Solving non-square linear system with R

Question

How to solve a non-square linear system with R : A X = B ?

(in the case the system has no solution or infinitely many solutions)

Example :

Answer 1

If the matrix A has more rows than columns, then you should use least squares fit.

If the matrix A has fewer rows than columns, then you should perform singular value decomposition. Each algorithm does the best it can to give you a solution by using assumptions.

Here's a link that shows how to use SVD as a solver:

http://www.ecse.rpi.edu/~qji/CV/svd_review.pdf

Let's apply it to your problem and see if it works:

Your input matrix A and known RHS vector B:

> A=matrix(c(0,1,-2,3,5,-3,1,-2,5,-2,-1,1),3,4,T)
> B=matrix(c(-17,28,11),3,1,T)
> A
     [,1] [,2] [,3] [,4]
[1,]    0    1   -2    3
[2,]    5   -3    1   -2
[3,]    5   -2   -1    1
> B
     [,1]
[1,]  -17
[2,]   28
[3,]   11

Let's decompose your A matrix:

> asvd = svd(A)
> asvd
$d
[1] 8.007081e+00 4.459446e+00 4.022656e-16

$u
           [,1]       [,2]       [,3]
[1,] -0.1295469 -0.8061540  0.5773503
[2,]  0.7629233  0.2908861  0.5773503
[3,]  0.6333764 -0.5152679 -0.5773503

$v
            [,1]       [,2]       [,3]
[1,]  0.87191556 -0.2515803 -0.1764323
[2,] -0.46022634 -0.1453716 -0.4694190
[3,]  0.04853711  0.5423235  0.6394484
[4,] -0.15999723 -0.7883272  0.5827720

> adiag = diag(1/asvd$d)
> adiag
          [,1]      [,2]        [,3]
[1,] 0.1248895 0.0000000 0.00000e+00
[2,] 0.0000000 0.2242431 0.00000e+00
[3,] 0.0000000 0.0000000 2.48592e+15

Here's the key: the third eigenvalue in d is very small; conversely, the diagonal element in adiag is very large. Before solving, set it equal to zero:

> adiag[3,3] = 0
> adiag
          [,1]      [,2] [,3]
[1,] 0.1248895 0.0000000    0
[2,] 0.0000000 0.2242431    0
[3,] 0.0000000 0.0000000    0

Now let's compute the solution (see slide 16 in the link I gave you above):

> solution = asvd$v %*% adiag %*% t(asvd$u) %*% B
> solution
          [,1]
[1,]  2.411765
[2,] -2.282353
[3,]  2.152941
[4,] -3.470588

Now that we have a solution, let's substitute it back to see if it gives us the same B:

> check = A %*% solution
> check
     [,1]
[1,]  -17
[2,]   28
[3,]   11

That's the B side you started with, so I think we're good.

Here's another nice SVD discussion from AMS:

http://www.ams.org/samplings/feature-column/fcarc-svd

Answer 2

Aim is to solve Ax = b

where A is p by q, x is q by 1 and b is p by 1 for x given A and b.

Approach 1: Generalized Inverse: Moore-Penrose https://en.wikipedia.org/wiki/Generalized_inverse

Multiplying both sides of the equation, we get

A'Ax = A' b

where A' is the transpose of A. Note that A'A is q by q matrix now. One way to solve this now multiply both sides of the equation by the inverse of A'A. Which gives,

x = (A'A)^{-1} A' b

This is the theory behind generalized inverse. Here G = (A'A)^{-1} A' is pseudo-inverse of A.

library(MASS)

ginv(A) %*% B

#          [,1]
#[1,]  2.411765
#[2,] -2.282353
#[3,]  2.152941
#[4,] -3.470588

Approach 2: Generalized Inverse using SVD.

@duffymo used SVD to obtain a pseudoinverse of A.

However, last elements of svd(A)$d may not be quite as small as in this example. So, probably one shouldn't use that method as is. Here's an example where none of the last elements of A is close to zero.

A <- as.matrix(iris[11:13, -5])    
A
#   Sepal.Length Sepal.Width Petal.Length Petal.Width
# 11          5.4         3.7          1.5         0.2
# 12          4.8         3.4          1.6         0.2
# 13          4.8         3.0          1.4         0.1

svd(A)$d
# [1] 10.7820526  0.2630862  0.1677126

One option would be to look as the singular values in cor(A)

svd(cor(A))$d
# [1] 2.904194e+00 1.095806e+00 1.876146e-16 1.155796e-17

Now, it is clear there is only two large singular values are present. So, one now can apply svd on A to get pseudo-inverse as below.

svda <- svd(A)
G = svda$v[, 1:2] %*% diag(1/svda$d[1:2]) %*% t(svda$u[, 1:2])
# to get x
G %*% B