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The first eigenvector of a Distance Matrix
is nearly constant

Stefan Steinerberger Department of Mathematics, University of Washington, Seattle, WA 98195, USA steinerb@uw.edu
Abstract.

Let x1,,xnx_{1},\dots,x_{n} be points in a metric space and define the distance matrix Dn×nD\in\mathbb{R}^{n\times n} by Dij=d(xi,xj){D}_{ij}=d(x_{i},x_{j}). The Perron-Frobenius Theorem implies that there is an eigenvector vnv\in\mathbb{R}^{n} with non-negative entries associated to the largest eigenvalue. We prove that this eigenvector is nearly constant in the sense that the inner product with the constant vector 𝟙n\mathbb{1}\in\mathbb{R}^{n} is large

v,𝟙12v2𝟙2\left\langle v,\mathbb{1}\right\rangle\geq\frac{1}{\sqrt{2}}\cdot\|v\|_{\ell^{2}}\cdot\|\mathbb{1}\|_{\ell^{2}}

and that each entry satisfies viv2/4nv_{i}\geq\|v\|_{\ell^{2}}/\sqrt{4n}. Both inequalities are sharp.

Key words and phrases:
Distance Matrix, Perron-Frobenius, Graph Distance Matrix.
2010 Mathematics Subject Classification:
51K05, 30L15, 31E05.
S.S. is supported by the NSF (DMS-2123224) and the Alfred P. Sloan Foundation.

1. Introduction

This paper is the result of trying to understand a curious phenomenon concerning the graph distance matrix (Problem 1, introduced below). We give a relatively simple argument (Proposition 1) which explains, in a particular setting, why one might expect to observe such a phenomenon. Proposition 1 is a conditional result and it is not a priori clear why or how often (meaning for ‘how many’ graphs) the condition should be satisfied. A surprisingly large number of times it is indeed satisfied for reasons that are not presently understood and we will refer to this as Problem 2. The main contribution of the paper, besides introducing these two problems, is to show that there is indeed a general result (see §1.2) in metric spaces (X,d)(X,d) which is of intrinsic interest and explains the second phenomenon ‘up to constants’. For many graphs that constant factor seems to rather close to 2\sqrt{2} (the largest it could be) and this is currently not understood: we refer to §1.3.

1.1. The first problem.

We start by explaining the first problem which first arose implicitly in [20]: let G=(V,E)G=(V,E) be a finite, connected graph on nn vertices V={v1,,vn}V=\left\{v_{1},\dots,v_{n}\right\} and let Dn×nD\in\mathbb{R}^{n\times n} denote the graph distance matrix defined by Dij=d(vi,vj){D}_{ij}=d(v_{i},v_{j}) where d(vi,vj)d(v_{i},v_{j}) is the graph distance between the vertices vi,vjv_{i},v_{j}.

Problem 1. The linear system of equations

Dx=𝟙=(1,1,,1)nDx=\mathbb{1}=(1,1,\dots,1)\in\mathbb{R}^{n}

tends to have a solution xnx\in\mathbb{R}^{n} for ‘most’ graphs. Why is that?

It is certainly the case that for ‘many’ graphs, the matrix DD simply happens to be invertible and this case is not particularly interesting. However, when the graph GG exhibits additional structure and/or symmetries, then this is often reflected in the non-invertibility of DD, however, even then the linear system seems to have a solution for a constant right-hand side. We can make this quantitative: the database of graphs in Mathematica 12 contains 7642 connected graphs with 3|V|503\leq|V|\leq 50 vertices. Among those, the matrix DD is not invertible for 3155 of these graphs (this large percentage is due to the fact that the database contains more ‘structured’ than ‘generic’ graphs). However, Dx=𝟙Dx=\mathbb{1} has a solution in all but 5 cases.
Using the Monte Carlo method, we estimate that for Erdős-Renyi graphs G(n,p)G(n,p) the likelihood of a linear system associated to G(50,p)G(50,p) not having a solution is (uniformly in pp) less than 1%1\% and decays rapidly for larger nn. The phenomenon arose in [20] where solutions of Dx=𝟙Dx=\mathbb{1} are used to construct a notion of curvature on graphs. Naturally, the effectiveness of such a notion hinges on whether a solution of Dx=𝟙Dx=\mathbb{1} typically exists. Our starting point is a spectral existence result.

Proposition 1.

