1
$\begingroup$

I want to call a matrix a Jacobi matrix (cause there may be different notions of Jacobi matrices) if it is a tridiagonal matrix with positive off-diagonal entries. Now, I read that the spectrum of such matrices is simple and interlacing. Although, I find quite many proofs of the fact that the spectrum is interlacing, I could not see that it is simple. Just in one paper, it was said that this would be an immediate consequence of the tridiagonal form of the linear system $$(A - \lambda I)v=0.$$

Thus, now it should be somehow possible to conclude from this that for such Jacobi matrices the nullspace is one-dimensional, but I don't see how.

Improve this question
$\endgroup$
3
  • $\begingroup$ What do you mean by "spectrum is interlacing"? $\endgroup$
    – Chris Godsil
    Jan 12, 2015 at 20:20
  • $\begingroup$ @ChrisGodsil it just means that if you take the upper (n-1)x(n-1) submatrix of $A$, then the eigenvalues between this matrix and $A$ are interlacing. So between any eigenvalues of $A$ is one of the submatrix, but I guess that this property is more or less unrelated to this question. $\endgroup$
    – Jiao Guo
    Jan 12, 2015 at 20:27
  • $\begingroup$ Because the spectrum of an nxn truncation of a Jacobi matrix has for eigenvalues the zeroes of a degree n orthogonal polynomial, read the literature on OPs. $\endgroup$
    – Adrien Hardy
    Jan 12, 2015 at 20:59

3 Answers 3

Reset to default
4
$\begingroup$

A standard reference is MR1908601 Gantmacher, F. P.; Krein, M. G. Oscillation matrices and kernels and small vibrations of mechanical systems.

Simplicity of eigenvalues is proved in the first paragraph of Chapter 2.

Improve this answer
$\endgroup$
1
$\begingroup$

In the case of finite or semi-infinite Jacobi matrix, the first entry of an eigenvector uniquely determine other entries since they are related by the tree-term recurrence. This is the reason why the multiplicity of an eigenvalue can not exceed 1.

As already mentioned by Christian Remling, this is not the case for both-infinite Jacobi operators. In this case, eigenvalues can be of multiplicity 2 (not only the absolutely continuous spectrum). Nevertheless, the multiplicity can not exceed 2. This is again closely related with the fact that the second order difference equation has 2 linearly independent solutions.

Improve this answer
$\endgroup$
0
$\begingroup$

For a problem on a half line or a bounded (= finite) interval (let's say with $n=0$ as its left endpoint) $$ (Ju)_n = \begin{cases} a_0 u_1 + b_0 u_0 & n=0\\ a_n u_{n+1} + a_{n-1} u_{n-1} + b_n u_n & n\ge 1 \end{cases} $$ it is easy to check directly that $\delta_0$ is a cyclic vector; just show by induction on $k$ that $\delta_0,\ldots , \delta_k$ are in the span of $J^m \delta_0$, $m=0,\ldots, k$.

A Jacobi operator on the whole line (on the Hilbert space $\ell^2(\mathbb Z)$) need not have simple spectrum. The absolutely continuous spectrum can have multiplicity $2$. For example, this happens for constant coefficients, say $a_n=1$, $b_n=0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I saw that this post is old, but I have a quick question, In what you stated, does it mean that any $\delta_k$ works as a cyclic vector? $\endgroup$
    – Tom Builder
    Jun 26, 2020 at 16:09
  • $\begingroup$ @TomBuilder: No, that's not true in general. For example, if there is an eigenvector $u$ with $u_k=0$, then $u$ is orthogonal to the cyclic subspace generated by $\delta_k$. $\endgroup$
    – Christian Remling
    Jun 26, 2020 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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