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How to solve for the non trivial solution to the homogeneous system of linear equation.

I tried with solve command but it gives only trivial solutions. eigen(A)$vector[,x] gives answer only for the square matrix i.e for evendetermined system.

Bernhard Barker
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ddivya
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    Please provide a reproducible example. – Roland Aug 21 '15 at 11:06
  • my equation is 2x+3y+4z=0,x+y+z=0.I need non trivial solution how do i get it using r program.if i have one more equation i will get square matrix where entries of the matrices are coefficients of the equation . Now eigen(A) gives eigen values and corresponding eigen vector ,so the eigen value which near zero and its corresponding eigen vector form the non trivial solution to the equation. But with two equation and three unknows i cant get a square matrix so i cant find eigen values, what shall i do in this case. – ddivya Aug 22 '15 at 13:56
  • You have an under-determined equation system. You should probably use a computer algebra system (you could use the Ryacas package). – Roland Aug 22 '15 at 14:29
  • can any1 tell me how to use the Ryacas package to solve it? please – Jorge Lopez Apr 02 '19 at 08:18
  • this is my matrix M<- matrix(c(1,2,3,2,-1,-2,3,1,1,4,0,2),3) – Jorge Lopez Apr 02 '19 at 08:18
  • how do we find the non trivial solution for this homogenous system? – Jorge Lopez Apr 02 '19 at 08:18

1 Answers1

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You could use the singular value decomposition. For ddivya, M <- rbind(c(2,3,4), 1). Then do, e.g.,

s <- La.svd(M, nv=ncol(M))
(v <- s$vt[ncol(M),])

Then any multiple of v is a solution; you can verify that M %*% v is zero. Both your systems have only more column than rows and full rank (no element of s$d are zero or close). Otherwise you'd find more basis vectors spanning your space of solutions at the bottom of $s$vt$.

Quigi
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