Assume a matrix in this form:
$$
\begin{bmatrix}
b & c & 0 & \dots & 0 & a \\
a & b & c & 0 & \dots & 0 \\
0 & a & b & c & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots & \ddots. & 0 \\
0 & \cdots & 0 & a & b & c \\
c & 0 & \cdots & 0 & a & b \\
\end{bmatrix}_{n \times n}
$$
I want to check that the eigenvalues are of the form
$$ \lambda_p = ae^{-2\pi ip/n}+b+ce^{2\pi ip/n}$$ with associated eigenvector $$ v_p = v_{p,j}
= e^{-2\pi ipj /n}$$ I only find information about determinant of tridiagonal Toeplitz matrix but not for this case. Any help is appreciated!
Best Answer
Simply multiply the matrix with each eigenvector and notice that you obtain
$$ \begin{bmatrix} b & c & 0 & \dots & 0 & a \\ a & b & c & 0 & \dots & 0 \\ 0 & a & b & c & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & \ddots. & 0 \\ 0 & \cdots & 0 & a & b & c \\ c & 0 & \cdots & 0 & a & b \\ \end{bmatrix} v_p = (ae^{-2\pi ip/n}+b+ce^{2\pi ip/n})v_p = \lambda_p v_p $$
directly from the matrix definition. It would be significantly harder to actually find $v_p$ and $\lambda_p$ without knowing them beforehand, but since you are given $n$ candidate solutions (with $1 \leq p \leq n$), one only needs to check that they indeed form eigenpairs.