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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.

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  • $\begingroup$ What do you mean by "spectrum is interlacing"? $\endgroup$
    – Chris Godsil
    Commented 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
    Commented 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
    Commented Jan 12, 2015 at 20:59

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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.

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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.

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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 $a_n=1$, $b_n=0$.

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  • $\begingroup$ 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
    Commented 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
    Commented Jun 26, 2020 at 17:07

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