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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!

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    $\begingroup$ Have you tried checking what happens if you multiply the vector with the matrix? $\endgroup$
    – user619894
    Commented Mar 26, 2021 at 20:45

2 Answers 2

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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.

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  • $\begingroup$ I thought there was proof or something, but you're right about knowing them beforehand! $\endgroup$
    – duster
    Commented Mar 26, 2021 at 20:54
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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...

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

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