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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.

David
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  • Please discuss what you have tried, did you find the moore penrose inverse? Can you use that to find the other? You need to explain where you are stuck so that people can help. – EHH Mar 31 '16 at 11:49
  • What is your definition of a generalized inverse? – Ben Grossmann Mar 31 '16 at 11:49
  • I have edited my question. Kindly have a look in to that@Omnomnomnom @EHH – David Mar 31 '16 at 18:07
  • @David. What is the trick in getting the generalized inverse after reducing the matrix in echelon form? – holala Sep 23 '20 at 22:16

2 Answers2

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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.

user1551
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  • Assume we have the given matrix $A$ in David's question above so after computing for the kernel, every vector in the kernel times its transpose plus $G$ will be a generalized inverse of $A$ based on your presentation right? So how do we explicitly compute $G$ for $A$? – holala Sep 23 '20 at 22:25
  • @holala Yes. To find $G$, you may use elementary row/column operations to write $A$ in the form of $P(I_r\oplus 0_{(m-r)\times(n-r)})Q$, where $P\in GL_m$ and $Q\in GL_n$ are invertible and $r$ is the rank of $A$. Then you may take $G=Q^{-1}(I_r\oplus 0_{(n-r)\times(m-r)})P^{-1}$. – user1551 Sep 23 '20 at 22:54
  • I am not conversant with $P(I_r\oplus 0_{(m-r)\times(n-r)})Q$ Could you provide an example for easy understanding? Probably an update to your presentation. – holala Sep 24 '20 at 00:30
  • @holala If you don't know elementary row/column operations, you may also perform a rank decomposition $A=U_0V_0^T$ where $U_0$ is $m\times r$ and $V_0$ is $n\times r$. Complete $U_0$ and $V_0$ to two invertible matrices $U$ and $V$. Then $A=U(I_r\oplus0_{(m-r)\times(n-r)})V^T$. – user1551 Sep 24 '20 at 00:34
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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:

  1. $\mathbf{A} \, \mathbf{A}^{\dagger} \mathbf{A} = \mathbf{A}$
  2. $\mathbf{A}^{\dagger} \mathbf{A} \, \mathbf{A} = \mathbf{A}^{\dagger}$
  3. $\left( \mathbf{A} \, \mathbf{A}^{\dagger} \right)^{*} = \mathbf{A} \, \mathbf{A}^{\dagger}$
  4. $\left( \mathbf{A}^{\dagger} \mathbf{A} \right)^{*} = \mathbf{A}^{\dagger} \mathbf{A}$

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

dantopa
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