Computing eigenvalue of the adjacency matrix of a path

Question

Let $A\in \{0,1\}^{n \times n}$ be the adjacency matrix of a path of length $n$, i.e. having ones on the two off-diagonals, and zeros elsewhere. How does one compute the eigenvalues of this?

I know that the answer is $2 \cos(\frac{\pi k}{n+1})$ for $k=1, \dotsc, n$.

When trying to write down the characteristic polynomial, it becomes quite hard. expanding is not possible. Use recursion might be, but I do not know how to get cosine out of recursion.

So I would appreciate any hint.

Answer 1

If we have a $n\times n$ tridiagonal Toeplitz matrix of the form: $$A = \begin{bmatrix} a & c & & & & \\ b & a & c &&\mathbf 0 \\ & b & a & c \\ &&\ddots&\ddots&\ddots& \\ &\mathbf 0&&&& \\ &&&&&&&\end{bmatrix},$$ its eigenvalues are given by the formula: $$ \lambda_k = a + 2 \sqrt{bc}\cdot \cos\left(\frac{ k\pi}{n+1}\right),\quad k=1,\ldots,n$$

I think this will help you for your specific case.