find two different generalized inverse of the given matrix Definition:
For a given matrix $A_{m\times n}$, a matrix $G_{n\times m}$ is said to be a generalized inverse of $A$, if it satisfies
$$AGA=A.$$

Question:
Find two different generalized inverse of the given matrix
$$\begin{pmatrix}
 1 & 0 &-1 & 2\\2 & 0 &-2 & 4 \\-1 & 1 & 1 & 3\\
-2 & 2 & 2 & 6
 \end{pmatrix}$$
Work done:
Since the echelon form of the matrix is,
$$
\left(\begin{array}{rrrr}
1 & 0 & -1 & 2 \\
0 & 1 & 0 & 5 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$$ rank is 2.
since there are two distinct $2\times 2$ minors, 
one of the generalized inverse is, 
$$\left(\begin{array}{rrrr}
0 & 0 & 0 & 0 \\
\frac 1 2  &0 & 0 & 0 \\
\frac 1 2 &  1  & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)$$
and the other one is,
$$\left(\begin{array}{rrrr}
0 & 0 & 0 & 0 \\
0 &0 & \frac 3 {10} & -\frac 4{10} \\
0&  0  & \frac 1 {10} & \frac 2 {10} \\
0 & 0 & 0 & 0
\end{array}\right)$$
Luckily we get two different solutions,

But if the question is to find 5 different generalized inverses, How to do that?

As we know there are plenty of generalized inverses are there for this given matrix, different possible ways are welcome.
Thanks in advance.
 A: If $AGA=A$, then $A(G+uv^T)A=A$ if $u\in\ker A$ or $v\in\ker A^T$. Note that when $A$ is not a nonsingular matrix (this includes the case where $A$ is not square), at least one of $A$ or $A^T$ has a nonzero nullspace. Therefore, if you can find one generalised inverse of $A$, you can find infinitely many others if the field is infinite.
By the way, the two matrices that you claim to be generalised inverses of your example $A$ do not seem to be correct.
A: The matrix and its Moore-Penrose pseudoinverse are
$$
\mathbf{A} = 
\left[
\begin{array}{rrrr}
 1 & 0 & -1 & 2 \\
 2 & 0 & -2 & 4 \\
 -1 & 1 & 1 & 3 \\
 -2 & 2 & 2 & 6 \\
\end{array}
\right],
\qquad
\mathbf{A}^{\dagger} = \frac{1}{40}
\left[
\begin{array}{rrrr}
 8 & 16 & -5 & -10 \\
 -2 & -4 & 3 & 6 \\
 -8 & -16 & 5 & 10 \\
 6 & 12 & 5 & 10 \\
\end{array}
\right].
$$
The pseudoinverse satisfies all four requirements: 


*

*$\mathbf{A} \, \mathbf{A}^{\dagger} \mathbf{A} = \mathbf{A}$

*$\mathbf{A}^{\dagger} \mathbf{A} \, \mathbf{A} = \mathbf{A}^{\dagger}$

*$\left( \mathbf{A} \, \mathbf{A}^{\dagger} \right)^{*} = \mathbf{A} \, \mathbf{A}^{\dagger}$

*$\left( \mathbf{A}^{\dagger} \mathbf{A} \right)^{*} = \mathbf{A}^{\dagger} \mathbf{A}$


The inverses presented in the question are problematic:
$$
\mathbf{G}_{1} = \left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 \frac{1}{2} & 0 & 0 & 0 \\
 \frac{1}{2} & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
 \mathbf{A} \, \mathbf{G}_{1} \mathbf{A} = 
\left(
\begin{array}{rrrr}
 -\frac{5}{2} & 0 & \frac{5}{2} & -5 \\
 -5 & 0 & 5 & -10 \\
 3 & 0 & -3 & 6 \\
 6 & 0 & -6 & 12 \\
\end{array}
\right)
\ne \mathbf{A}
$$
$$
\mathbf{G}_{2} = \left(
\begin{array}{cccr}
 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{3}{10} & -\frac{2}{5} \\
 0 & 0 & \frac{1}{10} & \frac{1}{10} \\
 0 & 0 & 0 & 0 \\
\end{array}
\right), \qquad
%
 \mathbf{A} \, \mathbf{G}_{2} \mathbf{A} = 
\left(
\begin{array}{rrrr}
 \frac{3}{10} & -\frac{3}{10} & -\frac{3}{10} &
   -\frac{9}{10} \\
 \frac{3}{5} & -\frac{3}{5} & -\frac{3}{5} & -\frac{9}{5}
   \\
 \frac{1}{5} & -\frac{1}{5} & -\frac{1}{5} & -\frac{3}{5}
   \\
 \frac{2}{5} & -\frac{2}{5} & -\frac{2}{5} & -\frac{6}{5}
\end{array}
\right)
\ne \mathbf{A}
$$
