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.
$\endgroup$ 42 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$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: $$ \mathbf{G}_{1} = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ \frac{1}{2} & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right), \qquad % \mathbf{A} \, \mathbf{G}_{1} \mathbf{A} = \left( \begin{array}{rrrr} -\frac{5}{2} & 0 & \frac{5}{2} & -5 \\ -5 & 0 & 5 & -10 \\ 3 & 0 & -3 & 6 \\ 6 & 0 & -6 & 12 \\ \end{array} \right) \ne \mathbf{A} $$ $$ \mathbf{G}_{2} = \left( \begin{array}{cccr} 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{10} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{10} & \frac{1}{10} \\ 0 & 0 & 0 & 0 \\ \end{array} \right), \qquad % \mathbf{A} \, \mathbf{G}_{2} \mathbf{A} = \left( \begin{array}{rrrr} \frac{3}{10} & -\frac{3}{10} & -\frac{3}{10} & -\frac{9}{10} \\ \frac{3}{5} & -\frac{3}{5} & -\frac{3}{5} & -\frac{9}{5} \\ \frac{1}{5} & -\frac{1}{5} & -\frac{1}{5} & -\frac{3}{5} \\ \frac{2}{5} & -\frac{2}{5} & -\frac{2}{5} & -\frac{6}{5} \end{array} \right) \ne \mathbf{A} $$
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