Suppose D0n×nD\in\mathbb{R}_{\geq 0}^{n\times n} has eigenvalues λ1>0λ2λn\lambda_{1}>0\geq\lambda_{2}\geq\dots\geq\lambda_{n} and eigenvector Dv=λ1vDv=\lambda_{1}v. If

1v,𝟙n2<|λ2|λ1λ2,1-\left\langle v,\frac{\mathbb{1}}{\sqrt{n}}\right\rangle^{2}<\frac{|\lambda_{2}|}{\lambda_{1}-\lambda_{2}},

then Dx=𝟙Dx=\mathbb{1} has a solution.

The right-hand side is positive but need not be large. Note that DD vanishes on the diagonal and thus the sum of the eigenvalues is 0. Given our assumption about the sign of the eigenvalues, this implies λ1=|λ2|++|λn|\lambda_{1}=|\lambda_{2}|+\dots+|\lambda_{n}|. We would thus expect that, generically, λ1|λ2|\lambda_{1}\gg|\lambda_{2}| which makes the right-hand side rather small. Hence, the applicability of Proposition 1 depends on whether v,𝟙/n\left\langle v,\mathbb{1}/\sqrt{n}\right\rangle is close to 1, whether the first eigenvector vv is ‘nearly constant’.

1.2. The Result

If Dn×nD\in\mathbb{R}^{n\times n} is a matrix with nonnegative entries, then the Perron-Frobenius theorem guarantees the existence of an eigenvector vnv\in\mathbb{R}^{n} with nonnegative entries associated to the largest eigenvector.

Theorem.

Let x1,,xnXx_{1},\dots,x_{n}\in X be points in a metric space (not all identical) and Dn×nD\in\mathbb{R}^{n\times n} denote the matrix given by Dij=d(xi,xj)D_{ij}=d(x_{i},x_{j}). Its Perron Frobenius eigenvector v0nv\in\mathbb{R}^{n}_{\geq 0}, normalized to unit length v=1\|v\|=1, satisfies

min1invi12nandv,𝟙n2.\min_{1\leq i\leq n}v_{i}\geq\frac{1}{2\sqrt{n}}\qquad\emph{and}\qquad\left\langle v,\mathbb{1}\right\rangle\geq\frac{\sqrt{n}}{\sqrt{2}}.
Refer to captionε\ll\varepsilon11
Figure 1. Sharpness: a star graph asymptotically for the first inequality, a metric space for the second inequality.

We note that, by Cauchy-Schwarz, all v0nv\in\mathbb{R}^{n}_{\geq 0} of unit length v=1\|v\|=1 satisfy

max1invi1nandv,𝟙n\max_{1\leq i\leq n}v_{i}\geq\frac{1}{\sqrt{n}}\qquad\mbox{and}\qquad\left\langle v,\mathbb{1}\right\rangle\leq\sqrt{n}

with equality if and only if all the entries of vv are 1/n1/\sqrt{n}.

The Theorem can be interpreted as saying that the Cauchy-Schwarz inequality is close to being attained (which happens if the vectors vv and 𝟙\mathbb{1} are linearly dependent, meaning that vv is close to a constant vector). The two inequalities in the Theorem are sharp: if we consider the star graph, then the minimum of the largest eigenvector is attained in the center and behaves asymptotically as described in the Theorem. For the second inequality we consider a metric space where n1n-1 points are distance ε\leq\varepsilon (for suitable ε\varepsilon depending on nn) from each other with an additional point at distance 11 from the first n1n-1 points. We refer to §2.3 for details.

1.3. The second problem.

Returning to our original motivation, the Theorem by itself does not quite explain the phenomenon introduced above. The constant 1/21/\sqrt{2}, while sharp, is not particularly close to 1. However, when returning to the graph setting, it does seem as if v,𝟙/n\left\langle v,\mathbb{1}/\sqrt{n}\right\rangle is usually quite close to 1 and often strikingly so. Among the 7642 connected graphs having 3n503\leq n\leq 50 vertices that are implemented in the Mathematica database, the average value of v,𝟙/n\left\langle v,\mathbb{1}/\sqrt{n}\right\rangle is 0.996. Path graphs have a comparatively small constant that can be computed exactly (we refer to prior work of Ruzieh & Powers [17]). It is a nice fact (see [17]) that the eigenvector behaves likes a hyperbolic cosine.

Proposition 2.

