2
$\begingroup$

on page 13 of the paper here there is a proof in theorem 4 that all eigenvalues of this tridiagonal matrix, which has strictly positive entries down the subdiagonals, are simple. Unfortunately, I don't get the argument. Apparently, it is almost immediate to the editor that $ker(J-\lambda I)$ must be one-dimensional for eigenvalues $\lambda$, where $J$ is the special tridiagonal matrix.

$\endgroup$
  • 2
    $\begingroup$ You've removed a bit of the context, actually: They require that the off-diagonal elements all be positive. (Which is rather natural, since otherwise the identity matrix is a big counter-example.) $\endgroup$ – Semiclassical Oct 16 '14 at 22:21
  • $\begingroup$ yes, this is why I am talking about 'this' and not 'any' tridiagonal matrix, but your are right: It is good that you pointed this out. Do you understand their argument? $\endgroup$ – user159356 Oct 16 '14 at 22:26
  • $\begingroup$ Not off the bat (though I should, since I've encountered this paper in my research before). My point was more that said context should be in the question rather than requiring the reader to look it up in the paper, especially since it does seem to be a simple linear algebra question. $\endgroup$ – Semiclassical Oct 16 '14 at 22:28
  • $\begingroup$ just out of curiosity, as you say that you know this paper. Do you know a fast way to actually show that the Whittaker Hill equation has this semifinite gap property? $\endgroup$ – user159356 Oct 16 '14 at 22:35
  • 2
    $\begingroup$ The link is not working and without it the question have no sense. Could anyone provide a reference or write the matrix? $\endgroup$ – Ernesto Iglesias Jun 15 '18 at 20:26
2
$\begingroup$

If you delete the first row and last column from an irreducible $n\times n$ tridiagonal matrix $T$, the resulting submatrix is triangular with non-zero diagonal entries. Hence it is invertible, and it follows that $\mathbb{rank}(T-\lambda I)$ is always at least $n-1$.

$\endgroup$
  • $\begingroup$ This seems to leave a gap: How does one show that a tridiagonal matrix with positive subdiagonal entries is irreducible? (I suppose it's a well-known result, but I don't recognize it off the top of my head.) $\endgroup$ – Semiclassical Oct 17 '14 at 1:12
  • $\begingroup$ @semiclassical: By irreducible I meant only that the subdiagonal entrier were nonzero. $\endgroup$ – Chris Godsil Oct 17 '14 at 10:57
  • $\begingroup$ Ah. I was mislead by Wikipedia's citation of 'irreducible matrix' on its irreducibility page which is rather different than that. Though I don't think irreducibility in your sense is necessarily enough: The paper cited by Tobias requires that the subdiagonals be strictly positive rather than nonzero. (I'll see if I can give a relevant counter-example in this vein.) $\endgroup$ – Semiclassical Oct 17 '14 at 12:04
  • $\begingroup$ Quite forgot to follow up on this, but here's a counter-example: the matrix $$T=\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$$ is irreducible in your sense, and the submatrix upon deleting the first row and last column is indeed triangular (indeed diagonal) with nonzero diagonal entries. But the characteristic polynomial $\det(T-\lambda I)=-\lambda^3$ has zero as a triple-root, and so this eigenvalue is not simple. $\endgroup$ – Semiclassical Oct 29 '14 at 17:10
  • $\begingroup$ @semiclassical: For me a tridiagonal matrix has its off-diagonal entries non-negative. This is a common convention. $\endgroup$ – Chris Godsil Oct 29 '14 at 18:11

Your Answer

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