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. -- View this message in context: http://r.789695.n4.nabble.com/Finding-solution-set-of-system-of-linear-equations-tp3541490p3541490.html Sent from the R help mailing list archive at Nabble.com.
Robert A LaBudde
2011-May-22 03:02 UTC
[R] Finding solution set of system of linear equations.
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.
>
>
>
>
>--
>View this message in context:
>http://r.789695.n4.nabble.com/Finding-solution-set-of-system-of-linear-equations-tp3541490p3541490.html
>Sent from the R help mailing list archive at Nabble.com.
>
>______________________________________________
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>and provide commented, minimal, self-contained, reproducible code.
===============================================================Robert A.
LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral at lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
Thanks Robert. That all seems to work. I also found the MASS::Null() function that gives the null space for the matrix(transpose) given as argument. I am still trying to appreciate the math behind the Moore-Penrose inverse matrix. If you have any suggestions for understanding how to use R to solve these kinds of systems, please send me the pointer. I think the R documentation for solve() should state right up front that it only solves non-singular systems, and it should point me to MASS::ginv() and how to use it to solve this obvious kind of problem. Better, there could be a generalized solve() which produces a particular solution, the null space, image space (basis thereof). This other solution (using ginv() and giA%*%A - I for the kernel) seems deeply embedded in a particular solution technique (and should be available), but the generalized solve_lin_sys(), as I suggested, seems generally quite useful.. Don -- View this message in context: http://r.789695.n4.nabble.com/Finding-solution-set-of-system-of-linear-equations-tp3541490p3542930.html Sent from the R help mailing list archive at Nabble.com.