R help - Apr 2012 - convex nonnegative basis vectors in nullspace of matrix

Dear all,

I want to explore the nullspace of a matrix S: I currently use the function
Null from the MASS package to get a basis for the null
space:
> S  = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S
> MASS::Null(t(S))
My problem is that I actually need a nonnegative basis for the null space of
S.
There should be a unique set of convex basis vectors spanning a vector space
in which each vector v satisfies sum (S %*%  v) == 0 and min(v)>=0. 
Is there maybe an R function that can calculate that for me?
I would appreciate any help, 
Thanks in advance,

Hannes


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On Wed, Apr 11, 2012 at 06:04:28AM -0700, capy_bara
wrote:
> Dear all,
> 
> I want to explore the nullspace of a matrix S: I currently use the function
> Null from the MASS package to get a basis for the null space:
> > S  = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S
> > MASS::Null(t(S))
> My problem is that I actually need a nonnegative basis for the null space
of
> S.
> There should be a unique set of convex basis vectors spanning a vector
space
> in which each vector v satisfies sum (S %*%  v) == 0 and min(v)>=0. 
Hi.

The null space of the above matrix has dimension 2. Its intersection
with nonnegative vectors in R^5 is an infinite cone. In order to restrict
it to a finite set, we can consider its intersection with the set
of vectors with the sum of coordinates equal to 1. Then, the solution
is a finite convex polytop and we can search for its vertices. The
following code searches for vertices in random directions and finds two
vertices. In this simple case, the polytop is in fact a line segment,
so we get its endpoints. These endpoints form a linear basis of the
original null space consisting of nonnegative vectors.

  library(lpSolve)
  S <- matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1))
  a <- MASS::Null(t(S))
  n <- nrow(a)
  a1 <- rbind(a, colSums(a))
  b <- rep(0, times=n+1)
  b[n+1] <- 1
  dir <- c(rep(">=", times=n), "==")
  sol <- matrix(nrow=100, ncol=n)
  for (i in seq.int(length=nrow(sol))) {
      crit <- rnorm(ncol(a))
      out <- lp(objective.in=crit, const.mat=a1, const.dir=dir, const.rhs=b)
      sol[i, ] <- a %*% out$solution
  }
  unique(round(sol, digits=10))

            [,1]      [,2]      [,3]      [,4]      [,5]
  [1,] 0.2500000 0.2500000 0.0000000 0.2500000 0.2500000
  [2,] 0.1642631 0.3357369 0.1714738 0.1642631 0.1642631

Use this with care, since for more complex cases, this method does not
guarantee that all vertices are found. So, it is not guaranteed that
every nonnegative vector in the null space is a nonnegative combination
of the obtained vectors.

Hope this helps.

Petr Savicky.

On Wed, Apr 11, 2012 at 06:04:28AM -0700, capy_bara
wrote:
> Dear all,
> 
> I want to explore the nullspace of a matrix S: I currently use the function
> Null from the MASS package to get a basis for the null space:
> > S  = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S
> > MASS::Null(t(S))
> My problem is that I actually need a nonnegative basis for the null space
of
> S.
> There should be a unique set of convex basis vectors spanning a vector
space
> in which each vector v satisfies sum (S %*%  v) == 0 and min(v)>=0. 
Hi.

In my previous solution, i forgot that lp() assumes all variables
nonnegative. So, the code was searching only a subset of the true
set of solutions. A better alternative is as follows.

  library(lpSolve)
  S <- matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1))
  a <- MASS::Null(t(S))
  a1 <- cbind(a, -a)
  n <- nrow(a1)
  a2 <- rbind(a1, colSums(a1))
  b <- rep(0, times=n+1)
  b[n+1] <- 1
  dir <- c(rep(">=", times=n), "==")
  sol <- matrix(nrow=100, ncol=n)
  for (i in seq.int(length=nrow(sol))) {
      crit <- rnorm(ncol(a))
      crit <- c(crit, -crit)
      out <- lp(objective.in=crit, const.mat=a2, const.dir=dir, const.rhs=b)
      sol[i, ] <- a1 %*% out$solution
  }
  unique(round(sol, digits=10))

       [,1] [,2] [,3] [,4] [,5]
  [1,] 0.00 0.50  0.5 0.00 0.00
  [2,] 0.25 0.25  0.0 0.25 0.25

Petr Savicky.