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 
$$\inf\_i\lvert\langle v\_i^{n},e\_1 \rangle \rvert^2$$



Rate of convergence for eigendecomposition
Even though this is not a circulant matrix, its eigenvalues and eigenvectors are known in closed form; see for instance [this Wikipedia article](https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#Pure_Dirichlet_boundary_conditions_2), 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.


