Deterministic Versus Stochastic Consensus Dynamics on Graphs

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.

A Appendix

1.1 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:

$$\begin{aligned} c>\max \limits _{i}\frac{b_{i}^{2}+a_{i}^{2}}{2a_{i}}, \end{aligned}$$

we have that:

$$\begin{aligned} \sqrt{(a_{i}-c)^{2}+b_{i}^{2}}<c\quad \text {for all } i. \end{aligned}$$

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:

$$\begin{aligned} \mathbf {s}(t)\rightarrow \bar{s}(0)\,\mathbf{1} \quad \text {as}\quad t\rightarrow +\infty . \end{aligned}$$

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:

$$\begin{aligned} \mathbf {s}(t) = e^{-tL}\mathbf {s_{0}} \end{aligned}$$

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:

$$\begin{aligned} \mathbf {s}(t) - \bar{s}(0)\mathbf{1} = e^{-tL}(\mathbf {s_{0}}-\bar{s}(0)\,\mathbf{1}). \end{aligned}$$

(35)

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

$$\begin{aligned} L = PDP^{-1} \end{aligned}$$

(36)

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:

$$\begin{aligned} \mathbf {y}(t):=P^{-1}(\mathbf {s}(t)-\bar{s}(0)\,\mathbf{1}), \end{aligned}$$

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

$$\begin{aligned} \mathbf {y}'(t)&= \frac{d}{dt}[P^{-1}(\mathbf {s}(t) - \bar{s}(0)\,\mathbf{1})] = D\mathbf {y}(t). \end{aligned}$$

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

$$\begin{aligned} y_{i}(t) = y_{i}(0)e^{-\lambda _{i}t}. \end{aligned}$$

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:

$$\begin{aligned} y_1(t) = \left\langle \mathbf{s}(t) - \bar{s}(0)\,\mathbf{1}\,,\; \frac{\mathbf{1}}{\Vert \mathbf{1}\Vert }\right\rangle = \bar{s}(t) - \bar{s}(0)= 0 \end{aligned}$$

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

1.2 A.2 Convergence of Linear Systems

Lemma

Given a linear system defined by:

$$\begin{aligned} \mathbf {x'}=A\mathbf {x}\,, \quad \mathbf {x}(0)=\mathbf {x}_{0}. \end{aligned}$$

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

$$\begin{aligned} \lim _{t\rightarrow +\infty }\mathbf {x}(t)=\mathbf {u} \end{aligned}$$

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

Proof

For ease of notation we will write

$$\begin{aligned} \text {Spec}(A) = \{\lambda _{1},\ldots ,\lambda _{d}\} \end{aligned}$$

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:

$$\begin{aligned} \mathbb {R}^{n}= \bigoplus _{i=1}^{d}E_{\lambda _{i}}. \end{aligned}$$

(37)

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

$$\begin{aligned} \mathbf {x}(t) = e^{At}\mathbf {x}_{0}. \end{aligned}$$

By (37) we may write:

$$\begin{aligned} \mathbf {x}_{0} = \mathbf{e}_{1}+\ldots +\mathbf{e}_{d} \end{aligned}$$

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

$$\begin{aligned} e^{tA_{|E_{\lambda _{1}}}}&=\text {Id}\quad \text {for each } t\in \mathbb {R}. \end{aligned}$$

Therefore we must have that

$$\begin{aligned} e^{At}\mathbf {x}_{0}&= e^{At}{} \mathbf{e}_{1}+\ldots +e^{At}{} \mathbf{e}_{d} \\&=e^{tA_{|E_{\lambda _{1}}}}\mathbf{e}_{1}+\ldots +e^{tA_{|E_{\lambda _{d}}}}{} \mathbf{e}_{d}\\&=\mathbf{e}_{1}+e^{tA_{|E_{\lambda _{2}}}}\mathbf{e}_{2}+\ldots +e^{tA_{|E_{\lambda _{d}}}}{} \mathbf{e}_{d}. \end{aligned}$$

We now claim that for any $i\ge 2$:

$$\begin{aligned} \lim _{t\rightarrow +\infty }e^{tA_{|E_{\lambda _{i}}}}{} \mathbf{e}_{i}=0. \end{aligned}$$

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

$$\begin{aligned} A_{|E_{\lambda _{i}}} = \lambda _{i}Id + N_i \end{aligned}$$

where $N_I$ is nilpotent. Consequently

$$\begin{aligned} e^{tA_{|E_{\lambda _{i}}}}e_{i}&= e^{t(\lambda _{i}\text {Id} +N_i)}{} \mathbf{e}_{i} \\&=e^{t\text {Re}(\lambda _{i})}e^{t\text {Im}(\lambda _{i})i}\left( Id+t N_i+\ldots +\frac{1}{k!}t^{k}N_i^{k}\right) \mathbf{e}_{i}\\&\le C_{1}e^{t\text {Re}(\lambda _{i})}\left( Id+tN_i+\ldots +\frac{1}{k!}t^{k}N_i^{k}\right) \mathbf{e}_{i} \end{aligned}$$

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

$$\begin{aligned} \left\| e^{tA_{|E_{\lambda _{i}}}}{} \mathbf{e}_{i}\right\| \le Ce^{(\text {Re}(\lambda _{i})+\epsilon )t}\rightarrow 0 \quad \text {as}\quad t\rightarrow +\infty . \end{aligned}$$

(38)

We deduce that for $i\ge 2$:

$$\begin{aligned} \lim _{t\rightarrow +\infty }e^{tA_{|E_{\lambda _{i}}}}{} \mathbf{e}_{i}=0 \end{aligned}$$

which implies

$$\begin{aligned} \lim _{t\rightarrow +\infty }e^{At}\mathbf {x_{0}} \;=\; \lim _{t\rightarrow +\infty } \mathbf{e}_{1}+e^{tA_{|E_{\lambda _{2}}}}\mathbf{e}_{2}+\ldots +e^{tA_{|E_{\lambda _{d}}}}{} \mathbf{e}_{d} \;=\; \mathbf{e}_{1}. \end{aligned}$$

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

1.3 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:

$$\begin{aligned} V=\uplus _{i}^{k}U_{i}. \end{aligned}$$

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:

$$\begin{aligned} (U_{m},U_{n})\in E^{*} \text { if there exists } u\in U_{m} \text { and } v\in U_{n}\quad \text {with}\quad (u,v)\in E. \end{aligned}$$

(39)

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:

$$\begin{aligned} U_{m}\le ^{*}U_{n}\quad \text {if}\quad (U_{m},U_{n})\in E^{*}. \end{aligned}$$

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. $\square$