Does AA^\dagger=I Imply A^\dagger is the Generalized Inverse of A?

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Discussion Overview

The discussion centers on whether the equation \( AA^\dagger = I \) implies that \( A^\dagger \) is the generalized inverse of \( A \). The context involves properties of rectangular matrices over the field of complex numbers and the implications of the Moore-Penrose pseudoinverse.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if \( AA^\dagger = I \) necessitates that \( A^\dagger \) is the generalized inverse of \( A \).
  • Another participant asserts that the answer is affirmative and references a Wikipedia page for support.
  • A different participant proposes a specific formulation for \( A^\dagger \) as \( A^*(AA^*)^{-1} \) and suggests that the fourth condition of the generalized inverse follows from this.
  • In contrast, a later reply argues against the implication, stating that the conditions in the Penrose definition require both \( AX \) and \( XA \) to be Hermitian, which does not trivially follow from the initial equation.

Areas of Agreement / Disagreement

Participants express differing views on whether \( A^\dagger \) must be the generalized inverse of \( A \) given \( AA^\dagger = I \). No consensus is reached, and multiple competing interpretations are presented.

Contextual Notes

The discussion highlights potential ambiguities in the definitions and properties of generalized inverses, particularly regarding the conditions required for \( A^\dagger \) to qualify as such.

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Hi,
Does the equation [tex]AA^\dagger=I[/tex] force [tex]A^\dagger[/tex] to be the generalized inverse of A? That is: [tex]AA^\dagger=I\Rightarrow A^\dagger\text{ is the generalized inverse of } A?[/tex] A is any rectangular matrix over the field of complex numbers. It is very easy to verify the first three properties, but I'm not sure about the last one
 
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It looks the answer is yes. The result can be found in http://en.wikipedia.org/wiki/Proofs_involving_the_Moore%E2%80%93Penrose_pseudoinverse" .
 
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Hi!
I think so. Actually i think we can write out what A† is.
It is A*(AA*)^(-1), and the 4th condition follows.
 
No.
In my question A\dagger was conjugate transpose and conditions 3 and 4 in Penrose's definition are just the requirement that AX and XA both should be hermitian, where X is the generalized inverse of A (in my case which is A\dagger and so 3-4 follows trivially).
 

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