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Unformatted text preview: c so it has 4 real eigenvalues
λ1 = 1, λ2 = λ3 =⅓, λ4 = ⅓ (thanks Matlab)
coincidence: λ2 = λ3 and λ4 =  λ2 (4cylce is a special graph)
noncoincidence (true for any lazy r.w. on an undirected connected graph):
– largest eigenvalue =1
– all others have absolute value <1
• left eigenvector corresponding to λ1 = 1?
• e1= (¼, ¼, ¼, ¼) is a left eigenvector for λ1
• Wow. Why would xt → e1 as t→∞? Proving xt → (¼, ¼, ¼, ¼) ?
• Recall
A= ⅓
⅓
0
⅓ ⅓0⅓
⅓⅓0
⅓⅓⅓
0⅓⅓ 2
3
1 4 • λ1 = 1, λ2 = λ3 =⅓, λ4 = ⅓
• e1= (¼, ¼, ¼, ¼)
• Proof (see also notes): choose eigenvectors e2, e3, e4 for λ2, λ3, λ4, so that {e1, e2, e3,
e4} is a basis of R4 (guaranteed by the spectral theorem, since A is symmetric)
• so x0= a1 e1 + a2 e2 + a3 e3+ a4 e4, for some a1, a2, a3 , a4
• Now xt= x0 At =
exact same proof for any
= a1 e1 At + a2 e2 At + a3 e3 At + a4 e4 At
lazy r.w. on any connected
= a1 e1 λ1t + a2 e2 λ2t + a3 e3 λ3t + a4 e4 λ4t
undirected graph.
→ a1 e1, as t→∞
• since e1= (¼, ¼, ¼, ¼) is a distribution, it must be that a1=1
• Hence xt→ (¼, ¼, ¼, ¼), as t→∞. Menu
• Random walks on graphs
• Markov Chains
• Examples:
– pagerank
– cardshuffling
– colorings General Setup
• Let G =(V, E) be a:
– (i) stronglyconnected weighted directed graph (i.e. there is a path from every vertex to every vertex), – with (ii) selfloop on every vertex (to avoid periodicities)
• The weight pij of every edge (i, j ) ∈ E represents a probability, namely:
• pij : probability of transitioning from i to j.
• Hence for all i: (i,j )∈E pij = 1.
• Transition matrix: A = [ pij ].
• In the example on the left: A= .9 .075 .025
.15 .8 .05
.25 .25 .5 General Setup
• Transition matrix: A = [ pij ].
• Still true that, if x0 is the distribution over vertices at time 0, then xt= x0 At.
• Claim: A...
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This note was uploaded on 09/07/2013 for the course EECS 18.410 taught by Professor Erikdemaine during the Spring '13 term at MIT.
 Spring '13
 ErikDemaine
 Algorithms

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