Simple Spectrum of Jacobi matrices 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.
 A: 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.
A: 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. 
A: 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 $a_n=1$, $b_n=0$.
