## Abstract

We study two agent based models of opinion formation—one stochastic in nature and one deterministic. Both models are defined in terms of an underlying graph; we study how the structure of the graph affects the long time behavior of the models in all possible cases of graph topology. We are especially interested in the emergence of a *consensus* among the agents and provide a condition on the graph that is necessary and sufficient for convergence to a consensus in both models. This investigation reveals several contrasts between the models—notably the convergence rates—which are explored through analytical arguments and several numerical experiments.

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## Acknowledgements

This work has been supported by the NSF Grants DMS-1515592 and RNMS11-07444 (KI-Net).

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Communicated by Irene Giardina.

## A Appendix

### A Appendix

### A.1 Results Concerning Symmetric and Strongly Connected Networks

### Proposition

The matrix *L* is a diagonal dominant matrix, its eigenvalues satisfy \(\text {Re}(\lambda _i)\ge 0\) and no eigenvalue \(\lambda _i\) is purely imaginary except zero.

### Proof

The diagonal entries of *L* satisfy \(\sigma _i=\sum _{j,j\ne i} a_{ij}\) where \(a_{ij}>0\), thus summing over each row will give zero and we deduce that *L* is diagonal dominant. By the Gershgorin disc theorem the eigenvalues \(\lambda _i\) are contained in the closed discs \(B(\sigma _i,\sigma _i)\) (see Fig. 3). Thus, the eigenvalues \(\lambda _i\) are either 0 or have a real part strictly positive. Moreover, 0 is always an eigenvalue of *L* as the constant vector \(\mathbf{1}=(1,\ldots ,1)^T\) is always an eigenvector of *L* associated with the eigenvalue \(\lambda =0\). \(\square \)

### Lemma

If *L* is irreducible then the eigenvalue 0 is simple.

### Proof

By Proposition 1 we have that the real parts of the eigenvalues of *L* are all nonnegative and there are no purely imaginary eigenvalues. We also note that 0 is an eigenvalue of L of at least multiplicity one because it is associated to \(\mathbf{1}\). Consider the matrix \(D = L-cId\) where \(c\in \mathbb {R}\) and notice that \(\lambda = a + ib\) is an eigenvalue of *L* if and only if \(\widehat{\lambda } = a - c + ib\) is an eigenvalue of *D*. Therefore, since 0 is an eigenvalue of *L* we must have that \(-c\) is an eigenvalue of *D*. If we choose *c* such that:

we have that:

Therefore \(-c\) is the eigenvalue of largest modulus of *D* and since *D* is irreducible we have by the Perron-Frobenius theorem that \(-c\) must be simple and since the eigenvalues of *D* are in a one to one correspondence with the eigenvalues of *L* we must have that 0 is a simple eigenvalue of *L* as desired. \(\square \)

### Corollary

Suppose *L* is irreducible and symmetric. Then the solution of the consensus model, \(\mathbf{s}(t)\), satisfies:

with \({\bar{s}}\) the average opinion (9). Moreover, \(|s_{i}(t)-\bar{s}(0)|\le Ce^{-\lambda _{2}t}\) where C depends only on the initial condition and \(\lambda _{2}\) is the second largest eigenvalue of *L*.

### Proof

The consensus model is a linear system, therefore its solution is given by:

where \(\mathbf {s_{0}} = (s_{1}(0),\ldots ,s_{N}(0))^{T}\). Note that since \(\mathbf{1}\) is an eigenvector of *L* corresponding to the eigenvalue 0, it is also an eigenvector of \( e^{-tL}\) corresponding to the eigenvalue of 1. Therefore we may write:

Also note that since *L* is diagonal dominant and symmetric that there exists *P* such that

where \(D = \text {diag}(0, \lambda _{2}, \lambda _{3},\ldots )\) and *P* is the matrix composed of the eigenvectors of *L*. Since *L* is symmetric these eigenvalues are real. By Proposition 1 and Lemma 1 there is exactly one zero eigenvalue and \(\lambda _{i}\) is strictly positive for \(2\le i\le n\). We now define:

this is the difference between the opinion at time t and the average opinion represented in the diagonal coordinate system. Notice that:

Therefore, since this is an uncoupled system of linear equations we must have that:

So for \(i\ge 2\) we must have that \(y_{i}(t)\rightarrow 0\) as \(t\rightarrow +\infty \) exponentially with rate at least \(\lambda _{2}\). To conclude, it remains to show that \(y_1(t) = 0\). Using that the eigenvectors of *L* form an orthogonal basis, the entry \(y_1(t)\) is given by:

since the mean value \(\bar{s}(t)\) is preserved over time. \(\square \)

### A.2 Convergence of Linear Systems

### Lemma

Given a linear system defined by:

Assume *A* has *d* distinct eigenvalues and a zero eigenvalue of multiplicity *m* with *m* linearly independent associated eigenvectors. If for all \(\lambda _{i}\in \text {Sp}(A)\) with \(i>m\) we have that \(Re(\lambda _{i})<0\) then

where \(\mathbf {u}\) is in the center subspace of *A*, \(E^{c}\).

