1
$\begingroup$

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!

$\endgroup$
1
  • 3
    $\begingroup$ Have you tried checking what happens if you multiply the vector with the matrix? $\endgroup$ Commented Mar 26, 2021 at 20:45

2 Answers 2

Reset to default
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I thought there was proof or something, but you're right about knowing them beforehand! $\endgroup$ Commented Mar 26, 2021 at 20:54
2
$\begingroup$

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...

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

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.