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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 $\endgroup$ May 22, 2013 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, 2013 at 0:02
  • $\begingroup$ Might be of interest: sciencedirect.com/science/article/pii/S037704270600015X $\endgroup$
    – anderstood
    Feb 7, 2017 at 0:26

5 Answers 5

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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.

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  • $\begingroup$ Do you have any references for the first property? Thanks for your help! $\endgroup$ May 23, 2013 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, 2013 at 17:11
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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.

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  • $\begingroup$ I wasn't aware of this. Thanks a lot! $\endgroup$ May 23, 2013 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. $\endgroup$ May 23, 2013 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 $\endgroup$ May 23, 2013 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? $\endgroup$ Jun 17, 2014 at 1:22
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According to doi:10.1016/S0024-3795(99)00114-7, the closed form for the eigenvalues of a tridiagonal Toepliz matrix of the form

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

is:

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

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    $\begingroup$ The one in the question is not a Toeplitz matrix. $\endgroup$ Dec 1, 2014 at 10:23
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    $\begingroup$ You are correct, I didn't notice... $\endgroup$
    – Sparkler
    Dec 1, 2014 at 15:19
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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 $$\mathcal{I}(m,n)=\{k\in\mathbb{N}^{m}\mid k_j+2\leq k_{j+1} \mbox{ for } 1\leq j \leq m-1,\; 1\leq k_1, \; k_m<n \}.$$

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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, 2014 at 22:05
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    $\begingroup$ Have you yourself even tried what you suggested? $\endgroup$
    – Hans
    Mar 15, 2018 at 18:19

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