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Re: sparse symmetric matrices

From: Bill Greene
Subject: Re: sparse symmetric matrices
Date: Sat, 11 Jun 2011 15:38:59 -0400
User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv: Gecko/20110303 Thunderbird/3.1.9

I guess I should have mentioned that I did try:

u = ks \ b'

This returns an incorrect solution, u.
I also tried to let chol() compute a permutation
[lt, err, p] = chol(ks, 'vector');
and this results in an incorrect solution, also.

I'm guessing all of these errors are related: I'm
creating only the upper triangle and, when permuted,
entries end up in both triangles.

I'm hoping someone has a good strategy for dealing with


On 6/11/2011 1:48 PM, David Bateman wrote:
On 06/11/2011 07:11 PM, Bill Greene wrote:
I'm trying to solve a linear system with a sparse,
symmetric matrix and have some questions I'm hoping
someone can help me with. I'm using Octave-3.2.4,
by the way.

I'm calling chol() directly and providing only entries
in the upper triangle of the matrix. This works fine
even though the Octave doc does not say specifically that
only the upper triangle is needed.

However, if I first call amd() to order the matrix, I get
a permutation that puts the entries in both upper and lower
triangles so chol() fails.

My questions are:

1) Is there a fundamentally better way to deal with this type
    of system (I strongly prefer storing only 1/2 the matrix)?
2) Is there a method (with good performance) that will
    rearrange my permuted sparse matrix to move all the lower
    entries to the upper triangle?

Here is a code sample that works:
K = [[0,0,2]; [0,1,-2]; [0,4, -1];
       [1,1,3]; [1,2,-2];
       [2,2,5]; [2,3,-3]; [3,3,10]; [3,4,4]; [4,4,10]];
  for i = 1:2
    K(:,i) += 1;
  ks = sparse(K(:,1), K(:,2), K(:,3));
  b = [0,1,0,0,0];
  lt = chol(ks);
  u = lt \ (lt'\b')

But if I replace the relevant lines with:
p = amd(ks)
lt = chol(ks(p,p));
x = lt \ (lt'\b(p)');
u = x(p);

I run into the problem described above.


Bill Greene
Why not just call

x = ks \ b;

and let Octave handle the recognition that the matrix is positive
definite, the permutation, the cholesky and the back substitution. If
the time needed to recognize that the matrix is positive definite is too
long. you can call matrix_type on ks before solving the problem to
bypass the automatic recognition of the matrix type


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