Rate of convergence for eigendecomposition

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 $$\inf_i\lvert\langle v_i^{n},e_1 \rangle \rvert^2$$

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.

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 $$1/n.$$

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