I just came across a matrix of the form
$A:=\begin{pmatrix}
0&-\frac{c_0}{b_0}&0&\cdots&0\\-\frac{a_1}{b_1}&0&-\frac{c_1}{b_1}&\cdots&0\\0&-\frac{a_2}{b_2}&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&-\frac{a_{N-1}}{b_{N-1}}\\0&\cdots&\cdots&-\frac{a_N}{b_N}&0
\end{pmatrix}$ for some N$\in \mathbb{Z}^+$ where $a_n=-\frac{1}{2}\alpha(\beta^2n^2-rn), \ b_n=1+\alpha(\beta^2n^2+r), and \ c_n=-\frac{1}{2}\alpha(\beta^2n^2+rn)\ $such that $\alpha, \beta,r$ are known real constants.
From the Gershgorin circle theorem, I know that its maximum eigenvalue must lie in the Gershgorin discs. However, despite it being quite sparse, I could not get an explicit formula for its maximum eigenvalue.
I have tried solving the Av=$\lambda$v equation, where v is an eigenvector and $\lambda$ is an eigenvalue, in which I obtain a recurrence relation, but I didn't have an initial boundary condition. The equation det($\lambda$I-A)=0, where I is the identity matrix, also gives me a complicated equation that I can't solve.
Can anyone tell me what I have missed or is this an impossible-to-solve problem?
Best Answer
Let me try to expand a little bit the problem (so it's too long for a usual comment).
Consider the determinant $D_N=D_N(\lambda;a_1,\dots,a_{N-1};b_1,\dots,c_{N-1})$ of the corresponding matrix $\lambda-A$. Expanding the determinant along the first row gives $$ D_N(\lambda;a_1,\dots,a_{N-1};b_1,\dots,b_{N-1}) =\lambda D_{N-1}(\lambda;a_2,\dots,a_{N-1};b_2,\dots,b_{N-1}) -a_1b_1D_{N-2}(\lambda;a_3,\dots,a_{N-1};b_3,\dots,b_{N-1}); $$ in other words, $$ D_N/D_{N-1}=\lambda-\frac{a_1b_1}{D_{N-1}/D_{N-2}} =\dots =\lambda-\frac{a_1b_1}{\lambda-\dfrac{a_2b_2}{\lambda-\dfrac{a_3b_3}{\dots -\dfrac{a_{N-1}b_{N-1}}{\lambda}}}}. $$ In order to get some information about the asymptotics of the zero(s) of $D_N(\lambda)/D_{N-1}(\lambda)$ one really have to have some knowledge about the $a_ib_i$, $i=1,2,\dots$. This reduces the problem to a problem for the related family of orthogonal polynomials and even Deift's book is too advanced, it is the best source on this.