Eigenvalues of Symmetric Tridiagonal Matrices 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?
 A: 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:
$$\lambda_{k}=a+2\sqrt{bc}\cos\left[\frac{k\pi}{(n+1)}\right], \quad k=1\cdots n $$
A: 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:


*

*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 $b_j$.

*The eigenvalues of $A$ and $A_{n-1}$ interlace.
A: 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.
A: 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
$$\mathcal{I}(m,n)=\{k\in\mathbb{N}^{m}\mid k_j+2\leq k_{j+1} \mbox{ for } 1\leq j \leq m-1,\; 1\leq k_1, \; k_m<n \}.$$
A: Mathematica gives to you the closed form that you want. All you have to do is use de recurrence package of the program 