Consider the path graph PnP_{n} on nn vertices. Then

limnv(Pn),𝟙n=2sinhccc+coshcsinhc=0.98261,\lim_{n\rightarrow\infty}\left\langle v(P_{n}),\frac{\mathbb{1}}{\sqrt{n}}\right\rangle=\frac{\sqrt{2}\sinh{c}}{\sqrt{c}\sqrt{c+\cosh{c}\sinh{c}}}=0.98261\dots,

where cc is the positive number satisfying ctanhc=1c\tanh{c}=1.

A search in the Mathematica database identifies the ‘sun’ graphs (also sometimes known as trampoline graph, see [9]) as having the smallest constant within the library. A computation shows that the limiting behavior of the constant for sun graphs is given by (1/2+1/5)1/20.973(1/2+1/\sqrt{5})^{1/2}\sim 0.973.

Refer to captionRefer to caption
Figure 2. The sun graph S8S_{8} and a broom/comet.

Motivated by the extremal metric space (see Fig. 1), it is natural to consider so-called broom graphs or comets [14, 23]. They are obtained by taking a star and attaching a path to the central vertex. Numerics suggest that they might lead to a relatively small inner product v,𝟙/n\left\langle v,\mathbb{1}/\sqrt{n}\right\rangle (certainly quite a bit smaller than the sun graphs and possibly even close to 1/21/\sqrt{2}), however, their leading eigenvector does not seem to be known. We conclude by formally stating the second problem.

Problem 2. If G=(V,E)G=(V,E) is an unweighted, connected graph on nn vertices and vv is the leading eigenvector of its distance matrix DD, then it seem for most graphs |v,𝟙|(1ε)n\left|\left\langle v,\mathbb{1}\right\rangle\right|\sim(1-\varepsilon)\sqrt{n} with a very small value of ε\varepsilon. Can this be made precise and/or quantified?

1.4. Related results

Let (X,d)(X,d) be a metric space and x1,,xnXx_{1},\dots,x_{n}\in X. The largest eigenvector of the distance matrix then describes the global maximum of the energy functional E:nE:\mathbb{R}^{n}\rightarrow\mathbb{R} given by

E(a1,,an)=i,j=1nd(xi,xj)aiaji=1nai2.E(a_{1},\dots,a_{n})=\frac{\sum_{i,j=1}^{n}d(x_{i},x_{j})a_{i}a_{j}}{\sum_{i=1}^{n}a_{i}^{2}}.

This is naturally related to the study of energy integrals of the form

E(μ)=XXd(x,y)𝑑μ(x)𝑑μ(y),E(\mu)=\int_{X}\int_{X}d(x,y)d\mu(x)d\mu(y),

where μ\mu is a probability measure. The study of such measures is classical, we refer to the 1958 paper of Björck [6] as well as Alexander [1, 2], Alexander & Stolarsky [3], Gross [13], Wolf [24, 25] and references therein. Both in the continuous [6] as well as in the discrete case [21] the optimal measure μ\mu tends to be localized in a rather small set contained in the boundary. We observe that by switching from probability measures (essentially L1L^{1}) to L2L^{2}, the geometry of the problem changes substantially: having mass localized is actually rather expensive and it is beneficial to spread things out. Our main result makes this precise. The distance matrix for graphs was introduced by Graham & Pollak [12]. Papendieck & Recht [16] studied the largest possible entry of the principal eigenvector. Das [10, 11] considered the problem of bounding the extremal entries of the principal eigenvectors of the distance matrix. The survey of Aouchiche & Hansen [4] lists several more results about the principal eigenvalue and principal eigenvector, we also refer to the books of Stanić [19], Stevanović [22]. Distance matrices in the Euclidean setting were introduced by Schoenberg [18] (see also Young & Householder [26]) and studied in [5, 7, 8, 26]. We caution that the classical Euclidean Distance Matrix (Menger [15]) is different and has Dij=xixj2D_{ij}=\|x_{i}-x_{j}\|^{2}.

2. Proofs

2.1. Proof of Proposition 1

Proof.

We consider the problem of maximizing the quadratic form Q:nQ:\mathbb{R}^{n}\rightarrow\mathbb{R}

Q(v)=v,DvQ(v)=\left\langle v,Dv\right\rangle

subject to the restriction

v,𝟙=1.\left\langle v,\mathbb{1}\right\rangle=1.

Since DD has at least two positive entries, we conclude via v=(1/n,,1/n)v=(1/n,\dots,1/n) that

sup𝟙,v=1Q(v)>0.\sup_{\left\langle\mathbb{1},v\right\rangle=1}Q(v)>0.

