Right inverse = Moore-Penrose inverse?

Let $X=A^*(AA^*)^{-1}$ right inverse for $A$

Definition Let $A$ be an $m \times n$ matrix. Matrix $X$ called pseudo-inverse or generalized inverse of $A$ if and only if $X$ satisfying following properties:
i. $AXA=A$
ii. $XAX=X$
iii. $(AX)^*=AX$
iv. $(XA)^*=XA$
with $A^*$ is transpose of $A$ and $X$ called Moore-Penrose inverse of $A$ if and only if $X$ satisfying all properties.

is right inverse satisfying all properties?
Proof:
Let $X=A^*(AA^*)^{-1}$ then

(iv)
$(XA)^*=(A^*(AA^*)^{-1}A)^*$
$=A^*((AA^*)^{-1})^*A$
$=A^*((AA^*)^*)^{-1}A$
$=A^*(AA^*)^{-1}A$
$=XA$

Next proof (iii)
$(AX)^*=(AA^*(AA^*)^{-1})^*$
$=((AA^*)^{-1})^*AA^*$
$=((AA^*)^*)^{-1}AA^*$
$=(AA^*)^{-1}AA^*$
I stuck on there

(ii)
$XAX=A^*(AA^*)^{-1}AA^*(AA^*)^{-1}$
$A^*(AA^*)^{-1}AA^*(AA^*)^{-1}=A^*((AA^*)^{-1}AA^*)(AA^*)^{-1}$
$=A^*(AA^*)^{-1}$
$=X$

(i)
$AXA=AA^*(AA^*)^{-1}A$
$=(AA^*(AA^*)^{-1})A$
$=A$

• You could notice that $M^{-1}M = I = MM^{-1}$, then run the previous lines in reverse with the order switched – Nick Alger Dec 8 '16 at 0:00
• Yeah, you right. I forget this one, haha. Thank's – Kris Dec 8 '16 at 0:04

The $m\times n$ matrix $A$ has a right inverse if and only if it has rank $m$. In this case $AA^*$ also has rank $m$, so it is invertible and $A^*(AA^*)^{-1}$ is a right inverse of $A$.

In this particular case it is indeed the Moore-Penrose pseudoinverse of $A$.

Properties 1, 2 and 3 hold for every right inverse of $A$. Indeed, if $AR=I$, then $ARA=IA=A$ and $RAR=RI=R$.

Property 3 is obvious: $AR=I$, so $(AR)^*=I=AR$.

Property 4 is specific of $X$: write $B=(AA^*)^{-1}$, so $X=A^*B$: $$(XA)^*=(A^*BA)^*=A^*B^*A$$ Since $AA^*$ is symmetric (or Hermitian if we're dealing with complex numbers and $A^*$ means the conjugate transpose), also its inverse is, so $B^*=B$. Hence $$(XA)^*=(A^*BA)^*=A^*B^*A=A^*BA=XA$$

A dual result holds when $A$ has a left inverse; in this case $A^*A$ is invertible and the Moore-Penrose pseudoinverse of $A$ is $(A^*A)^{-1}A^*$.

This is important because if $A$ is a generic matrix and $A=LR$ is a full rank decomposition, that is, $L$ has a left inverse and $R$ has a right inverse (both with the same rank of $A$), then the pseudoinverse $A^+$ of $A$ is $A^+=R^+L^+$, with the pseudoinverses of $L$ and $R$ computed as before.