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).