Social Media Mining: An Introduction

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CUUS2079-03 CUUS2079-Zafarani 978 1 107 01885 3 January 13, 2014 16:45

3.1 Centrality 55

v 1 v 2 v 3

v 3

v 1

v 4 v 2 v 5

(a) A three node graph (b) A five node graph

Figure 3.2. Eigenvector Centrality Example.

AssumingCe=[u 1 u 2 u 3 ]T,

⎡

⎣

0 −λ 10

10 −λ 1

010 −λ

⎤

⎦

⎡

⎣

u 1

u 2

u 3

⎤

⎦=

⎡

⎣

0

0

0

⎤

⎦. (3.12)

SinceCe =[000]T, the characteristic equation is

det(A−λI)=

∣∣

∣∣

∣∣

0 −λ 10

10 −λ 1

010 −λ

∣∣

∣∣

∣∣=^0 , (3.13)

or equivalently,

(−λ)(λ^2 −1)−1(−λ)= 2 λ−λ^3 =λ(2−λ^2 )= 0. (3.14)

So the eigenvalues are(−

√

2 , 0 ,+

√

√ 2). We select the largest eigenvalue:

2. We compute the corresponding eigenvector:

⎡

⎢⎣

0 −

√

21 0

10 −

√

21

010 −

√

2

⎤

⎥⎦

⎡

⎣

u 1

u 2

u 3

⎤

⎦=

⎡

⎣

0

0

0

⎤

⎦. (3.15)

AssumingCevector has norm 1, its solution is

Ce=

⎡

⎣

u 1

u 2

u 3

⎤

⎦=

⎡

⎣

√^1 /^2

2 / 2

1 / 2

⎤

⎦, (3.16)

which denotes that nodev 2 is the most central node and nodesv 1 andv 3

have equal centrality values.