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 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