Consider the following adjacency matrix of a complete graph:
$$A=(e^{-|i-j|})\_{1\leq i\neq j\leq n}$$
with 0 on the diagonal. Let $D=diag\{d\_1,...,d\_n\}$ be the degree matrix where $d\_i=\sum\_{j\neq i}e^{-|i-j|}$. Then $L=D-A$ is the Laplacian. Let $L^\dagger$ be the Moore-Penrose inverse of the Laplacian. I'm interested in the following quantity
$$a\_{ij}=|(e\_1-e\_2)^TL^\dagger(e\_i-e\_j)|$$
where $e\_i=(0,0...,0,1,0,...0)$ with 1 on the ith coordinate. I conjecture that $a\_{ij}$ will decay exponentially when both $i$ and $j$ moves away from 1 and 2. Something like $a\_{ij}\leq C \_1e^{-C\_2\min\{i,j\}}$ where $C\_1,C\_2$ are some constants. From the physics point of view, $a\_{ij}$ is the voltage potential difference between $i$ and $j$. It is intuitive that when they are far away from the source, 1 and 2, they should be very small given the structure of the graph.


In fact, my simulation shows that as long as $i,j\neq1,2$, $a\_{ij}$ suddenly becomes extremely close to 0. There seems to be no decay, but an acute cut. This phenomenon holds for slight perturbation of $A$, keeping the decaying property. 


Is this conjecture true? How can we prove it? What is the rate of decay?


Another quantity that is also interesting is
$$\sum\_{i\neq j}e^{-|i-j|}a\_{ij}$$
which is the weighted average of potential differences. How can we bound this? For this quantity, I conjecture it is bounded by some constant instead of growing with $n$. The physical meaning of this quantity is the sum of all currents in each edge. 


(Update)


Enlightened by the discussion with @Abdelmalek Abdesselam below. We have the Neuman series representation:
$$a\_{ij}=|(e\_1-e\_2)^TD^{-1/2}\sum\_{k\geq0}\left(D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}\right)^kD^{-1/2}(e\_i-e\_j)|$$
where $J$ is the matrix of all 1s and $\alpha$ is some constant to be chosen. We want to choose $\alpha$ such that the power of the matrix decays fast. How can we achieve this and bound the entries of $D^{-1/2}AD^{-1/2}-\alpha D^{1/2}JD^{1/2}$? A possible choice is $\alpha=1/tr(D)$.



Exponential decay of voltage potential difference
**Edit:** This turns out to be quite simple. Observe that $a\_{1i} / a\_{2i} = q$ does not depend on $i \in \{3, 4, \ldots, n\}$. Thus, if $x\_1 = 1$, $x\_2 = -q$ and $x\_i = 0$ for $i \in \{3, 4, \ldots, n\}$, then we clearly have $L x = c e\_1 - c e\_2$, where $c = \sum\_{i=3}^n a\_{1i}$. It follows that $$L^\dagger (e\_1 - e\_2) = c^{-1} x + \operatorname{const}.$$


I leave the previous version of this answer below, as it provides a way to evaluate $L^\dagger$ explicitly.




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I cannot say I understand what is *really* going on here, but at least I have a proof that $a\_{ij} = 0$ when $i, j \ge 3$. (I leave my previous comment/answer, as it contains some related stuff that is not included here.)




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*Notation:* Every sum is a sum over $\{1, 2, \ldots, n\}$. We write $q = e^{-1}$ (and actually any $q \in (0, 1)$ will work). Given a vector $x = (x\_i)$ we write
$$ \Delta x\_i = x\_{i+1} + x\_{i-1} - 2 x\_i $$
if $1 < i < n$.