We will now argue that the maximum is attained. Using the symmetry of DD and the Spectral Theorem, we can decompose any vector into eigenvectors of DD

v=v,v1v1+i=2nv,vivi.v=\left\langle v,v_{1}\right\rangle v_{1}+\sum_{i=2}^{n}\left\langle v,v_{i}\right\rangle v_{i}.

Recalling that λ1>0λ2λn\lambda_{1}>0\geq\lambda_{2}\geq\dots\geq\lambda_{n}, we have

Q(v)\displaystyle Q(v) =v,v12λ1+i=2nλiv,vi2v,v12λ1+λ2i=2nv,vi2\displaystyle=\left\langle v,v_{1}\right\rangle^{2}\lambda_{1}+\sum_{i=2}^{n}\lambda_{i}\left\langle v,v_{i}\right\rangle^{2}\leq\left\langle v,v_{1}\right\rangle^{2}\lambda_{1}+\lambda_{2}\sum_{i=2}^{n}\left\langle v,v_{i}\right\rangle^{2}
=v,v12λ1+λ2(v2v,v12)=(λ1λ2)v,v12+λ2v2.\displaystyle=\left\langle v,v_{1}\right\rangle^{2}\lambda_{1}+\lambda_{2}\left(\|v\|^{2}-\left\langle v,v_{1}\right\rangle^{2}\right)=(\lambda_{1}-\lambda_{2})\left\langle v,v_{1}\right\rangle^{2}+\lambda_{2}\|v\|^{2}.

We can decompose the leading eigenvector v1v_{1} (corresponding to λ1\lambda_{1}) into its projection on 𝟙/n\mathbb{1}/\sqrt{n} and a remaining vector v1nv_{1}^{*}\in\mathbb{R}^{n}

v1=v1,𝟙n𝟙n+v1.v_{1}=\left\langle v_{1},\frac{\mathbb{1}}{\sqrt{n}}\right\rangle\frac{\mathbb{1}}{\sqrt{n}}+v_{1}^{*}.

The Pythagorean theorem implies that v1v_{1}^{*} has mean value 0 and thus

1=v12=v1,𝟙n2+v12.1=\|v_{1}\|^{2}=\left\langle v_{1},\frac{\mathbb{1}}{\sqrt{n}}\right\rangle^{2}+\|v_{1}^{*}\|^{2}.

Then

v,v12=(v1,𝟙nv,𝟙n+v,v1)2\left\langle v,v_{1}\right\rangle^{2}=\left(\left\langle v_{1},\frac{\mathbb{1}}{\sqrt{n}}\right\rangle\left\langle v,\frac{\mathbb{1}}{\sqrt{n}}\right\rangle+\left\langle v,v_{1}^{*}\right\rangle\right)^{2}

and since the sum of entries of vv is 11, this implies

v,𝟙n=v,𝟙1n=1n\left\langle v,\frac{\mathbb{1}}{\sqrt{n}}\right\rangle=\left\langle v,\mathbb{1}\right\rangle\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}}

and hence, for vv restricted to the hyperplane x1++xn=1x_{1}+\dots+x_{n}=1,

v,v12=(v1,𝟙n1n+v,v1)2.\left\langle v,v_{1}\right\rangle^{2}=\left(\left\langle v_{1},\frac{\mathbb{1}}{\sqrt{n}}\right\rangle\frac{1}{\sqrt{n}}+\left\langle v,v_{1}^{*}\right\rangle\right)^{2}.

Note that the first term is merely a constant, independent of vv, and thus

v,v12v,v12+𝒪(v+1).\left\langle v,v_{1}\right\rangle^{2}\leq\left\langle v,v_{1}^{*}\right\rangle^{2}+\mathcal{O}(\|v\|+1).

Using the bound from above and the Cauchy-Schwarz inequality,

Q(v)\displaystyle Q(v) (λ1λ2)v,v12+λ2v2\displaystyle\leq(\lambda_{1}-\lambda_{2})\left\langle v,v_{1}\right\rangle^{2}+\lambda_{2}\|v\|^{2}
(λ1λ2)v,v12+λ2v2+𝒪(v+1)\displaystyle\leq(\lambda_{1}-\lambda_{2})\left\langle v,v_{1}^{*}\right\rangle^{2}+\lambda_{2}\|v\|^{2}+\mathcal{O}(\|v\|+1)
[(λ1λ2)v12+λ2]v2+𝒪(v+1).\displaystyle\leq\left[(\lambda_{1}-\lambda_{2})\|v_{1}^{*}\|^{2}+\lambda_{2}\right]\cdot\|v\|^{2}+\mathcal{O}(\|v\|+1).

