# Generalized inverse

```
knitr::opts_chunk$set(
warning = FALSE,
message = FALSE,
fig.height = 5,
fig.width = 5
)
options(digits=4)
par(mar=c(5,4,1,1)+.1)
```

In matrix algebra, the inverse of a matrix is defined only for *square* matrices,
and if a matrix is *singular*, it does not have an inverse.

The **generalized inverse** (or *pseudoinverse*)
is an extension of the idea of a matrix inverse,
which has some but not all the properties of an ordinary inverse.

A common use of the pseudoinverse is to compute a 'best fit' (least squares) solution to a system of linear equations that lacks a unique solution.

`library(matlib)`

Construct a square, *singular* matrix [See: Timm, EX. 1.7.3]

```
A <-matrix(c(4, 4, -2,
4, 4, -2,
-2, -2, 10), nrow=3, ncol=3, byrow=TRUE)
det(A)
```

The rank is 2, so `inv(A)`

wont work

`R(A)`

In the echelon form, this rank deficiency appears as the final row of zeros

`echelon(A)`

`inv()`

will throw an error

`try(inv(A))`

A **generalized inverse** does exist for any matrix,
but unlike the ordinary inverse, the generalized inverse is not unique, in the
sense that there are various ways of defining a generalized inverse with
various inverse-like properties. The function `matlib::Ginv()`

calculates
a *Moore-Penrose* generalized inverse.

`(AI <- Ginv(A))`

We can also view this as fractions:

`Ginv(A, fractions=TRUE)`

### Properties of generalized inverse (Moore-Penrose inverse)

The generalized inverse is defined as the matrix $A^-$ such that

- $A
*A^-*A = A$ and - $A^-
*A*A^- = A^-$

```
A %*% AI %*% A
AI %*% A %*% AI
```

In addition, both $A * A^-$ and $A^- * A$ are symmetric, but
neither product gives an identity matrix, `A %*% AI != AI %*% A != I`

```
zapsmall(A %*% AI)
zapsmall(AI %*% A)
```

## Rectangular matrices

For a *rectangular matrix*, $A^- = (A^{T} A)^{-1} A^{T}$
is the generalized inverse of $A$
if $(A^{T} A)^-$ is the ginv of $(A^{T} A)$ [See: TIMM: EX 1.6.11]

```
A <- cbind( 1, matrix(c(1, 0, 1, 0, 0, 1, 0, 1), nrow=4, byrow=TRUE))
A
```

This $4 \times 3$ matrix is not of full rank, because columns 2 and 3 sum to column 1.

```
R(A)
(AA <- t(A) %*% A)
(AAI <- Ginv(AA))
```

The generalized inverse of $A$ is $(A^{T} A)^- A^{T}$, `AAI * t(A)`

`AI <- AAI %*% t(A)`

Show that it is a generalized inverse:

```
A %*% AI %*% A
AI %*% A %*% AI
```