solve() only works for nonsingular systems of equations.

Use a generalized inverse for singular systems:

> A<- matrix(c(1,2,1,1, 3,0,0,4, 1,-4,-2,-2, 0,0,0,0), ncol=4, byrow=TRUE)
> A
     [,1] [,2] [,3] [,4]
[1,]    1    2    1    1
[2,]    3    0    0    4
[3,]    1   -4   -2   -2
[4,]    0    0    0    0
> b<- c(0,2,2,0)  #rhs
> b
[1] 0 2 2 0
>
> require('MASS')
> giA<- ginv(A) #M-P generalized inverse
> giA
           [,1]          [,2]        [,3] [,4]
[1,]  0.6666667  1.431553e-16  0.33333333    0
[2,]  0.3333333 -1.000000e-01 -0.03333333    0
[3,]  0.1666667 -5.000000e-02 -0.01666667    0
[4,] -0.5000000  2.500000e-01 -0.25000000    0
>
> require('Matrix')
> I<- as.matrix(Diagonal(4))  #order 4 identity matrix
> I
     [,1] [,2] [,3] [,4]
[1,]    1    0    0    0
[2,]    0    1    0    0
[3,]    0    0    1    0
[4,]    0    0    0    1
>
> giA%*%b   #particular solution
              [,1]
[1,]  6.666667e-01
[2,] -2.666667e-01
[3,] -1.333333e-01
[4,] -2.220446e-16
> giA%*%A - I   #matrix for parametric homogeneous solution
              [,1] [,2] [,3]          [,4]
[1,]  0.000000e+00  0.0  0.0  5.551115e-16
[2,]  3.469447e-17 -0.2  0.4  4.024558e-16
[3,]  4.510281e-17  0.4 -0.8  2.706169e-16
[4,] -3.330669e-16  0.0  0.0 -7.771561e-16


At 09:34 PM 5/21/2011, dslowik wrote:
I have a simple system of linear equations to solve for X, aX=b:
> a
     [,1] [,2] [,3] [,4]
[1,]    1    2    1    1
[2,]    3    0    0    4
[3,]    1   -4   -2   -2
[4,]    0    0    0    0
> b
     [,1]
[1,]    0
[2,]    2
[3,]    2
[4,]    0

(This is ex Ch1, 2.2 of Artin, Algebra).
So, 3 eqs in 4 unknowns. One can easily use row-reductions to find a
homogeneous solution(b=0) of:
X_1 = 0, X_2 = -c/2, X_3 = c,  X_4 = 0

and solutions of the above system are:
X_1 = 2/3, X_2 = -1/3-c/2,  X_3 = c, X_4 = 0.

So the Kernel is 1-D spanned by X_2 = -X_3 /2, (nulliity=1), rank is 3.

In R I use solve():
> solve(a,b)
Error in solve.default(a, b) :
  Lapack routine dgesv: system is exactly singular

and it gives the error that the system is exactly singular, since it seems
to be trying to invert a.
So my question is:
Can R only solve non-singular linear systems? If not, what routine should I
be using? If so, why? It seems that it would be simple and useful enough to
have a routine which, given a system as above, returns the null-space
(kernel) and the particular solution.




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