$\exists$ left inverse of a square matrix $\Rightarrow$ $\exists$ right inverse? I have proved that if a square $n$ x $n$ matrix $A$ has a right and left inverse, then these are equal and form an inverse matrix of $A$.
However I'm interested in the following implication:
Suppose a left inverse $B$ of a square $n$ x $n$ matrix $A$ exist. Does this imply that a right inverse $C$ of $A$ exist ?
Also, if this is true - is the implication also true in the case of a right inverse $B$ ?
Thanks.
 A: Yes, If $A$ has a left-inverse $B$, then $BA = I$, and so $A$ is injective (as a linear map). But it is a linear operator on a finite dimensional space, so it is also surjective, and so it has a right-inverse.
A: Suppose that the matrix $A$ has a left inverse $B$, such that $BA=I$. The determinant of a matrix product is the product of the factors' determinants, and so
$\left\lvert BA \right\rvert=\left\lvert B \right\rvert \left\lvert A \right\rvert=\left\lvert I \right\rvert=1$.
Thus,both the inverse and the original matrix must be non-singular (in the sense of having non-null determinant). Since $A$ is non-singular, we can build
$$D=\frac{1}{\left\lvert A \right\rvert}A^{adj}$$
where $A^{adj}$ is the adjoint (transposed) matrix or cofactors of $A$. It's not difficult to check that $A D=I$. So, $A$ has a right inverse given by $D$.
Now, assume $A$ has a right inverse $R$, so that $AR=I$. Again, by properties of the determinant of a matrix product, it must be the case that both $A$ and $R$ are non-singular. Thus, we can find the right-inverse $R^{-1}$ for $R$ as we found $D$ before. Then,
$$A R =I\\
R A R =R I=R\\
R A R R^{-1}=R R^{-1}=I\\
R A=I
$$
and so, $R$ is also a left inverse.
