What can I say about the eigenvalues and eigenvectors of the tridiagonal matrix $T$ given as $T = \begin{pmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-1} \\ & & & c_{n-1} & a_n \end{pmatrix}$.

If I set $a_i = 0$, do you know any previous results?

I know some results for simple cases like constant elements or symmetric matrices, but I would like to know if there are any results for more general cases.

  • $\begingroup$ Possible duplicate of How can I calculate eigenvalues of a tridiagonal matrix? $\endgroup$ – Benjamin Steinberg Mar 2 '18 at 0:07
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    $\begingroup$ I believe that the references mentioned in that post are related to symmetric tridiagonal matrices. $\endgroup$ – Guilherme Mar 2 '18 at 0:30
  • $\begingroup$ I'll retract my vote. $\endgroup$ – Benjamin Steinberg Mar 2 '18 at 1:12
  • $\begingroup$ I'm not sure it may be of any help, but if you further assume that $T$ is normal, which requires that $|c_i|=|b_i|$ for all $i$, then you can split $T$ as $T=S+A$ with $S$ (resp. $A$) tridiagonal symmetric (resp. antisymmetric) with $A,S$ diagonalizable in the same basis. This simplifies quite a lot the problem. In the non-normal case, the eigenvectors may not be orthogonal and I fear no particular structure would arise. $\endgroup$ – Adrien Hardy Mar 2 '18 at 1:30
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    $\begingroup$ If $b_i$ and $c_i$ have the same sign for all $i$, then I think you can apply a diagonal rescaling to make it symmetric => it has real eigenvalues. $\endgroup$ – Federico Poloni Mar 2 '18 at 11:20

It seems that your question has already been answered here:

Eigenvalues of Symmetric Tridiagonal Matrices

No results for general tridiagonal matrices.

  • $\begingroup$ Thank you very much for your answer, but the difference between my question and this one is that I am not considering symmetric matrices. $\endgroup$ – Guilherme Mar 6 '18 at 17:18
  • $\begingroup$ @Guilherme Sure, but if there is no general results for symmetric tridiagonal matrices, a fortiori there is no results for non-symmetric ones, isn't it? $\endgroup$ – Fabrice Pautot Mar 6 '18 at 20:14
  • $\begingroup$ You are correct. I missed that point. Thanks for pointing this. $\endgroup$ – Guilherme Mar 7 '18 at 18:08

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