# Computing eigendecomposition of a Hermitian matrix that is almost unitary

I have a dense Hermitian matrix that is approximately unitary, so it has eigenvalues that are $\sim \pm1$. I would like to compute all the eigenvectors corresponding to the $+1$ eigenvalue (not necessarily a small fraction of the total number of eigenvalues). Is there some more efficient way to accomplish this than tridiagonal QR?

• Look for the kernel of $A-I$? – Federico Poloni Apr 8 '14 at 10:50
• Now that I re-read your question, it is not clear to me whether you mean "the eigenvalues that are exactly +1", or "those that are closer to +1 than to -1 and hopefully come from +1 eigenvalues of the unperturbed unitary matrix". Could you clarify? – Federico Poloni Apr 9 '14 at 20:04
• I mean the latter. The eigenvalues "should" be clustered around $\pm1$, but practically, get spread out a bit, while still respecting the upper bound (but not the lower). Typically, of the set closer to $+1$, the vast majority of them are almost exactly $1$. – Victor Liu Apr 9 '14 at 21:41

1) perform a few iterations of the matrix sign iteration $A\mapsto \frac{1}{2}(A+A^{-1})$; the eigenvectors are unchanged, while the eigenvalues converge quadratically to $\pm 1$. When $A-A^{-1}$ is small enough, stop the iteration.
2) compute the kernel of $A-I$.