Eigenvectors and eigenvalues of Tridiagonal matrix with varying diagonal elements

Question

is it possible to analytically evaluate the eigenvectors and the eigenvalues of a tridiagonal $n\times n$ matrix of the form :

\begin{pmatrix} 1 & b & 0 & ... & 0 \\\ b & 2 & b & & ... \\\ 0 & b & 3 & ... & 0 \\\ ... & & ... & & b \\\ 0 & ... & 0 & b & n \end{pmatrix}

Where $b$ is a constant.

Any help is appreciated.

Answer 1

The case $b=0$ is easy, and therefore not considered.

Suppose that $u_n$ is an approximation of $u(x_j)$ with $x_j=j/(n+1))$, where $u(0)=0$ and $u(1)=0$. A taylor expansion gives $$ u(x_{j+1})=u(x_{j})+\frac{1}{n+1}u^{\prime}(x_{j})+\frac{1}{2(n+1)^2}u^{(2)}(x_j)+\frac{1}{6(n+1)^3}u^{(3)}(x_j)+\frac{1}{24(n+1)^4}u^{(4)}(\zeta_j) $$ $$ u(x_{j-1})=u(x_{j})-\frac{1}{n+1}u^{\prime}(x_{j})+\frac{1}{2(n+1)^2}u^{(2)}(x_j)-\frac{1}{6(n+1)^3}u^{(3)}(x_j)+\frac{1}{24(n+1)^4}u^{(4)}(\zeta_j) $$ Therefore $$ b u(x_{j+1}) + b u(x_{j-1}) -2 b u(x_{j}) = \frac{b}{(n+1)^2}u^{(2)}(x_j) + O(\frac{1}{n^4}). $$ $$ b u(x_{j+1}) + b u(x_{j-1}) +j u(x_{j}) = \frac{b}{(n+1)^2}u^{(2)}(x_j) +(2b+ (n+1)x_j) u(x_j)+ O(\frac{1}{n^4}). $$ So an eigenpair of the matrix is an approximation (for $n$ large) of an eigenpair of $$ \epsilon^{-2} b u^{\prime\prime} + (2b +\epsilon^{-1} x) u = \lambda u $$ with $u(0)=u(1)=0$, on $(0,1)$, and $\epsilon=1/(n+1)$.

Multiplying out by $\epsilon^2/b$, and writing $\mu=\epsilon^2\lambda/b$ we find $$ u^{\prime\prime}+\left(2\epsilon^2 + \frac{\epsilon}{b}x\right)u=\mu u. $$ So the first order term is negligible for the smallest eigenvalues, (not for the ones of order $\epsilon^{-1}$), and therefore if $b<0$, the first eigenvalues are (up to a mistake in the previous lines) $$ \lambda_{k}=-\frac{k^2}{n^2}\pi^2 b\left(1+0\left(\frac{1}{n}\right)\right),\mbox{ for } \frac{k}{n} \ll 1. $$

Answer 2

Experiment with Mathematica indicates that even for the special case of $a_j = j$ there is no simple expression for the eigenvalues. Determinant is probably not hard, but I did not try.