving concepts that I haven't covered yet.
The statement in the title has been given as a corollary to the following theorem in Linear Algebra by Hoffman and Kunze.
Theorem: For an $n\times n$ matrix $A$, the following are equivalent.
(1) $A$ is invertible.
(2) The homogeneous system $AX=0$ has only the trivial solution $X=0$.
(3) The system of equations $AX=Y$ has a solution $X$ for each $n\times 1$ matrix $Y$.
How does the title statement follow from the above theorem? I am looking for an argument that does not involve the concepts of rank, row/column spaces, linear maps, determinants and adjoints.
I have two approaches in mind:
(1) I am willing to use the fact that if $A$ has a left inverse $B$ and a right inverse $C$, then $B=C$. So, I realize it is enough to show that a square matrix that has a left inverse also has a right inverse.
(2) Stage 1: As done in the text, argue that if $A$ has a left inverse, $A$ is invertible (I want an argument that uses the above mentioned theorem). Stage 2: Then I can argue as follows: Suppose $C$ is a right inverse for $A$. Then $C$ has a left inverse. Therefore, $C$ is invertible (from stage 1). Now since $A=C^{-1}$, it follows that $C$ is the inverse of $A$ showing $A$ is invertible.
A square matrix with either a left or right inverse is invertible.
linear algebramatrices
Best Answer
Theo Bendit has already answered your question in the comments. I am just writing this to expand on his answer. This post is made community wiki.
Note that if $A$ has a left inverse, then the second statement in the question is satisfied. Namely, the statement that $AX=0$ if and only if $X=0$. To prove this, simply multiply both sides of the equation $AX=0$ by the left inverse $B$. Then we have that $(BA)X=B0\implies X=0$. Thus $AX=0$ implies $X=0$. Clearly $X=0$ is a solution, so it is the unique solution. Because this is equivalent to the statement that $A$ is invertible, there exists a right inverse of $A$ as well.
Now suppose that $A$ has a right inverse $B$. Then the third statement is satisfied. Namely, for each $n\times 1$ matrix $Y$, there exists an $n\times 1$ matrix $X$ such that $AX=Y$. To prove this, let $Y$ be an $n\times 1$ matrix. Then $Y=(AB)Y=A(BY)$. Thus, if we let $X=BY$, then we have that $AX=Y$. Since this statement is equivalent to the statement that $A$ is invertible, there exists a left inverse of $A$.
I hope this helps!