Real Matrix Inverses

Why

Let $A \in \R ^{m \times n}$ and define $f: \R ^n \to \R ^m$ by $f(x) = Ax$. Then $f$ is a linear function from $\R ^{n}$ to $\R ^{m}$. Conversely, suppose $g: \R ^n \to \R ^m$ is a linear function. Then there exists a matrix $B \in \R ^{m \times n}$ so that $g(z) = Bz$. Does this function have an inverse?

Derivation

If $A \in \R ^{m \times n}$, with $m \neq n$, then the inverse of $f$ can not exist. For a square matrix $A \in \R ^{n \times n}$, $B \in \R ^{n \times n}$ is a left inverse if $BA = I$. In other words, $B$ is a left inverse element of $A$ in the algebra of matrices with the operation of multiplication. $C \in \R ^{n \times n}$ is a right inverse if $AC = I$.

Definition

We call a square matrix $A$ invertible if there is $B \in \R ^{n \times n}$ so that $BA = I$.

Now suppose that $A \in \R ^{n \times n}$. Of course, the inverse may not exist. Consider, for example if $A$ was the $n$ by $n$ matrix of zeros. If there exists a matrix $B$ so that $BA = I$ we call $B$ the left inverse of $A$ and likewise if $AC = I$ we call $C$ the right inverse of $A$. In the case that $A$ is square, the right inverse and left inverse coincide.