Therefore, if

(λ1λ2)v12+λ2<0,\left(\lambda_{1}-\lambda_{2}\right)\|v_{1}^{*}\|^{2}+\lambda_{2}<0,

then there exists a global maximum that is attained because the quadratic form tends to -\infty in all directions of the hyperplane v,𝟙=1\left\langle v,\mathbb{1}\right\rangle=1. Recalling that

v12=1v1,𝟙n2\|v_{1}^{*}\|^{2}=1-\left\langle v_{1},\frac{\mathbb{1}}{\sqrt{n}}\right\rangle^{2}

we arrive at the desired condition. Let now vmaxv_{\max} denote the location of such a maximum. It maximizes a function (Q(v)Q(v)) subject to a constraint (vmax,𝟙=1\left\langle v_{\max},\mathbb{1}\right\rangle=1) and therefore, for some Lagrange multiplier λ\lambda\in\mathbb{R},

Q(vmax)=2Dvmax=λ𝟙.\nabla Q(v_{\max})=2Dv_{\max}=\lambda\cdot\mathbb{1}.

We note that since Dvmax,vmax=Q(vmax)>0\left\langle Dv_{\max},v_{\max}\right\rangle=Q(v_{\max})>0, we have Dvmax0Dv_{\max}\neq 0 and thus λ0\lambda\neq 0. After rescaling, this implies the existence of a solution of Dx=𝟙Dx=\mathbb{1}. ∎

2.2. Proof of the Theorem

Proof.

We start by noting that since not all the points x1,,xnXx_{1},\dots,x_{n}\in X are identical, there exists at least a pair which has positive distance and thus the matrix DD has at least two positive entries (this is to ensure that we do not argue about the eigenvalues of the zero matrix). In particular, λ1>0\lambda_{1}>0. Let v=(v1,,vn)nv=(v_{1},\dots,v_{n})\in\mathbb{R}^{n} denote the 2\ell^{2}-normalized Perron-Frobenius eigenvector all of whose entries are nonnegative. Then, for an arbitrary point xkx_{k}, we have

λ1=Dv,v\displaystyle\lambda_{1}=\left\langle Dv,v\right\rangle =i,j=1nd(xi,xj)vivji,j=1n(d(xi,xk)+d(xk,xj))vivj\displaystyle=\sum_{i,j=1}^{n}d(x_{i},x_{j})v_{i}v_{j}\leq\sum_{i,j=1}^{n}(d(x_{i},x_{k})+d(x_{k},x_{j}))v_{i}v_{j}
=(j=1nvj)i=1nd(xk,xi)vi+(i=1nvi)j=1nd(xk,xj)vj\displaystyle=\left(\sum_{j=1}^{n}v_{j}\right)\sum_{i=1}^{n}d(x_{k},x_{i})v_{i}+\left(\sum_{i=1}^{n}v_{i}\right)\sum_{j=1}^{n}d(x_{k},x_{j})v_{j}
=2(j=1nvj)i=1nd(xk,xi)vi.\displaystyle=2\left(\sum_{j=1}^{n}v_{j}\right)\sum_{i=1}^{n}d(x_{k},x_{i})v_{i}.

Using the kk-th row of the equation Dv=λ1vDv=\lambda_{1}v, we note that

i=1nd(xk,xi)vi=(Dv)k=λ1vk.\sum_{i=1}^{n}d(x_{k},x_{i})v_{i}=(Dv)_{k}=\lambda_{1}v_{k}.

Altogether, we conclude

0<λ12v1λ1vk=2λ1vkv,𝟙0<\lambda_{1}\leq 2\|v\|_{\ell^{1}}\cdot\lambda_{1}v_{k}=2\lambda_{1}v_{k}\cdot\left\langle v,\mathbb{1}\right\rangle

and thus, since λ1>0\lambda_{1}>0,

vk12v,𝟙.v_{k}\geq\frac{1}{2\left\langle v,\mathbb{1}\right\rangle}.

Then, using the Cauchy-Schwarz inequality,

vk12v,𝟙12v𝟙=12n.v_{k}\geq\frac{1}{2\left\langle v,\mathbb{1}\right\rangle}\geq\frac{1}{2\|v\|\cdot\|\mathbb{1}\|}=\frac{1}{2\sqrt{n}}.

