# 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