Calculate inverse of a non-square matrix in 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