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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?

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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.

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  • $\begingroup$ 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

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