From: T.Moore@massey.ac.nz (Terry Moore)
Newsgroups: sci.math
Subject: Re: Definition of Generalized Inverse of a Matrix ?
Date: Tue, 04 Nov 1997 13:26:32 +1200

In article <63k95d$neu@mcmail.CIS.McMaster.CA>,
kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) wrote:

> In article <39fdbb.62836e@mizar.ursa.fi>,
> Paul T. Karch  <Paul.T..Karch@mizar.ursa.fi> wrote:
> >From: karch@nauticom.net (Paul T. Karch)
> >
> >Hello all.
> >
> >I can't find the definition of a generalized matrix inverse in my books and I
> >don't have access to a math library.  So my question is: what is its
definition
> >?
> 
> A matrix X is said to be a generalized inverse to a matrix A if the
> following two equations hold:
> 
>   A * X * A = A    and    X * A * X = X
> 
> A matrix can have many generalized inverses. If, in addition, two more
> equations hold:

There are other definitions. As someone has already posted, the
weakest definition is A * X * A = A. This is sufficient for
solving linear equations. To solve Az = b, if it is consistent,
use z = Xb. This is a solution because, if there is z' such that Az' = b,
then Az = AXb = AXAz' = Az' = b.
You can prove consistency by showing that z is a solution, i.e.
show that AXb = b.
[The solution is not unique unless A has full column rank in which case
X is a left inverse (XA = I).]
The general solution is z = Xb + (I - XA)u where u is arbitrary.
As AXb = b you just need to show that A(I - XA)u = 0 which is
obvious, and that there are no other solutions, which isn't.
You have to show that for any z' with Az' = b, there is a
solution, u', to z' = Xb + (I - XA)u'. To do this find a
generalised inverse of I - XA and check consistency as
above. [I - XA is its own generalised inverse].

Among the extra conditions sometimes used, some make
the solution Xb shortest, some give a closest approximation
to the solution of inconsistent equations [there could be
more than one], and some do both.

-- 

Terry Moore, Statistics Department, Massey University, New Zealand.

Theorems! I need theorems. Give me the theorems and I shall find the
proofs easily enough. Bernard Riemann