Suppose I have the symmetric tridiagonal matrix:

$ \begin{pmatrix} a & b_{1} & 0 & ... & 0 \\\ b_{1} & a & b_{2} & & ... \\\ 0 & b_{2} & a & ... & 0 \\\ ... & & ... & & b_{n-1} \\\ 0 & ... & 0 & b_{n-1} & a \end{pmatrix} $

All of the entries can be taken to be positive real numbers and all of the $a_{i}$ are equal. I know that when the $b_{i}$'s are equal (the matrix is uniform), there are closed-form expressions for the eigenvalues and eigenvectors in terms of cosine and sine functions. Additionally, I know of the recurrence relation:

$det(A_{n}) = a\cdot det(A_{n-1}) - b_{n-1}^{2}\cdot det(A_{n-2})$

Additionally, since my matrix is real-symmetric, I know that its eigenvalues are real.

Is there anything else I can determine about the eigenvalues? Furthermore, is there a closed-form expression for them?

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    $\begingroup$ you might as well take $a=0$, since this is just an additive constant for each eigenvalue $\endgroup$ – Carlo Beenakker May 22 '13 at 23:30
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    $\begingroup$ There are older questions on MO dealing with eigenvalues of symmetric tridiagonal; no closed form. You might want to ask on the scicomp stackexchange. $\endgroup$ – Suvrit May 23 '13 at 0:02
  • $\begingroup$ Might be of interest: sciencedirect.com/science/article/pii/S037704270600015X $\endgroup$ – anderstood Feb 7 '17 at 0:26

The type of matrix you have written down is called Jacobi matrix and people are still discovering new things about them basically their properties fill entire bookcases at a mathematics library. One of the reasons is the connection to orthogonal polynomials. Basically, if $\{p_n(x)\}_{n\geq 0}$ is a family of orthogonal polynomials, then they obey a recursion relation of the form $$ b_n p_{n+1}(x) + (a_n- x) p_n(x) + b_{n-1} p_{n-1}(x) = 0. $$ You should be able to recognize the form of your matrix from this.

As far as general properties of the eigenvalues, let me mention two:

  1. The eigenvalues are simple. In fact one has $\lambda_j - \lambda_{j-1} \geq e^{-c n}$, where $c$ is some constant that depends on the $b_j$.

  2. The eigenvalues of $A$ and $A_{n-1}$ interlace.

  • $\begingroup$ Do you have any references for the first property? Thanks for your help! $\endgroup$ – FlamingWilderbeest May 23 '13 at 13:58
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    $\begingroup$ No. The proofs are simple enough that I don't know where they can be found. math.caltech.edu/Szego.html might contain these things. The main reason I'm hesitating posting a proof is that most proofs are heavy on notation, e.g. introduce orthogonal polynomials. $\endgroup$ – Helge May 23 '13 at 17:11

Amongst the polynomials that can arise as characteristic polynomials of tridiagonal matrices with zero diagonal, one finds the Hermite polynomials. Schur showed that Hermite polynomials of even degree are irreducible and that their Galois groups are not solvable. Hence there can be no closed form expression for the zeros in terms of the $b_i$'s in general.

  • $\begingroup$ I wasn't aware of this. Thanks a lot! $\endgroup$ – FlamingWilderbeest May 23 '13 at 1:28
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    $\begingroup$ Another argument is: it's easy to reduce any symmetric matrix to tridiagonal with similarity transforms. So if this problem were easy to solve, all symmetric eigenproblems would be. $\endgroup$ – Federico Poloni May 23 '13 at 6:40
  • $\begingroup$ Well, I didn't mean to imply that this would be easy to solve. It is too hopeful to expect a closed-form expression for the eigenvalues in terms of the bi's. However, knowing anything about what they look like would be tremendously useful to me $\endgroup$ – FlamingWilderbeest May 23 '13 at 14:57
  • $\begingroup$ Closed form expression built from arithmetic operations and radicals. When all the $b_i$ are equal we also do not get a closed form expressions of this type or do we? $\endgroup$ – Vít Tuček Jun 17 '14 at 1:22

According to doi:10.1016/S0024-3795(99)00114-7, the closed form for the eigenvalues of a tridiagonal Toepliz matrix of the form

$$ \begin{bmatrix}a & b\\ c & a & b\\ & \ddots & a & \ddots \\ & & & \ddots & \\ & & & c & a \end{bmatrix} $$


$$\lambda_{k}=a+2\sqrt{bc}\cos\left[\frac{k\pi}{(n+1)}\right], \quad k=1\cdots n $$

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    $\begingroup$ The one in the question is not a Toeplitz matrix. $\endgroup$ – Federico Poloni Dec 1 '14 at 10:23
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    $\begingroup$ You are correct, I didn't notice... $\endgroup$ – Sparkler Dec 1 '14 at 15:19

Withnout loss of generality, one can put $a=0$. For sure, there is no closed-form (or explicit) formula for the eigenvalues in general. However, at least the characteristic polynomial of $A_n$ can be written explicitly in temrs of $b_k$'s: $$\det(\lambda-A_n)=\lambda^{n}+\sum_{m=1}^{\lfloor\frac{n}{2}\rfloor}(-1)^{m}\left(\sum_{k\in\mathcal{I}(m,n)}b_{k_{1}}^{2}b_{k_{2}}^{2}\dots b_{k_{m-1}}^{2}b_{k_{m}}^{2}\right)\lambda^{n-2m}$$ where $$\mathcal{I}(m,n)=\{k\in\mathbb{N}^{m}\mid k_j+2\leq k_{j+1} \mbox{ for } 1\leq j \leq m-1,\; 1\leq k_1, \; k_m<n \}.$$


Mathematica gives to you the closed form that you want. All you have to do is use de recurrence package of the program

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    $\begingroup$ Chris Godsil's answer seems to suggest otherwise $\endgroup$ – Yemon Choi Jun 16 '14 at 22:05
  • $\begingroup$ Have you yourself even tried what you suggested? $\endgroup$ – Hans Mar 15 '18 at 18:19

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