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:
[ /(A-B/DC) -/(A-B/DC)B/D ]
[ -(D-CA'B)C/A /(D-C/AB) ]
To prove this we multiply out:
[ /(A-B/DC) -/(A-B/DC)B/D ] [ A B ]
[ -/(D-C/AB)C/A /(D-C/AB) ] [ C D ]
Which is:
[ /(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 ]
Which reduces to:
[ /(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 ]
So, clearest form for a right inverse is:
[ /(A-B/DC) -/AB/(D-C/AB) ] (Note the repeat of the diagonal elements
[ -/DC/(A-B/DC) /(D-C/AB) ] in each column)
and clearest form for a left inverse is:
[ /(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
:)
--- Or if you prefer the futz and rescale view:
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 by John Halleck, All Rights Reserved. This snapshot was last modified on February 11th, 2008