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?

  • $\begingroup$ Look for the kernel of $A-I$? $\endgroup$ Apr 8, 2014 at 10:50
  • $\begingroup$ 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? $\endgroup$ Apr 9, 2014 at 20:04
  • $\begingroup$ 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 $1$. $\endgroup$
    – Victor Liu
    Apr 9, 2014 at 21:41

1 Answer 1


A thing you might try:

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 $A-I$.

More information on the matrix sign iteration on the book N. Higham, Functions of Matrices.

This method trades off eigensolves for QR factorizations; it might be faster than QR, depending on the size of the matrices, of the perturbation, and other machine-related factors difficult to quantify.


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