Decay of eigenfunctions for Laplacian Consider the discrete second derivative with Dirichlet boundary conditions on $\mathbb C^n$.
Its eigendecomposition is fully known:
see wikipedia
It seems like the largest eigenvalue $\lambda_1$ is one with a fast decaying eigenfunction, by this I mean that at the first coordinate $\vert v_{1,1} \vert \le Cn^{-3/2}.$ The first $1$ indicates the eigenfunction, the second one the coordinate.
A priori there is no reason to have this type of decay, at the first coordinate, I guess.
Is there a way to prove this without(!) using that the eigenfunctions are explicitly known?-Thus, can one show this directly from the matrix?
 A: OK, so here is an extended version of my comment.

By Perron–Frobenius, the first eigenvector $v_1$ is positive, and thus $L v_1 = \lambda_1 v_1 \geqslant 0$ (here $-L$ is the matrix of the discrete Laplacian with Dirichlet boundary conditions, so that $L \geqslant 0$). Let us write $v_{1,0} = v_{1,n+1} = 0$. It follows that $v_1$ is a concave function on $\{0,1,\ldots,n+1\}$. In particular:
$$ v_{1,i} \leqslant i v_{1,1} , $$
and hence
$$1 = \|v_1\|_2^2 \leqslant v_{1,1}^2 \sum_{i = 1}^n i^2 \le C_1 v_{1,1}^2 n^3.$$
In particular,
$$ v_{1,1} \geqslant C_1^{-1/2} n^{-3/2} . \tag{1} $$

The proof of the converse inequality is slightly more complicated.
First, we consider an auxiliary vector
$$ x_i = C_2 i (n + 1 - i) ,$$
with $C_2$ such that $$(L x)_i = 1 .$$ Then $\langle L x, x\rangle \leqslant C_3 n^3$ and $\|x\|_2^2 \geqslant C_4 n^4$, which implies that $\lambda_1 \le C_5 n^{-1}$.
We define another auxiliary vector
$$ u_i = C_2 \lambda_1 i^2 ,$$
with $C_2$ such that $L u_i = -\lambda_1$. Set $$w = v_1 + v_{1,1} u.$$ Then $$(L w)_i = \lambda_1 (v_{1,i} - v_{1,1}) \geqslant 0$$ as long as $v_{1,i} \geqslant v_{1,1}$; say, for $i = 1, 2, \ldots, m$. It follows that $w$ is a convex function on $\{0, 1, \ldots, m + 1\}$, with $w_0 = 0$ and $w_1 = (1 + C_2 \lambda_1) v_{1,1}$. In particular, $w_{m+1} \geqslant (m + 1) w_1$, that is, $$v_{1,m+1} + C_2 \lambda_1 v_{1,1} (m + 1)^2 \geqslant (m + 1) (1 + C_2 \lambda_1) v_{1,1} .$$
In other words,
$$v_{1,m+1} \geqslant (m + 1) v_{1,1} (1 + C_2 \lambda_1 - C_2 \lambda_1 (m + 1)) .$$
By definition of $m$, we either have $m = n$ or $v_{1,m+1} < v_{1,1}$; in the latter case, it follows that $(m + 1) > 1 / (C_2 \lambda_1) > C_6 n$. In either case, $m \geqslant C_6 n$.
For $i = 1, 2, \ldots, C_6 n$ we therefore have
$$ v_{1,i} = w_i - v_{1,1} C_2 \lambda_1 i^2 \geqslant i w_1 - v_{1,1} C_2 \lambda_1 i^2 = i (1 + C_2 \lambda_1) v_{1,1} - v_{1,1} C_2 \lambda_1 i^2 , $$
that is,
$$ v_{1,i} \geqslant i v_{1,1} (1 + C_2 \lambda_1 - C_2 \lambda_1 i) \geqslant i v_{1,1} (1 - C_7 i / n) . $$
Therefore,
$$ 1 = \|v_1\|_2^2 \geqslant v_{1,1}^2 \sum_{i = 1}^{C_8 n} i^2 (1 - i / (C_8 n))^2 \geqslant v_{1,1}^2 C_9 n^3 $$ 
(with $C_8 = \min\{C_6, 1 / C_7\}$). We conclude that
$$ v_{1,1} \leqslant C_1^{-1/2} n^{-3/2} . \tag{2} $$

Estimates (1) and (2) give the desired bound. The above argument should extend to generators of symmetric "uniformly elliptic" random walks on $\{1, 2, \ldots, n\}$. I suppose a much more clean argument can be given, feel free to edit. Finally, similar results are surely known for more general graphs with boundaries, but I do not have a reference off the top of my head.

Regarding totally positive matrices, I may have been too optimistic. Still, let me recommend Ando's 1987 paper Totally positive matrices. Section 6 of this paper gives a nice summary of the properties of eigenvectors of TP matrices; for example, Theorem 6.3 asserts that the nodes of the eigenvectors (extended to piecewise linear functions) are interlacing.
A: I denote $\Delta^{(n)}$ the discrete Laplacien on $\mathbb{C}^n$ (ie the $n\times n $ matrix with diagonal $-2$ and upper/lower diagonal $1$.) 
Consider the scaling $ v \rightarrow \phi(t) = \sqrt{n} v(\lfloor tn\rfloor)$ defined on $\mathbb{C}^n\rightarrow L^2([0,1])$. 
I think we have the following heuristic 
$$ n^2\Delta^{(n)} \rightarrow \Delta$$
$$ v_1^{(n)} \rightarrow \phi_1$$
with $\Delta$ the continuous Laplacien with Dirichlet boundary conditions and $\phi_1$ its first eigenvector and then $$v_{1,1}^{(n)}\approx \frac{1}{\sqrt{n}}\phi_1 (\frac{1}{n})\approx \frac{1}{n^{3/2}}\phi_1'(0) $$
