Let $\mathbf{A}$ be an $m \times n$ matrix. A matrix $\mathbf{G}$ of order $n \times m$ is said to be a generalized inverse (or a g-inverse) of $\mathbf{A}$ if $\mathbf{A G A}=\mathbf{A}$.

If $\mathbf{A}$ is square and nonsingular, then $\mathbf{A}^{-\mathbf{1}}$ is the unique g-inverse of $\mathbf{A}$. Otherwise, A has infinitely many g-inverses, as we will see shortly.

1.1. Let $\mathbf{A}, \mathbf{G}$ be matrices of order $m \times n$ and $n \times m$ respectively. Then the following conditions are equivalent:
(i) $\mathbf{G}$ is a g-inverse of $\mathbf{A}$.
(ii) For any $\mathbf{y} \in \mathcal{C}(\mathbf{A}), \mathbf{x}=\mathbf{G} \mathbf{y}$ is a solution of $\mathbf{A} \mathbf{x}=\mathbf{y}$.

ProoF. (i) $\Rightarrow$ (ii). Any $\mathbf{y} \in \mathcal{C}(\mathbf{A})$ is of the form $\mathbf{y}=\mathbf{A z}$ for some $\mathbf{z}$. Then $\mathbf{A}(\mathbf{G y})=\mathbf{A G A z}=\mathbf{A z}=\mathbf{y}$
(ii) $\Rightarrow$ (i). Since $\mathbf{A G y}=\mathbf{y}$ for any $\mathbf{y} \in \mathcal{C}(\mathbf{A})$ we have $\mathbf{A G A z}=\mathbf{A z}$ for all $\mathbf{z}$. In particular, if we let $\mathbf{z}$ be the $i$ th column of the identity matrix, then we see that the $i$ th columns of AGA and $\mathbf{A}$ are identical. Therefore, $\mathbf{A G A}=\mathbf{A}$.

Let $\mathbf{A}=\mathbf{B C}$ be a rank factorization. We have seen that $\mathbf{B}$ admits a left inverse $\mathbf{B}{\ell}^{-}$, and $\mathbf{C}$ admits a right inverse $\mathbf{C}{\mathbf{r}}^{-} .$Then $\mathbf{G}=\mathbf{C}{\mathbf{r}}^{-} \mathbf{B}{\ell}^{-}$is a g-inverse of $\mathbf{A}$, since
$$
\mathbf{A G A}=\mathbf{B C}\left(\mathbf{C}{\mathbf{r}}^{-} \mathbf{B}{\ell}^{-}\right) \mathbf{B C}=\mathbf{B C}=\mathbf{A}
$$

Alternatively, if $\mathbf{A}$ has rank $r$, then by $7.3$ of Chapter 1 there exist nonsingular matrices $\mathbf{P}, \mathbf{Q}$ such that
$$
\mathbf{A}=\mathbf{P}\left[\begin{array}{cc}
\mathbf{I}{r} & \mathbf{0} \ \mathbf{0} & \mathbf{0} \end{array}\right] \mathbf{Q} $$ It can be verified that for $a n y \mathbf{U}, \mathbf{V}, \mathbf{W}$ of appropriate dimensions, $$ \left[\begin{array}{cc} \mathbf{I}{r} & \mathbf{U} \
\mathbf{V} & \mathbf{W}
\end{array}\right]
$$
is a g-inverse of
$$
\left[\begin{array}{ll}
\mathbf{I}{r} & \mathbf{0} \ \mathbf{0} & \mathbf{0} \end{array}\right] $$ Then $$ \mathbf{G}=\mathbf{Q}^{-1}\left[\begin{array}{cc} \mathbf{I}{r} & \mathbf{U} \
\mathbf{V} & \mathbf{W}
\end{array}\right] \mathbf{P}^{-1}
$$
is a g-inverse of $\mathbf{A}$. This also shows that any matrix that is not a square nonsingular matrix admits infinitely many g-inverses.

Another method that is particularly suitable for computing a g-inverse is as follows. Let $\mathbf{A}$ be of rank $r$. Choose any $r \times r$ nonsingular submatrix of $\mathbf{A}$. For convenience let us assume
$$
\mathbf{A}=\left[\begin{array}{ll}
\mathbf{A}{11} & \mathbf{A}{12} \
\mathbf{A}{21} & \mathbf{A}{22}
\end{array}\right]
$$