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Suppose the equations $Ax=b$ are compatible, and the Moore-Penrose inverse of $A$ is known as $A^{\dagger}$. In order to calculate the Minimum Norm Solution $$x = A^{(1,4)}b$$ I take the advantage of $A^{\dagger} \in A\{1,4\}$, and thus, the Minimum Norm Solution becomes$$x = A^{\dagger}b$$

Furthurmore, if the equations $Ax=b$ are NOT compatible, I calculate the Minimum Norm Least Square Solution by $$x = A^{(1,3)}b= A^{\dagger}b$$ as also $A^{\dagger} \in A\{1,3\}$.

Do you agree these two replacements?

$\bf{Appendix}$

$\begin{equation} \mathbf{AXA} = \mathbf{A} \space \space \space \space \space \space \space \space (1) \\ \mathbf{XAX} = \mathbf{X} \space \space \space \space \space \space \space \space (2) \\ \mathbf{(AX)^H} = \mathbf{AX} \space \space \space \space \space \space \space \space (3) \\ \mathbf{(XA)^H} = \mathbf{XA} \space \space \space \space \space \space \space \space (4) \end{equation}$

$\mathbf{X}$ is called the Moore-Penrose inverse of $\mathbf{A}$

{1,3}-inverses include the matrices $\mathbf{X}$ which satisfy the equations (1) and (3) {1,4}-inverses include the matrices $\mathbf{X}$ which satisfy the equations (1) and (4)

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  • $\begingroup$ Why does nobody answer me? Didn't I make it clear? $\endgroup$ – John-Annual Feb 8 '14 at 13:03
  • $\begingroup$ Can you provide an explanation or link to the $\{1,3\}$- and $\{1,4\}$-inverses? $\endgroup$ – Roland Feb 8 '14 at 13:16
  • $\begingroup$ @Roland Sorry I forgot to mention the definition. I put it in the text. $\endgroup$ – John-Annual Feb 17 '14 at 12:51
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The question is unclear.

The pseudoinverse matrix delivered by the singular value decomposition satisfies all $4$ properties.

How are you proposing this change be made? Where does your matrix come from?

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