This implies, in particular, v,𝟙/n1/2\left\langle v,\mathbb{1}/\sqrt{n}\right\rangle\geq 1/2. However, one can do a little bit better: by noting that 1kn1\leq k\leq n was arbitrary, we also obtain by summation over kk that

v,𝟙n2v,𝟙\left\langle v,\mathbb{1}\right\rangle\geq\frac{n}{2\left\langle v,\mathbb{1}\right\rangle}

implying v,𝟙2n/2\left\langle v,\mathbb{1}\right\rangle^{2}\geq n/2 as desired. ∎

Remark. The proof suggests that the only way for the argument to be close to sharp is if the triangle inequality is close to being an equation for most cases. It seems interesting, albeit nontrivial, to see whether one can obtain quantitative improvements in settings where this is not the case.

2.3. Examples.

We will now discuss three relevant examples.
Star Graph. The star graph GnG_{n} is comprised of a central vertex vv which has an edge to nn other vertices. The symmetry of the graph is reflected in its leading eigenvector which we assume to have values aa in the center and bb in all other vertices. This leads to the problem of maximizing

n(n1)b2+naba2+nb2max.\frac{n(n-1)b^{2}+nab}{a^{2}+nb^{2}}\rightarrow\max.

A computation shows that the optimal choice for aa is

a=12n+5161n3/2+a=\frac{1}{2\sqrt{n}}+\frac{5}{16}\frac{1}{n^{3/2}}+\dots

This is not too surprising: for the star graph, our argument is essentially optimal.
A metric space. As for the second inequality, let us suppose we have n+1n+1 elements of which nn are all distance ε\leq\varepsilon from each other with one element at distance 11. The eigenvalue problem then consists of maximizing the expression

Dv,vv,v=nab+𝒪(n2a2ε)na2+b2max,\frac{\left\langle Dv,v\right\rangle}{\left\langle v,v\right\rangle}=\frac{nab+\mathcal{O}(n^{2}a^{2}\varepsilon)}{\sqrt{na^{2}+b^{2}}}\rightarrow\max,

where aa is the value in n1n-1 points that are distance ε\leq\varepsilon from each other and bb is the value in the single point that is far away. For ε=0\varepsilon=0, this is maximized for a=1/2na=1/\sqrt{2n} and b=1/2b=1/\sqrt{2}. Basic stability theory implies that this persists for ε\varepsilon sufficiently small. Thus

v,𝟙n+1=(1+o(1))12.\left\langle v,\frac{\mathbb{1}}{\sqrt{n+1}}\right\rangle=(1+o(1))\cdot\frac{1}{\sqrt{2}}.
Path Graphs: Proof of Proposition 2.

We use an exact expression for the first eigenvector of the distance matrix of PnP_{n} due to Ruzieh & Powers [17]: they show that the entries vkv_{k} for 1kn1\leq k\leq n can be taken as

vk=cosh((kn+12)θ),v_{k}=\cosh{\left(\left(k-\frac{n+1}{2}\right)\theta\right)},

where θ\theta is the positive solution of

tanh(θ2)tanh(nθ2)=1n.\tanh\left(\frac{\theta}{2}\right)\tanh\left(\frac{n\theta}{2}\right)=\frac{1}{n}.

A Taylor series expansion shows that for xx around 0

tanhx=xx33+𝒪(x4)\tanh{x}=x-\frac{x^{3}}{3}+\mathcal{O}(x^{4})

and thus, as nn\rightarrow\infty, we see that θn1\theta\sim n^{-1}. Making the ansatz θ=(2c)/n\theta=(2c)/n for some unspecified constant cc, we see that

1n\displaystyle\frac{1}{n} =tanh(θ2)tanh(nθ2)\displaystyle=\tanh\left(\frac{\theta}{2}\right)\tanh\left(\frac{n\theta}{2}\right)
=tanh(cn)tanh(c)=cntanh(c)c3n3tanh(c)+\displaystyle=\tanh\left(\frac{c}{n}\right)\tanh\left(c\right)=\frac{c}{n}\tanh\left(c\right)-\frac{c^{3}}{n^{3}}\tanh\left(c\right)+\dots

Therefore, as nn\rightarrow\infty, we have that the implicit constant cnc_{n} in the ansatz θn=(2cn)/n\theta_{n}=(2c_{n})/n converges to the solution of the equation ctanh(c)=1c\cdot\tanh(c)=1. This solution is approximately c1.2c\sim 1.2. The limiting profile of the eigenvector is given by coshx\cosh{x} for cxc-c\leq x\leq c which, after some computation, implies the desired statement. ∎

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