Let $T$ be a real symmetric tridiagonal matrix. Then the divide-and-conquer eigenvalue algorithm, as detailed in any standard text, goes by subdividing

$$ T = \begin{bmatrix} T_1 & 0 \\ 0 & T_2 \end{bmatrix} + \rho vv^T,$$ after which we can solve the eigenvalue problem for $T_1, T_2$, and then solve the full eigenvalue problem by accounting for the rank-$1$ shift $\rho vv^T$ using a root-finding technique.

However, now let $T$ be a complex Hermitian tridiagonal matrix. Then if we attempt to do the same strategy, $$ T = \begin{bmatrix} T_1 & 0 \\ 0 & T_2 \end{bmatrix} + S,$$ where $S$ is Hermitian but is not rank-$1$ (it appears to have rank $2$), so it seems like the same root-finding technique will fail, unless I'm mistaken. How can we adapt the algorithm to work with the complex Hermitian case?


There is no reason to ever have to give thought to the Hermitian tridiagonal eigenvalue problem, as it is trivial to define a unitary diagonal similarity transformation which reduces it to the real symmetric tridiagonal eigenvalue problem.

| cite | improve this answer | |
  • $\begingroup$ Ha, I must admit this did not cross my mind, even though I was aware of it. Thanks for the note. $\endgroup$ – Christopher A. Wong Apr 17 '13 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.