4
$\begingroup$

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

Please help me to proof (iii)!

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

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.