# Find eigenvalues and eigenvectors of particular Toeplitz matrix

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!

• Have you tried checking what happens if you multiply the vector with the matrix? Commented Mar 26, 2021 at 20:45

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.

• I thought there was proof or something, but you're right about knowing them beforehand! Commented Mar 26, 2021 at 20:54

In the case you don't know beforehand the eigenvectors, you can refer to the fact that your matrix is circulant. The eigenvectors of a $$n \times n$$ circulant matrix are the columns of the Discrete Fourier matrix $$F_n$$ (this is also explained in the referenced Wikipedia article) and it is precisely those which are given to you...

• Why downvoting this answer ? Doesn't it bring a discovering methodology for the eigenvectors, therefore for the eigenvalues ? Commented Aug 1, 2023 at 19:52