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Generalized inverse

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.

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)

## [1] 0

The rank is 2, so inv(A) won’t work

## [1] 2

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

##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    0    0    1
## [3,]    0    0    0

inv() will throw an error

## Error in Inverse(X, tol = sqrt(.Machine$double.eps), ...) : 
##   X is numerically singular

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.

##         [,1] [,2]    [,3]
## [1,] 0.27778    0 0.05556
## [2,] 0.00000    0 0.00000
## [3,] 0.05556    0 0.11111

We can also view this as fractions:

##      [,1] [,2] [,3]
## [1,] 5/18    0 1/18
## [2,]    0    0    0
## [3,] 1/18    0  1/9

The generalized inverse is defined as the matrix \(A^-\) such that

• \(A * A^- * A = A\) and
• \(A^- * A * A^- = A^-\)

##      [,1] [,2] [,3]
## [1,]    4    4   -2
## [2,]    4    4   -2
## [3,]   -2   -2   10

##         [,1] [,2]    [,3]
## [1,] 0.27778    0 0.05556
## [2,] 0.00000    0 0.00000
## [3,] 0.05556    0 0.11111

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

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    1    0    0
## [3,]    0    0    1

##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    0    0    0
## [3,]    0    0    1

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

##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    1    1    0
## [3,]    1    0    1
## [4,]    1    0    1

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

## [1] 2

##      [,1] [,2] [,3]
## [1,]    4    2    2
## [2,]    2    2    0
## [3,]    2    0    2

##      [,1] [,2] [,3]
## [1,]  0.5 -0.5    0
## [2,] -0.5  1.0    0
## [3,]  0.0  0.0    0

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

Show that it is a generalized inverse:

##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    1    1    0
## [3,]    1    0    1
## [4,]    1    0    1

##      [,1] [,2] [,3] [,4]
## [1,]  0.0  0.0  0.5  0.5
## [2,]  0.5  0.5 -0.5 -0.5
## [3,]  0.0  0.0  0.0  0.0

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