3
$\begingroup$

Consider the discrete Dirichlet Laplacian on a set of cardinality $n.$ For example the Dirichlet Laplacian $\Delta_D$ on a set of cardinaltity 4 is the matrix $$\Delta_D := \left( \begin{matrix} 2 & -1 & \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ & & -1 &2 \end{matrix}\right)\in \mathbb C^{4 \times 4}.$$

This matrix is self-adjoint. It has an eigendecomposition with eigenvectors $(v_i).$

In particular, we can decompose the first unit vector in its eigenbasis $$1 = \sum_{i=1}^n \lvert\langle v_i^{n},e_1 \rangle \rvert^2.$$

Clearly, as $n$ tends to infinity the convergence of the series implies that the smallest object $\inf_i\lvert\langle v_i^{n},e_1 \rangle \rvert^2$ decays faster than $1/n.$

I would like to know: Can one find the asymptotics of $$\inf_i\lvert\langle v_i^{n},e_1 \rangle \rvert^2$$

$\endgroup$
2
$\begingroup$

Even though this is not a circulant matrix, its eigenvalues and eigenvectors are known in closed form; see for instance this Wikipedia article, which gives an expression for the eigenvectors and eigenvalues of $-\frac{1}{h^2}\Delta_D$, using your notation. In particular, $$\langle v_i^n, e_1 \rangle = \sin \frac{i\pi}{n+1}. $$

Hence, if I don't make mistakes, $\langle v_1^n, e_1 \rangle^2 = \sin^2 \frac{\pi}{n+1} \sim \frac{\pi^2}{n^2}$, which should answer your question.

$\endgroup$
-1
$\begingroup$

The eigenvectors are the eigenvectors of a circulant matrix, as described in this Wikipedia article. You can see that all of your squared dot products are exactly $1/n.$

$\endgroup$
2
  • $\begingroup$ But the Laplacian is not circulant? $\endgroup$ – Federico Poloni Mar 26 '19 at 12:41
  • $\begingroup$ not a circulant matrix. $\endgroup$ – Denis Serre Mar 26 '19 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.