Eigenvalue Deflation (Wielandt or Hotelling) I am doing a project on eigenvalue deflation techniques and I wanted to include some examples of deflation giving poor results (results with high accumulated error). 
Ideally the examples would be able to be fixed using the shifted inverse power method. If you can help me by telling me what sort of matrix might have a high accumulated error, or by giving an actual example, that would be absolutely wonderful! 
Thank you in advance for your help (or for simply reading this post)!
 A: Let your matrix be $A = \lambda u u^T + \mu v v^T$, and assume that $\lambda \gg \mu$. That is, assume your matrix is rank 2, where the only two nonzero eigenvalues are $\lambda$ and $mu$, and one is much bigger than the other.
Assuming you used the power method to compute an eigenpair, you should compute $\tilde \lambda$ and $\tilde u$ (different than the "real" values due to numerical error). If you use Hotelling deflation, you will be left with a deflated matrix
$$ A' = \lambda u u^T - \tilde{\lambda}\tilde{u}\tilde{u}^T + \mu vv^T$$
Even if you computed the eigenvector exactly ($\tilde u = u$), then you will be left with
$$ A' = (\lambda - \tilde{\lambda})u u^T + \mu vv^T$$
Now if $\mu/\lambda \sim \epsilon$, where $\epsilon$ is the machine precision, then $|\lambda-\tilde{\lambda}| \approx |\mu|$, so you will compute $\mu$ to almost no bits of precision since $A'$ is drowned out by the numerical noise due to the deflation process.
You can generalize this to a matrix for which shifted inverse power iteration will work by choosing a random full rank symmetric matrix, but make the dominant eigenvalue about $1/\epsilon$ times bigger than any other.
More generally, matrices with strongly graded eigenvalues will suffer tremendously from this kind of deflation since each time you subtract off the next leading eigenspace, the remaining deflated matrix will be corrupted by a factor of approximately the ratio of eigenvalue magnitudes (when sorted).
