Question

I have written a linear solver employing Householder reflections/transformations in ANSI C which solves Ax=b given A and b. I want to use it to find the eigenvector associated with an eigenvalue, like this:

(A-lambda*I)x = 0

The problem is that the 0 vector is always the solution that I get (before someone says it, yes I have the correct eigenvalue with 100% certainty).

Here’s an example which pretty accurately illustrates the issue:

Given A-lambda*I (example just happens to be Hermitian):

1 2 0 | 0
2 1 4 | 0
0 4 1 | 0

Householder reflections/transformation will yield something like this

# # # | 0
0 # # | 0
0 0 # | 0

Back substitution will find that solution is {0,0,0}, obviously.

Answer 1

It’s been a while since I’ve written an eigensolver, but I seem to recall that the trick was to refactor it from (A - lambda*I) * x = 0 to A*x = lambda*x. Then your Householder or Givens steps will give you something like:

# # # | #
0 # # | #
0 0 1 | 1

…from which you can back substitute without reaching the degenerate 0 vector. Usually you’ll want to deliver x in normalized form as well.

My memory is quite rusty here, so I’d recommend checking Golub & Van Loan for the definitive answer. There are quite a few tricks involved in getting this to work robustly, particularly for the non-symmetric case.