This is a math geek page...

Normally when one is dealing with inverses, one is dealing with matrices that have real numbers as their elements. This allows much simplier forms. But I work with more general fields, so the forms are a little different, and a little more complex.

This page is seriously under construction.

Here are some examples:

Using /A to mean A's inverse, binding as tighter than other operators. (So A/BC is A * inv(B) * C) Consider a two by two matrix: [ A B ] [ C D ] If we assume that the elements are themselves matrices, then it has the RIGHT inverse of: [ /(A - B/DC) -/AB/(D - C/AB) ] [ -/DC/(A - B/DC) /(D - C/AB) ] (This can be seen by inspection, multiplying it back out.) [ A B ] [ /(A - B/DC) -/AB/(D - C/AB) ] [ C D ] [ -/DC/(A - B/DC) /(D - C/AB) ] equals [ A/(A-B/DC)-B/DC/(A-B/DC) -A/AB/(D-C/AB)+B/(D-C/AB) ] [ C/(A-B/DC)-D/DC/(A-B/DC) -C/AB/(D-C/AB)+D/(D-C/AB) ] The diagonal elements: A/(A-B/DC)-B/DC/(A-B/DC) = (A-B/DC)/(A-B/DC) = I -C/AB/(D-C/AB)+D/(D-C/AB) = D/(D-C/AB)-C/AB/(D-C/AB) = (D-C/AB)/(D-C/AB) = I The off diagonal elements: -A/AB/(D-C/AB)+B/(D-C/AB) = B/(D-C/AB)-A/AB/(D-C/AB) = B/(D-C/AB)-B/(D-C/AB) = 0 C/(A-B/DC)-D/DC/(A-B/DC) = C/(A-B/DC)-C/(A-B/DC) = 0 Which is just [ I 0 ] [ 0 I ] Which proves that is the inverse.

And the LEFT inverse is:

To prove this we multiply out:[ /(A-B/DC) -/(A-B/DC)B/D ] [ -(D-CA'B)C/A /(D-C/AB) ]

[ /(A-B/DC) -/(A-B/DC)B/D ] [ A B ] [ -/(D-C/AB)C/A /(D-C/AB) ] [ C D ]

[ /(A-B/DC)A-/(A-B/DC)B/DC /(A-B/DC)B-/(A-B/DC)B/DD ] [ -/(D-C/AB)C/AA+/(A-B/DC)C -/(D-C/AB)C/AB+/(D-C/AB)D ]

[ /(A-B/DC)(A-B/DC) /(A-B/DC)B-/(A-B/DC)B ] [ -(D-C/AB)C+/(D-C/AB)C -(D-C/AB)(C/AB+D) ]

Which simply reduces to [ I 0 ] which shows it is a left inverse. [ 0 I ]

[ /(A-B/DC) -/AB/(D-C/AB) ] (Note the repeat of the diagonal elements [ -/DC/(A-B/DC) /(D-C/AB) ] in each column)

[ /(A-B/DC) -/(A-B/DC)B/D ] (Note the repeat of the diagonal elements [ -/(D-C/AB)C/A /(D-C/AB) ] in each row.)

Although they are formally the same using the (: obvious :) identities: /AB/(D-C/AB) = /(A-B/DC)B/D /DC/(A-B/DC) = /(D-C/AB)C/A :)

Right Inverse: [ I -/AB ] [ /(A-B/DC) 0 ] [ -/DC I ] [ 0 /(D-C/AB) ] Left Inverse: [ /(A-B/DC) 0 ] [ I -B/D ] [ 0 /(D-C/AB) ] [ -C/A I ]

-------------------------------------------------------- Special cases: Upper triangular: (Both left and right inverse have the same "natural" form) inv ( [ A B ] ) = [ /A -/AB/C ] [ 0 C ] [ 0 /C ] Lower triangular: inv ( [ A 0 ] ) = [ /A 0 ] [ B C ] [ -/CB/A /C ]

This page is http://www.cc.utah.edu/~nahaj/math/matrix.inverse.notes.html © Copyright 2003-2010 by John Halleck, All Rights Reserved. This snapshot was last modified on August 19th, 2011 And the underlying file was last modified on February 11th, 2008