9.3 Consensus for Individuals with No Stubbornness
9.3.2 Consensus for a Class of Logic Matrices
As discussed in Section 2.3.3, in many opinion dynamics problems, it is desirable to scale the opinions to be in some predefined interval, with[−1, 1]being popular. With x and xi defined as in Section 9.2.1, define R = {x : xik ∈ [−1, 1],∀i ∈ I,∀k ∈ J }. It will now be shown that, under the assumption given below, x(0) ∈ R ⇒ x(t) ∈ R, ∀t≥0 and Eq. (9.6) is satisfied.
Assumption 9.2. (a) The ith diagonal entry of the Laplacian matrix L, associated with
G = (V,E,A), satisfies lii ≤ 1,∀i ∈ I. (b) The kth diagonal of the logic matrix C satisfies
ckk >0, ∀k∈ J, andkCk∞ =1.
It is clear that Assumption 9.2 places constraints, including, separately, the size of the allowable oscillations, on both the logic matrix C and the network topology
L. This was alluded to in Remark 9.3, which showed that conditions for consensus were tied to bothC andL, and not just one or the other. Note also that ckk >0 is a reasonable assumption as it simply implies that thekth topic is positively dependent on itself. No restrictions are placed on the signs of the off-diagonal entries ofC, i.e. how two different topics are coupled. First, it will be shown thatRis an invariant set for the system Eq. (9.5), and then it will be shown that consensus is always reached under Assumptions 9.1 and 9.2.
Lemma 9.1. Suppose that Assumption 9.2 holds forCandG = (V,E,A). Suppose further that each individual’s opinion changes according to Eq. (9.2). Then, ifx(0)∈ R, the opinion vectorx(t)∈ Rfor all t≥0, where Rwas defined above Assumption 9.2.
Proof. The proof is presented in Appendix Section 9.7.
To conclude Section 9.3, it will be shown that all networks that contain a directed spanning tree and satisfy Assumptions 9.1 and 9.2 will reach a consensus of opinions. Theorem 9.2. Suppose that Assumptions 9.1 and 9.2 hold for a givenCandG = (V,E,A). Then, with each individual’s opinions evolving according to Eq. (9.2), for all initial conditions,
the social network will reach a consensus value on all topics with an exponential convergence rate if and only ifG has a directed spanning tree. The consensus value is given in Eq. (9.12)
Proof. The necessity of the directed spanning tree was explained in the proof of The-
orem 9.1. Before sufficiency is established, some properties will first be established regarding the eigenvalues of the matrix M =−Ind+ (In− L)⊗C. Consider a given
l∈ {1, . . . ,nd}. Thelth diagonal entry ofM ismll =−1+ (1−∑j∈Niaij)ckk for some
i∈ I andk ∈ J. The off-diagonal entries of thelth row,mlj, are given by(1−lii)ckp for allq∈ I,p∈ J,p6= k, andaiqckp for allq∈ I,p ∈ J. From Assumption 9.2, one has 0< ckk≤ 1 and∑j∈Ni aij = lii ≤1∀i∈ I ⇒0≤1−∑j∈Niaij ≤1. It follows that
mll ≤0 for all l ∈ {1, . . . ,nd}. Turning to Geršgorin disc theory (see Theorem A.3), define Rl(M) =∑ndj=1,j6=l|mlj|, i.e. the sum of the absolute values of the off-diagonal entries of thelth row ofM. One can then verify that
Rl(M) =|1−lii| d
∑
p=1,p6=k |ckp|+∑
j∈Ni aijcˆk = (1−lii) d∑
p=1,p6=k |ckp|+liicˆk, (9.23)where ˆck = ∑dp=1|ckp|is the sum of the absolute values of the entries in the kth row ofC. Note that 0≤1−lii≤1. Thus,
mll+Rl(M) =−1+ (1−lii)ckk+ (1−lii) d
∑
p=1,p6=k |ckp|+liicˆk = −1+ (1−lii) d∑
p=1,p6=k |ckp|+ckk +liicˆk (9.24)From Assumption 9.2, there holds kCk∞ = 1 and ckk > 0, which implies that ∑d
p=1,p6=k|ckp|+ckk =cˆk ≤1. It follows that
mll+Rl(M) =−1+cˆk ≤0. (9.25) This implies thatmll ≤ −Rl(M), and that this holds for alll∈ {1, . . . ,nd}. Thus, the Geršgorin discs ofM are all in the left half-plane. Specifically, the discs are either in the open left half-plane (mll <−Rl(M)) or touch the imaginary axis at the origin but do not enclose it (mll =−Rl(M), including the possibility thatmll =0). This implies that the eigenvalues of M either have strictly negative real part, or are equal to zero. It will now be established that if G has a directed spanning tree, and Assump- tion 9.2 holds, then consensus is achieved. As in the proof of Theorem 9.1, define A = (In− L)⊗C, with eigenvalue and eigenvector pair λi(A) and vi. Define µk,
uk and ϕl,wl as an eigenvalue and eigenvector pair of (In− L)andC, respectively,
k ∈ I andl∈ J. As in the proof of Theorem 9.1, one concludes that ifG contains a directed spanning tree, then λ1(M) = −1+λ1(A) = 0 is an eigenvalue of M, with
associated right eigenvector v1 = 1n⊗ζ. Above, it was established that the eigen-
values of M either have strictly negative real part, or are equal to zero. Note also that µ1 = 1 ⇒ λi(A) = µ1ϕl 6= 1, ∀l ∈ {2, . . . ,d} becauseRe(ϕl) < 1 according to
Assumption 9.1. This implies that proving λi(A) = µkϕl 6= 1 ∀k ∈ {2, . . . ,n} and
l∈ J is equivalent to satisfying Condition 9.11 in Theorem 9.1.
Because L has a single zero eigenvalue and all other eigenvalues have positive real part, it follows from Assumption 9.2 thatRe(µk)<1 fork 6=1. This implies that λi(A) =µkϕ16=1, for allk6=1, since ϕ1=1 according to Assumption 9.1. Consider
nowk ∈ {2, . . . ,n}andl ∈ {2, . . . ,d}. Becauselii ≤ 1, it follows from Theorem A.3 that |µk| ≤1. Because cll > 0,∀l∈ J,kCk∞ = 1, and ϕ1 = 1 is simple eigenvalue, then for l 6= 1, the lth Geršgorin disc of C is situated at cll with radius 1−cll. It follows that |ϕl| < 1 for l = {2, . . . ,d}. Thus |λi| = |µkϕl| ≤ |µk||ϕl| < 1 for all
k ∈ {2, . . . ,n}and l ∈ {2, . . . ,d}. In other words, Condition 9.11 of Theorem 9.1 is satisfied. The final consensus value is computed as in the proof of Theorem 9.1.