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I would like to know if there is any way to find an analytical expression for the eigen values of the following matrix.

$$ A^h = \frac{1}{h^4} \begin{pmatrix} 5&-4&1&&&&&&\\ -4&6&-4&1&&&&\bigcirc&\\ 1&-4&6&-4&1&&&&\\ &\ddots & \ddots & \ddots & \ddots&&& \\ &&\ddots & \ddots & \ddots & \ddots&& \\ &&&&1&-4&6&-4&1\\ &\bigcirc&&&&1&-4&6&-4\\ &&&&&&1&-4&5\\ \end{pmatrix} $$

I have heard that it is possible through DFT, but I am not sure how to proceed with that. The size of the matrix is $N \times N$ and $h = \frac{1}{N}$.

Thanks!

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  • $\begingroup$ Does this come from cubic splines? $\endgroup$ Nov 13, 2015 at 22:18
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    $\begingroup$ Well, it actually came from the discretization of the bi-harmonic equation with a 5 point finite difference stencil. $\endgroup$
    – Mathnoob
    Nov 13, 2015 at 22:38

1 Answer 1

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Let $h=1$ for simplicity (rescaling is trivial). Note that $A^h=B^2$, where $B$ is the symmetric tridiagonal Toeplitz matrix with non-zero elements $(-1,2,-1)$, i.e. $$ B= \begin{bmatrix} 2 & -1 & 0 & 0 & \dots \\ -1 & 2 & -1 & 0 &\dots \\ 0 & -1 & 2 & -1 & \dots \\ & & \ddots & \ddots & \ddots \end{bmatrix}. $$ The eigenvalues of symmetric tridiagonal Toeplitz matrices are well known and can be found in many texts. For $B$, the eigenvalues are $$ \lambda_k(B) = 2 \left( 1-\cos{\left( \frac{k \pi}{N+1} \right)} \right). $$ It immediately follows that the eigenvalues of $A^h$ are given by $$ \lambda_k(A) = \lambda_k^2(B) = 4 \left( 1-\cos{\left( \frac{k \pi}{N+1} \right)} \right)^2. $$

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  • $\begingroup$ So simple when you see the trick! Very nice. And the eigenvalues of the tridiagonal matrix are indeed easy (there is a three term recurrence for the determinant). $\endgroup$ Nov 14, 2015 at 0:33
  • $\begingroup$ Great observation. Thanks! $\endgroup$
    – Mathnoob
    Nov 14, 2015 at 19:07
  • $\begingroup$ Also, if possible could you please give me some books or sources which tell about how to derive the eigen values of toeplitz matrix ? $\endgroup$
    – Mathnoob
    Nov 14, 2015 at 19:33

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