### Proof

For ease of notation we will write

where \(m_{i}\) is the algebraic multiplicity of \(\lambda _{i}\). We will also write that \(\lambda _{1} = 0\). We know that given \(\lambda _{i}\) that we may find \(m_{i}\) linearly independent generalized eigenvectors of *A*, \(\{\mathbf{v}_{\lambda _{i}}^{1},\ldots ,\mathbf{v}_{\lambda _{i}}^{m_{i}}\}\). The generalized eigenspace corresponding to \(\lambda _{i}\) is given by \(E_{\lambda _{i}} = \text {span}\{\mathbf{v}_{\lambda _{i}}^{1},\ldots ,\mathbf{v}_{\lambda _{i}}^{m_{i}}\}\) and by the generalized eigenspace decomposition theorem

we can find a basis of \(\mathbb {R}^{n}\) consisting of generalized eigenvectors of *A*. Therefore, we may write:

We know by the fundamental theorem of linear systems that the solution to the system is given by:

By (37) we may write:

where for each \(1\le i \le d\) we have \(\mathbf{e}_{i}\in E_{\lambda _{i}}\). Notice that since we are given that there are \(m_{1}\) distinct eigenvectors associated with \(\lambda _{1} = 0\) that for each \(w\in E_{\lambda _{1}}\) we have that \(A\mathbf{w}=0\). Consequently we have that \(A_{|E_{\lambda _{1}}}=0\) which implies that \(A_{|E_{\lambda _{1}}}=0\) for each \(t\in \mathbb {R}\). Taking the matrix exponential of both sides yields

Therefore we must have that

We now claim that for any \(i\ge 2\):

Since we are writing \(\mathbf {x_{0}}\) with respect to the basis of generalized eigenvectors of *A* we have that

where \(N_I\) is nilpotent. Consequently

with \(C_1>0\). Therefore, since \(\text {Re}(\lambda _{i})<0\) and every coordinate of \((Id+tN_i+\ldots +\frac{1}{k!}t^{k}N_i^{k})\mathbf{e}_{i}\) is polynomial in *t* we must have that there exist \(C>0\) and \(0<\epsilon <-\text {Re}(\lambda _{i})\) such that

We deduce that for \(i\ge 2\):

which implies

Since \(\mathbf{e}_{1}\in E^{c}\) we have that \(\lim _{t\rightarrow +\infty }\mathbf {x}(t)\in E^{c}\) as desired. \(\square \)

### A.3 Decomposition into Strongly Connected Components

### Lemma

If L is the Laplacian of a directed graph \(G=(V,E)\) then by relabeling vertices L can be represented:

### Proof

We may partition the vertex set of G, *V*, into its strongly connected components \(\{U_{1},\dots ,U_{k}\}\), so that:

If we consider the set \(\{U_{1},\dots ,U_{k}\}\) as the vertex set of a new graph \(G^{*}\) with edge set \(E^{*}\) given by:

Then the graph \(G^{*}\) is a directed acyclic graph as *G* has been partitioned into strongly connected components; if a cycle existed all vertices included in the cycle would represent the same strongly connected component of *G* by (39) contradicting the partition of *V* into strongly connected components. Therefore there exists a topological ordering \(\le ^{*}\) on the vertex set of \(G^{*}\). This ordering is given by:

We only need to label the vertices of *V* in such a way that they respect the topological ordering. Then, this labeling of *V* produces *L* in the desired form. \(\square \)

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Weber, D., Theisen, R. & Motsch, S. Deterministic Versus Stochastic Consensus Dynamics on Graphs.
*J Stat Phys* **176, **40–68 (2019). https://doi.org/10.1007/s10955-019-02293-5

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### Keywords

- Dynamical systems
- Agent based modeling
- Stochastic modeling
- Deterministic modeling
- Network dynamics
- Consensus dynamics