Numerical Linear Algebra - Finding the eigenvector associated with a known eigenvalue 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.
 A: It looks to me you're not doing inverse iteration properly. Usually, one already has a pivoted(!) LU decomposition of the matrix $\mathbf A-\lambda\mathbf I=\mathbf P^\top\mathbf L\mathbf U$ already available when doing this; I'll assume that as a given. (I am not assuming that your matrix is Hermitian/symmetric; one can definitely do better than this general method I am about to describe in that case.)
The traditional way of doing inverse iteration, due to Jim Wilkinson, builds up an initial eigenvector estimate by solving the system $\mathbf U\mathbf x_0=\mathbf e$, where $\mathbf e=(1 \dots 1)^\top$. In this operation, it is assumed that any zero diagonal elements in $\mathbf U$ have already been replaced by quantities around the size of $\|\mathbf A\|\varepsilon$, where $\|\cdot\|$ is any convenient norm and $\varepsilon$ is machine epsilon. One then proceeds with the cycle of solving $(\mathbf A-\lambda\mathbf I)\mathbf x_{k+1}=\mathbf x_k$ (using the LU decomposition, of course!) and then normalizing the $\mathbf x_{k+1}$ so obtained with some convenient vector norm. In practice, the method usually converges within two to three iterations, and a fourth iteration is rarely needed, if ever.
One might think it is perverse to be solving a system whose underlying matrix is nearly singular; it turns out that the growth of the unnormalized $x_k$ is in fact vital for the method to work in inexact arithmetic. See Ilse Ipsen's article on this method for details.
A: I have coded this before, but cannot comment on posts yet, so must 'answer'. If you are trying to do QR factorization with the Householder reflections, I think you might have something wrong in your code. Find a problem you can solve using them by hand first. Mathematica uses these reflections in QRDecomposition, so you might be able to use it to check your work as well. I hope that you get things working soon.
