Download another version

aa362, April 2nd, 2007
Lecture 28 – Spectral Analysis of Simple Graphs
A matrix A can be written as:
 a11

= M
a
 m1
A = VT D V
K a1n   e1
0   a11 K am1 



O M 
O
 M O M 


L amn   0
en   a1n L amn 
= Left EigenVector * EigenValues * Right Eigenvector
This decomposition can be used when say you have a matrix of documents to word
occurrences. Such a matrix will have huge dimensions. However, you can simplify the
problem by decomposing that matrix, using Singular Value Decomposition, into
“concepts” represented by eigenvalues.
Lemma 1
Let G be a connected regular graph (where a regular matrix is one in which all vertices
have same degree) of degree d. Let A be the adjacency matrix of this graph. Then the
eigenvalues of A satisfy:
d = λ 1 > λ 2 ≥ Kλ n ≥ − d
Where
λ1 > λ2
Because the graph is connected
While
λn ≥ −d
Equality happens in this expression only if G is a bipartite graph.
Proof
Let U = (1, 1, …., 1) be an eigenvector of A with eigenvalue d.
Therefore:
AU = dU
Let X be another eigenvector of A. By the way any constant times an eigenvector is also
an eigenvector. So, by another we mean , an eigenvector that is no proportional to U.
aa362, April 2nd, 2007
Let Xmax be the maximum component of X. Let S = { I | Xi = Xmax}
In other words S is a subset of vertices of G where the value assigned to that vertex by X
is Xmax. There is at least one vertex outside S that will be connected to a vertex j inside S
because the graph is connected. The value assigned by X to such a vertex(ices) is < Xmax.
Now multiply X with jth row of A. The jth row of A contains d 1’s.
(A)(X) = λ (X) = λ Xj
Since there is at least 1 ‘d’ in Aj that was outside of S, therefore:
A Xj < d Xmax
λ Xj < d Xmax
Since Xj = Xmax
λ<d
We have proven one half of the lemma. The other half of the proof can be proven
similarly using bipartite graphs.
Lemma 2
If G is not connected and has exactly k components then:
d = λ 1 = λ 2 = Kλ k > λ k + 1 K
Proof by Example
Lets say we have two components:
B
A=  1
 0
0

B2 
There are no edges between B1 and B2. Suppose I have an eigenvector for B1 then we can
get an eigenvector for A. How? Say the eigenvector for B1 is simply (x1, x2, x3), then we
can extend it with zeroes and multiply it with A to get an eigenvector for A.
aa362, April 2nd, 2007
 B1

 0
 x1 
 x1 
 
 
 x2 
 x2 
 x3 
0   x3 
=
λ


 

B2   0 
0 
M 
M 
 
 
0 
0 
 x1 
 
 x2 
 x3 
Then   is the eigenvector for A.
0 
M 
 
0 
Eigenvalues of a graph that contains k, instead of two, components can be computed by
taking the union as a multiset of eigenvalues of the individual components.
From our earlier lemma, for each of these blocks, you are gonna have one of the
eigenvalues be d. For k components, you are going to have k copies of this eigenvalue.
Hence, the proof.
Lemma 3
Let G1 and G2 be graphs where G1 ⊆ G2 which means that G2 is obtained by adding edges
to G1. The max eigenvalue of G2 is at least as large as the max eigenvalue of G1.
Proof
Let A1 and A2 be adjacency matrices. Let λi(Ai) be the max eigenvalue of Ai.
Let v1 be the eigenvector associated with λ1(A1).
v1 has non-negative coordinated because A1 only has 1’s and 0’s in it.
Now we know that
Av = λ v
v Av = λ vTv
vTAv = λ
T
T
λ1(A1) = v Av ≤ v A2 v (Since A2 has a few more 1’s than A1)
T
T
λ1(A2) = max| x|= 1 x A2 x ≥ v A2 v ≥ λ 1 ( A1 )
T
aa362, April 2nd, 2007
Star Graphs
Graphs in which a single vertex has edges to all other vertices each one of which only
have degree 1. The corresponding adjacency matrix A for such a graph would be:
 0 1 1L1 


1 0 0L 0 

A=
 1 0 0L 0 


M

An eigenvalue and eigenvector can be found for this matrix by making the following
guess:
 n − 1  n − 1 
 n − 1

 


 0 1 1L1  
  n − 1
1


 1
 =  n − 1 = n − 1 1

 1 0 0L 0   1

 


 1 0 0L 0  

 M
M


 M


M




1
  n − 1 
1

 n − 1


1


 is the eigenvector and
Therefore, 1


M

1



n − 1 is the eigenvalue for this matrix.
Since the rank of A is 2, because all rows except the first one are identical, therefore,
there are n – 2 eigenvalues that are zero (trivial) and 2 non-trivial eigenvalues one of
which is n − 1 . To calculate the other we can do the following calculation:
 n − 1   − (n − 1) 
 n − 1

 


 0 1 1L1  
  n− 1 
−1


 −1
 =  n − 1  = − n − 1 − 1

 1 0 0L 0   − 1

 


 1 0 0L 0  

 M
M


 M


M




 −1
  n − 1 
−1

Therefore, − n − 1 is the other eigenvalue.
Lemma 4
Let G be a connected, undirected, irregular graph. Let dmin and dmax be its min and max
degrees. Then
aa362, April 2nd, 2007
max{d min , d max } ≤ λ 1 ≤ min{d max , 2 | E |}
Let A be the adjacency matrix and U = (1, 1, …1)
AU = degree vector ≥ dmin U
AU = degree vector ≥ min degree vector
UTAU ≥ dmin UTU
xT Ax u T Au
λ1 = max T ≥ T ≥ d min
x x
u u
dmin ≤ λ1
To prove: d max ≤ λ1
Let Gstar be star formed by highest degree vertex in G. Then from discussion on star
graphs:
λ1(Gstar) =
d max
What is the relationship between λ1(Gstar) and λ1(G)? Since one can only add edges to Gstar
to get G, therefore, λ1 can only increase.
Therefore:
λ1(G) ≥ λ1(Gstar) =
d max