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

[NaturePaper]

Hi,
Does the equation $$AA^\dagger=I$$ force $$A^\dagger$$ to be the generalized inverse of A? That is: $$AA^\dagger=I\Rightarrow A^\dagger\text{ is the generalized inverse of } A?$$ 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

 

[NaturePaper]

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" .

 

[td21]

Hi!
I think so. Actually i think we can write out what A† is.
It is A*(AA*)^(-1), and the 4th condition follows.

 

[NaturePaper]

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).

 

[blue_raver22]

.

I can confirm that the equation AA^\dagger=I does indeed force A^\dagger to be the generalized inverse of A. This is known as the Moore-Penrose inverse, named after mathematicians Eliakim Hastings Moore and Roger Penrose. The last property, known as the orthogonality property, can be verified by using the definition of the generalized inverse and substituting it into the equation AA^\dagger=I. This property is important as it ensures that the generalized inverse is unique and can be used in a wide range of applications in linear algebra and data analysis.