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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]
## [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
## [,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]
## [,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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