# Partitioned matrix inverse notes

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.

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  ]
```

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