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Given a vector $(x\_i)$, we have
$$ Lx\_i = \sum\_j q^{|i-j|} (x\_i - x\_j) = b\_i x\_i - \sum\_j q^{|i-j|} x\_j ,$$
where
$$ b\_i = \sum\_j q^{|i-j|} = \frac{1 + q - q^i - q^{n+1-i}}{1 - q} $$
Therefore, when $1 < i < n$, we have
$$ \begin{aligned}
 \Delta Lx\_i & = \Delta (b x)\_i - \sum\_j (q^{|i-j+1|}+q^{|i-j-1|}-2q^{|i-j|}) x\_j \\
 & = \Delta (b x)\_i - (q + q^{-1} - 2) \sum\_j q^{|i-j|} x\_j + (q^{-1} - q) x\_i \\
 & = \Delta (b x)\_i + (q + q^{-1} - 2) L x\_i - (q + q^{-1} - 2) b\_i x\_i + (q^{-1} - q) x\_i \\
 & = (q + q^{-1} - 2) L x\_i + b\_{i+1} x\_{i+1} + b\_{i-1} x\_{i-1} - ((q + q^{-1}) b\_i - (q^{-1} - q)) x\_i .
\end{aligned} $$
A short calculation shows that
$$ b\_{i+1} + b\_{i-1} = ((q + q^{-1}) b\_i - (q^{-1} - q)) $$
(which looks somewhat miraculous, but there must be some insightful explanation for that). Thus,
$$ \Delta Lx\_i = (q + q^{-1} - 2) L x\_i + b\_{i+1} (x\_{i+1} - x\_i) + b\_{i-1} (x\_{i-1} - x\_i) . $$
Suppose that $x\_i = L^\dagger y\_i$ for some vector $(y\_i)$ such that $\sum\_i y\_i = 0$. Then $L x\_i = L L^\dagger y\_i = y\_i$. Write $c = q + q^{-1} - 2$. We then have
$$ \Delta y\_i - c y\_i = b\_{i+1} (x\_{i+1} - x\_i) + b\_{i-1} (x\_{i-1} - x\_i) . $$
In particular, the following claim follows.



> 
> **Proposition 1:** If $1 < i < n$, $y\_{i-1} = y\_i = y\_{i+1} = 0$ and $x\_i = x\_{i+1}$, then $x\_{i-1} = x\_i$.
> 
> 
> 


The above result will serve as an induction step. To initiate the induction, we need to study the $i = n$, which is slightly different. In that case:
$$ \begin{aligned}
 Lx\_{n-1} - Lx\_n & = b\_{n-1} x\_{n-1} - b\_n x\_n - \sum\_j (q^{|n-j-1|}-q^{|n-j|}) x\_j \\
 & = b\_{n-1} x\_{n-1} - b\_n x\_n - (q^{-1} - 1) \sum\_j q^{|n-j|} x\_j + (q^{-1} - q) x\_n \\
 & = b\_{n-1} x\_{n-1} - b\_n x\_n + (q^{-1} - 1) L x\_n - (q^{-1} - 1) b\_n x\_n + (q^{-1} - q) x\_n \\
 & = (q^{-1} - 1) L x\_n + b\_{n-1} x\_{n-1} - (q^{-1} b\_n - (q^{-1} - q)) x\_n .
\end{aligned} $$
This time we have
$$ q^{-1} b\_n - (q^{-1} - q) = b\_{n-1} , $$
and hence
$$ Lx\_{n-1} - Lx\_n = (q^{-1} - 1) L x\_n + b\_{n-1} (x\_{n-1} - x\_n) . $$
Again we consider $x\_i = L^\dagger y\_i$ for some vector $(y\_i)$ such that $\sum\_i y\_i = 0$, and we write $d = q^{-1} - 1$. We then have
$$ (y\_{n-1} - y\_n) - d y\_n = b\_{n-1} (x\_{n-1} - x\_n) . $$
As a consequence, we have the following result.



> 
> **Proposition 2:** If $y\_{n-1} = y\_n = 0$, then $x\_{n-1} = x\_n$.
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For $y = e\_1 - e\_2 = (1, -1, 0, 0, \ldots)$, we immediately obtain the desired result.



> 
> **Corollary:** If $y = e\_1 - e\_2$ and $x = L^\dagger y$, then $x\_3 = x\_4 = x\_5 = \ldots = x\_n$. Consequently, $a\_{ij} = x\_i - x\_j = 0$ whenever $i, j \ge 3$.
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> 




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Another consequence of the above result is that if $L^\dagger = (u\_{ij})$, then $$u\_{i+1,j+1}-u\_{i,j+1}-u\_{i+1,j}+u\_{i,j} = 0$$ whenever $i + 1 < j$ or $j + 1 < i$. Moreover, it should be relatively easy to use Propositions 1 and 2 to evaluate $u\_{ij}$ explicitly, and in particular to prove that
$$ u\_{ij} = v\_{\max\{i,j\}} + v\_{\max\{n+1-i,n+1-j\}} + \tfrac{1}{4} |i - j| $$
when $i \ne j$, where $(v\_i)$ is an explicitly given vector (in terms of products/ratios of $b\_i$, I guess).




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Final remark: there is a corresponding result in continuous variable: the Green function for the operator $$L f(x) = \int\_0^1 e^{-q |x - y|} (f(x) - f(y)) dy$$ has zero mixed second-order derivative. The proof follows exactly the same line, and is in fact somewhat less technical.


