Calculate inverse of a non-square matrix in R - r

Question

I'm pretty new to the R language and trying to find out how you can calculate the inverse of a matrix that isn't square. (non-square? irregular? I'm unsure of the correct terminology).
From my book and a quick google search, (see source), I've found you can use solve(a) to find the inverse of a matrix if a is square.
The matrix I have created is, and from what I understand, not square:
> matX<-matrix(c(rep(1, 8),2,3,4,0,6,4,3,7,-2,-4,3,5,7,8,9,11), nrow=8, ncol=3);
> matX
[,1] [,2] [,3]
[1,] 1 2 -2
[2,] 1 3 -4
[3,] 1 4 3
[4,] 1 0 5
[5,] 1 6 7
[6,] 1 4 8
[7,] 1 3 9
[8,] 1 7 11
>
Is there a function for solving a matrix of this size or will I have to do something to each element? as the solve() function gives this error:
Error in solve.default(matX) : 'a' (8 x 3) must be square
The calculation I'm trying to achieve from the above matrix is: (matX'matX)^-1
Thanks in advance.

Answer 1

ginv ginv in the MASS package will give the generalized inverse of a matrix. Premultiplying the original matrix by it will give the identity:
library(MASS)
inv <- ginv(matX)
# test it out
inv %*% matX
## [,1] [,2] [,3]
## [1,] 1.000000e+00 6.661338e-16 4.440892e-15
## [2,] -8.326673e-17 1.000000e+00 -1.110223e-15
## [3,] 6.938894e-17 8.326673e-17 1.000000e+00
As suggested in the comments this can be displayed in a nicer way using zapsmall:
zapsmall(inv %*% matX)
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
The inverse of matX'matX is now:
tcrossprod(inv)
## [,1] [,2] [,3]
## [1,] 0.513763935 -0.104219636 -0.002371406
## [2,] -0.104219636 0.038700372 -0.007798748
## [3,] -0.002371406 -0.007798748 0.006625269
solve however, if you aim is to calculate the inverse of matX'matX you don't need it in the first place. This does not involve MASS:
solve(crossprod(matX))
## [,1] [,2] [,3]
## [1,] 0.513763935 -0.104219636 -0.002371406
## [2,] -0.104219636 0.038700372 -0.007798748
## [3,] -0.002371406 -0.007798748 0.006625269
svd The svd could also be used and similarly does not require MASS:
with(svd(matX), v %*% diag(1/d^2) %*% t(v))
## [,1] [,2] [,3]
## [1,] 0.513763935 -0.104219636 -0.002371406
## [2,] -0.104219636 0.038700372 -0.007798748
## [3,] -0.002371406 -0.007798748 0.006625269
ADDED some additional info.

Answer 2

You can do what's called a "Moore–Penrose pseudoinverse". Here's a function exp.matthat will do this for you. There is also an example outlining it's use here.
exp.mat():
#The exp.mat function performs can calculate the pseudoinverse of a matrix (EXP=-1)
#and other exponents of matrices, such as square roots (EXP=0.5) or square root of
#its inverse (EXP=-0.5).
#The function arguments are a matrix (MAT), an exponent (EXP), and a tolerance
#level for non-zero singular values.
exp.mat<-function(MAT, EXP, tol=NULL){
MAT <- as.matrix(MAT)
matdim <- dim(MAT)
if(is.null(tol)){
tol=min(1e-7, .Machine$double.eps*max(matdim)*max(MAT))
}
if(matdim[1]>=matdim[2]){
svd1 <- svd(MAT)
keep <- which(svd1$d > tol)
res <- t(svd1$u[,keep]%*%diag(svd1$d[keep]^EXP, nrow=length(keep))%*%t(svd1$v[,keep]))
}
if(matdim[1]<matdim[2]){
svd1 <- svd(t(MAT))
keep <- which(svd1$d > tol)
res <- svd1$u[,keep]%*%diag(svd1$d[keep]^EXP, nrow=length(keep))%*%t(svd1$v[,keep])
}
return(res)
}
example of use:
source("exp.mat.R")
X <- matrix(c(rep(1, 8),2,3,4,0,6,4,3,7,-2,-4,3,5,7,8,9,11), nrow=8, ncol=3)
iX <- exp.mat(X, -1)
zapsmall(iX %*% X) # results in identity matrix
[,1] [,2] [,3]
[1,] 1 0 0
[2,] 0 1 0
[3,] 0 0 1