4
$\begingroup$

I would like to solve the following equation involving a determinant of an $n \times n $ matrix: $$ \begin{vmatrix} 2\cos \theta & -1 & 0 & \cdots & 0 \\ -1 & 2\cos \theta & -1 & \cdots & 0 \\ 0 & -1 & 2\cos \theta & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0&\cdots& 2\cos \theta \end{vmatrix} =0$$ where $0 \le \theta \le \pi $. I have made a few tries for small matrices, and it seems that the solution is $\theta=\frac{k\pi}{n+1}$ for $k=1, 2, \cdots, n$

However, I cannot prove a generalized case (for arbitrary $n$). How can I prove the generalized case?

Thank you.

$\endgroup$

2 Answers 2

1
$\begingroup$

if $d_n$ is the determinant of the $n \times n$ matrix, it is not hard to verify that it satisfies the recurrence $d_n = 2 \cos \theta \, d_{n-1} - d_{n-2}$ subject to $d_{-1} = 0, d_0 = 1$.

The characteristic equation is $\lambda^2 - 2 \cos \theta \lambda +1 = 0$ which has roots $\lambda = e^{\pm \theta}$.

First note that if $\cos \theta = 1$ then $d_n = n$ and if $\cos \theta = -1$ then $d_n = (-1)^{n+1} n$.

The general solution (for $|\cos \theta| < 1$) is of the form $n \mapsto b e^{i n \theta} + c e^{-i n \theta}$, and solving for the initial conditions gives $d_n = { \sin ((n+1)\theta) \over \sin \theta}$.

Solving $d_n = 0$ gives the required solutions.

$\endgroup$
0
$\begingroup$

Let $a_n$ be the value of the $n \times n$ determinant in question, we have: $a_n = 2\cos \theta \cdot a_{n-1}- a_{n-2}$. Using the characteristic equation approach we have: $x^2 - 2\cos \theta x + 1 = 0\implies \triangle'=\cos ^2\theta-1 = -\sin^2 \theta\implies x = \dfrac{\cos \theta \pm i\sin \theta}{2}$. Can you continue here?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .