Download Eigenvalues of regular graphs

3/1/2015
Ma/CS 6b
Class 23: Eigenvalues in Regular Graphs
By Adam Sheffer
Recall: The Spectrum of a Graph

Consider a graph 𝐺 = 𝑉, 𝐸 and let 𝐴 be
the adjacency matrix of 𝐺.
◦ The eigenvalues of 𝐺 are the eigenvalues of 𝐴.
◦ The characteristic polynomial 𝜙 𝐺; 𝜆 is the
characteristic polynomial of 𝐴.
◦ The spectrum of 𝐺 is
𝜆1 , … , 𝜆𝑡
𝑠𝑝𝑒𝑐 𝐺 =
,
𝑚1 , … , 𝑚𝑡
where 𝜆1 , … , 𝜆𝑡 are the eigenvalues of 𝐴 and
𝑚𝑖 is the multiplicity of 𝜆𝑖 .
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Example: Spectrum
𝐴=
0
1
0
1
1
0
1
0
0
1
0
1
1
0
1
0
𝑣1
𝑣2
𝑣4
𝑣3
𝜆 −1 0 −1
−1 0
det 𝜆𝐼 − 𝐴 = det −1 𝜆
0 −1 𝜆 −1
−1 0 −1 𝜆
= 𝜆2 𝜆 − 2 𝜆 + 2 .
𝑠𝑝𝑒𝑐 𝐶4 =
0 2 −2
2 1 1
Slight Change of Notation
Instead of multiplicities, let 𝜆1 , … , 𝜆𝑛 be
the not necessarily distinct eigenvalues of
𝑛.
2 −1
 For example, if the spectrum is
,
2 2
we write 𝜆1 = 𝜆2 = 2 and 𝜆3 = 𝜆4 = −1
(instead of 𝜆1 = 2, 𝑚1 = 2, 𝜆2 = −1, 𝑚2
= 2).

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Recall: The Spectral Theorem

Theorem. Any real symmetric 𝑛 × 𝑛
matrix has real eigenvalues and 𝑛
orthonormal eigenvectors.
◦ By definition, any adjacency matrix 𝐴 is
symmetric and real.
◦ The algebraic and geometric multiplicities are
the same in this case.
◦ We have 𝜙 𝐴; 𝜆 = 𝑛𝑖=1 𝜆 − 𝜆𝑖 .
◦ The multiplicity of an
eigenvalue 𝜆 is
𝑛 − 𝑟𝑎𝑛𝑘 𝜆𝐼 − 𝐴 .
More Examples

We already derived the following:
◦ 𝑠𝑝𝑒𝑐 𝐾𝑛 =
◦ 𝑠𝑝𝑒𝑐 𝐾𝑛,𝑚

−1
𝑛−1
𝑛−1
1
0
=
𝑚+𝑛−2
𝑚𝑛
1
− 𝑚𝑛
1
Our next goal is to study regular graphs.
◦ Can you come up with an eigenvector of any
regular graph with 𝑛 vertices?
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Eigenvalues of Regular graphs

If 𝐴 is the adjacency matrix of a 𝑑-regular
graph, then any row of 𝐴 contains exactly
𝑑 1’s.
◦ Thus, the vector 1𝑛 = 1,1, … , 1 is an
eigenvector of 𝐴 with eigenvalue 𝑑.

Theorem. Let 𝐺 be a connected graph.
The eigenvalue of 𝐺 of largest absolute
value is the maximum degree if and only
if 𝐺 is regular.
Proof
𝐴 – 𝑛 × 𝑛 adjacency matrix of a graph 𝐺.
 Δ 𝐺 – the maximum degree of 𝐺.
 𝑥 = 𝑥1 , … , 𝑥𝑛 – eigenvector of
eigenvalue 𝜆 of largest absolute value.
 𝑥𝑗 = max |𝑥𝑖 | .

𝑖
𝜆 𝑥𝑗 =
𝐴𝑥
𝑗
=
𝑥𝑖
𝑣𝑖 ∈𝑁 𝑣𝑗
≤ deg 𝑣𝑗 𝑥𝑗 ≤ Δ 𝐺 𝑥𝑗 ,
 So 𝜆 ≤ Δ 𝐺 .
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

Δ 𝐺 – the maximum degree of 𝐺.
We proved that the absolute value of any
eigenvalue of 𝐴 is at most Δ 𝐺 , using
𝜆 𝑥𝑗 =
𝐴𝑥
𝑗
=
𝑥𝑖 ≤ deg 𝑣𝑗 𝑥𝑗 ≤ Δ 𝐺 𝑥𝑗 .
𝑣𝑖 ∈𝑁 𝑣𝑗

For equality to hold, we need
◦ deg 𝑣𝑗 = Δ 𝐺 .
◦ 𝑥𝑖 = 𝑥𝑗 for each 𝑣𝑖 ∈ 𝑁 𝑣𝑗 .

That is, 𝑣𝑗 and all of its neighbors are of degree
Δ 𝐺 . Repeating the same argument for a
neighbor 𝑣𝑖 implies that 𝑣𝑖 ’s neighbors are also
of degree Δ 𝐺 . We continue to repeat the
argument to obtain that the graph is regular.
Multiplicity
The previous proof also shows that any
eigenvector 𝑥1 , … , 𝑥𝑛 of the eigenvalue
𝑑 satisfies 𝑥1 = 𝑥2 = ⋯ = 𝑥𝑛 .
 Thus, the space of eigenvectors of the
eigenvalue 𝑑 is of dimension 1 (that is, 𝑑
has multiplicity 1).

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The Spectrum of the Petersen
Graph

The Petersen graph 𝐺 = 𝑉, 𝐸 is a 3regular graph with 10 vertices.
◦ We know that it has eigenvalue 3 with
eigenvector 110 .

To find the other eigenvalues, we notice
some useful properties:
◦ If 𝑢, 𝑣 ∈ 𝐸 then 𝑢 and 𝑣 have
no common neighbors.
◦ If 𝑢, 𝑣 ∉ 𝐸 then 𝑢 and 𝑣 have
exactly one common neighbor.
The Adjacency Matrix

The number of neighbors shared by 𝑣𝑖
and 𝑣𝑗 is 𝐴2 𝑖𝑗 . That is
𝐴2

𝑖𝑗
3,
= 0,
1,
if 𝑖 = 𝑗,
if 𝑖 ≠ 𝑗 and 𝑣𝑖 , 𝑣𝑗 ∈ 𝐸,
if 𝑖 ≠ 𝑗 and 𝑣𝑖 , 𝑣𝑗 ∉ 𝐸.
That is, 𝐴2 + 𝐴 − 2𝐼 = 1𝑛×𝑛 .
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The Additional Eigenvalues






We have 𝐴2 + 𝐴 − 2𝐼 = 1𝑛×𝑛 .
Since 3 is an eigenvalue with eigenvector 1𝑛 ,
the other eigenvectors are orthogonal to 1𝑛 .
Thus, for an eigenvector 𝑣 of eigenvalue 𝜆 ≠ 3:
1𝑛×𝑛 𝑣 = 0𝑛 .
That is, 𝐴2 + 𝐴 − 2𝐼 𝑣 = 0𝑛 .
If 𝑣 is an eigenvector of 𝜆 then we have
𝜆2 𝑣 + 𝜆𝑣 − 2𝑣 = 0𝑛 .
Thus, the additional eigenvalues of 𝐴 are 1, −2.
3 1 −2
𝑠𝑝𝑒𝑐 𝐺 =
.
1 𝑚2 𝑚3
The Multiplicities
1 −2
.
𝑚 2 𝑚3
10
 Recall that 𝑖=1 𝜆𝑖 = 𝑡𝑟𝑎𝑐𝑒 𝐴 = 0.
 That is, 3 + 𝑚2 − 2𝑚3 = 0.
 Combining this with 𝑚2 + 𝑚3 = 9 and
𝑚2 , 𝑚3 ≥ 0, we obtain the unique
solution 𝑚2 = 5, 𝑚3 = 4.
3 1 −2
 𝑠𝑝𝑒𝑐 𝐺 =
.
1 5 4

𝑠𝑝𝑒𝑐 𝐺 =
3
1
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Petersen Graph and 𝐾10

Problem. Can we partition the edges of
𝐾10 into three disjoint sets, such that
each set forms a Petersen graph?
10
= 45 edges, and the Petersen
2
graph has 15.
◦ In 𝐾10 every vertex is of degree 9, and in the
Petersen graph 3.
◦ 𝐾10 has
Disproof






Assume, for contradiction, that the partition
exists, and let 𝐴, 𝐵, 𝐶 be the adjacency matrices
of the three copies of the Petersen graph.
The adjacency matrix of 𝐾10 is 110×10 − 𝐼.
That is, 𝐴 + 𝐵 + 𝐶 = 110×10 − 𝐼.
𝑉𝐴 , 𝑉𝐵 – the vector subspaces of eigenvectors
corresponding to the eigenvalue 1 in 𝐴 and 𝐵.
We know that dim 𝑉𝐴 = dim 𝑉𝐵 = 5.
Since both 𝑉𝐴 and 𝑉𝐵 are orthogonal to 110 , they
are not disjoint (otherwise we would have a set
of 11 orthogonal vectors in ℝ10 ).
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Disproof (cont.)
𝐴, 𝐵, 𝐶 – the adjacency matrices of the three
copies of the Petersen graph in 𝐾10 .
 𝐴 + 𝐵 + 𝐶 = 1𝑛×𝑛 − 𝐼.
 𝑉𝐴 , 𝑉𝐵 – the vector subspaces of eigenvectors
corresponding to the eigenvalue 1 in 𝐴 and 𝐵.
 𝑉𝐴 and 𝑉𝐵 are not disjoint.
 Let 𝑧 ∈ 𝑉𝐴 ∩ 𝑉𝐵 . Since every vector in 𝑉𝐴 and 𝑉𝐵
is orthogonal to 1𝑛 , so is 𝑧.
 We have
𝐶𝑧 = 1𝑛×𝑛 − 𝐼 − 𝐴 − 𝐵 𝑧 = 0 − 𝑧 − 𝐴𝑧 − 𝐵𝑧
= −3𝑧.
 Contradiction since -3 is not an eigenvalue of 𝐶.

Four Things You Did Not Know
About The Petersen Graph
It has 15 edges and 2000 spanning trees.
 It is the smallest 3-regular graph of girth 5
(this is called a 3,5 -cage graph).
 It likes gardening, ballet, and building
airplane models.
 It has gotten divorced three times.

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Moore Graphs

A Moore graph is a graph that is 𝑑regular, of diameter 𝑘, and whose
number of vertices is
𝑘−1
𝑑−1 𝑖.
1+𝑑
𝑖=0

As can be easily checked, this is the
minimum possible number of vertices of
any graph of diameter 𝑘 and minimum
degree 𝑑.
Examples of Moore Graphs

A Moore graph is a graph that is 𝑑-regular, of
diameter 𝑘, and whose number of vertices is
𝑘−1
𝑑−1 𝑖.
1+𝑑
𝑖=0

What Moore graphs do we know?
◦ The Petersen graph is 3-regular, of diameter
2, and contains 1 + 3 1𝑖=0 3 − 1 𝑖 = 10
vertices.
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Examples of Moore Graphs

A Moore graph is a graph that is 𝑑-regular, of
diameter 𝑘, and whose number of vertices is
𝑘−1
𝑑−1 𝑖.
1+𝑑
𝑖=0


The Petersen graph is 3-regular, of diameter 2,
and contains 1 + 3 1𝑖=0 3 − 1 𝑖 = 10 vertices.
Is there a Moore graph that is 2-regular and of
diameter 2? 𝐶5
Moore Graphs of Diameter 2 and
Girth 5
Recall that the girth of a graph is the
length of the shortest cycle in it.
 Theorem. There exist 𝑑-regular Moore
graphs with diameter 2 and girth 5 only
for 𝑑 = 2,3,7, and possibly 57.

◦ The case of 𝑑 = 57 is an open problem.
◦ If it exists, it has 3250 vertices, and 92,625
edges.
The case of 𝑑 = 7
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
𝐺 = 𝑉, 𝐸 – a 𝑑-regular graph of diameter 2,
girth 5, and with
1
𝑉 =1+𝑑
𝑑−1
𝑖
= 1 + 𝑑2 .
𝑖=0

Since the girth is five, if 𝑣𝑖 , 𝑣𝑗 ∈ 𝐸 then 𝑣𝑖 and
𝑣𝑗 have no common neighbors.

Since the diameter is two, if 𝑣𝑖 , 𝑣𝑗 ∉ 𝐸 then 𝑣𝑖
and 𝑣𝑗 have exactly one common neighbor.
Thus, the adjacency matrix 𝐴 satisfies

𝐴2

𝑖𝑗
𝑑,
= 0,
1,
if 𝑖 = 𝑗,
if 𝑖 ≠ 𝑗 and 𝑣𝑖 , 𝑣𝑗 ∈ 𝐸,
if 𝑖 ≠ 𝑗 and 𝑣𝑖 , 𝑣𝑗 ∉ 𝐸.
That is, we have 𝐴2 + 𝐴 − 𝑑 − 1 𝐼 = 1𝑛×𝑛 .
Proof (cont.)
𝐺 = 𝑉, 𝐸 – a 𝑑-regular graph of diameter 2,
girth 5, and with 𝑉 = 1 + 𝑑 2 .
 𝐴2 + 𝐴 − 𝑑 − 1 𝐼 = 1𝑛×𝑛 .
 𝜆 ≠ 𝑑 – an eigenvalue of 𝐴 with eigenvector 𝑣.
Since 𝜆 ≠ 𝑑, 𝑣 is orthogonal to 1𝑛 . Thus
𝐴2 + 𝐴 − 𝑑 − 1 𝐼 𝑣 = 0𝑛 ,
or
2
𝜆 𝑣 + 𝜆𝑣 − 𝑑 − 1 𝑣 = 0.
 This implies that 𝜆2 + 𝜆 − 𝑑 + 1 = 0, so
1± 1+4 𝑑−1
1 ± 4𝑑 − 3
𝜆2,3 = −
=−
.
2
2
𝑑 𝜆2 𝜆3
 We thus have 𝑠𝑝𝑒𝑐 𝐺 =
.
1 𝑚2 𝑚3

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Finding the Multiplicities

We have 𝜆2 = −
1+ 4𝑑−3
𝑠𝑝𝑒𝑐 𝐺 =



2
𝑑
1
1− 4𝑑−3
, 𝜆3 = −
𝜆2 𝜆3
.
𝑚2 𝑚3
2
,
0 = 𝑡𝑟𝑎𝑐𝑒 𝐴 = 𝑑 + 𝜆2 𝑚2 + 𝜆3 𝑚3
𝑚2 + 𝑚3 𝑚3 − 𝑚2
=𝑑−
+
4𝑑 − 3.
2
2
Since 𝑚2 + 𝑚3 = 𝑛 − 1 = 𝑑 2 , we have
𝑑 2 − 2𝑑 = 𝑚3 − 𝑚2 4𝑑 − 3.
This can happen if either 𝑚2 = 𝑚3 or
4𝑑 − 3 = 𝑠 2 for some integer 𝑠.
If 𝑚2 = 𝑚3 than 𝑑 2 − 2𝑑 = 0, implying 𝑑 = 2.
The Case of 4𝑑 − 3 = 𝑠 2



𝑑 2 − 2𝑑 = 𝑚3 − 𝑚2 4𝑑 − 3.
Assume that 4𝑑 − 3 = 𝑠 2 for some integer 𝑠.
That is, 𝑑 = 𝑠 2 + 3 /4.
Substituting into the above equation:
𝑠 2 + 3 2 2𝑠 2 + 6
−
= 𝑚3 − 𝑚2 𝑠.
16
4
Setting 𝑚3 − 𝑚2 = 2𝑚3 − 𝑑 2 , we have
𝑠 4 + 6𝑠 2 + 9 − 8𝑠 2 − 24 = 32𝑚3 𝑠 − 𝑠 5 − 6𝑠 3 − 9𝑠.


𝑠 5 + 𝑠 4 + 6𝑠 3 − 2𝑠 2 + 9 − 32𝑚3 𝑠 = 15.
So 𝑠 must divide 15, and we get 𝑠 ∈ ± 1,3,5,15 ,
which implies 𝑑 ∈ 1,3,7,57 .
The case 𝑑 = 1 leads to 𝐾2 . Not a Moore graph.
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Recall: Independent Sets
Consider a graph 𝐺 = 𝑉, 𝐸 . An
independent set in 𝐺 is a subset 𝑉 ′ ⊂ 𝑉
such that there is no edge between any
two vertices of 𝑉′.
 Finding a maximum independent set in a
graph is a major problem in theoretical
computer science.

◦ No polynomial-time algorithm
is known.
Past Bounds
Let 𝐺 = (𝑉, 𝐸) be a graph.
 Already proved:

◦ 𝐺 has an independent set of size at least
𝑣∈𝑉
1
.
1 + deg 𝑣
𝑑
◦ If 𝐸 = 𝑉 ⋅ , then 𝐺 has an independent
2
set of size at least 𝑉 /2𝑑.
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An Upper Bound

Theorem. For a 𝑑-regular graph
𝐺 = 𝑉, 𝐸 with smallest (most negative)
eigenvalue 𝜆𝑛 , the size of the largest
𝑛
independent set of 𝐺 is at most
.
1−𝑑/𝜆𝑛

Example.
0 2 −2
.
2 1 1
4
◦ So at most 2 = 2.
◦ 𝑠𝑝𝑒𝑐 𝐶4 =
1−
𝑣1
𝑣2
𝑣4
𝑣3
−2
Recall: The Rayleigh Quotient

The Rayleigh quotient is 𝑅 𝐴, 𝑥 = 𝑅(𝑥)
=

𝑥 𝑇 𝐴𝑥
𝑥𝑇𝑥
for 𝑛 × 𝑛 matrix 𝐴 and 𝑥 ∈ ℝ𝑛 .
Lemma. Let 𝐴 be a real symmetric 𝑛 × 𝑛
matrix. Then 𝑅 𝑥 attains its maximum
and minimum at eigenvectors of 𝐴. (We do
not prove the lemma.)

Question. What is 𝑅 𝑥 when 𝑥 is an
eigenvector of eigenvalue 𝜆?
◦
𝑥 𝑇 𝐴𝑥
𝑥𝑇𝑥
=
𝑥 𝑇 𝜆𝑥
𝑥𝑇𝑥
= 𝜆.
◦ Thus, the min and max values of 𝑅 𝑥 are the
min and max eigenvalues of 𝐴.
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



𝜆𝑛 – the most negative eigenvalue of 𝐺.
𝑆 – a largest independent set of 𝐺.
1𝑆 = 𝑥1 , … , 𝑥𝑛 – a vector with 𝑥𝑖 = 1 if 𝑣𝑖 ∈ 𝑆
(otherwise 𝑥𝑖 = 0).
𝑦 = 𝑛1𝑆 − 1𝑛 ⋅ 𝑆 .
𝑦 𝑇 𝐴𝑦 = 𝑛2 ⋅ 1𝑆𝑇 𝐴1𝑆 − 2 𝑆 𝑛 ⋅ 1𝑆𝑇 𝐴1𝑛 + 𝑆

2
⋅ 1𝑇𝑛 𝐴1𝑛 .
Since 𝑆 is an independent set, we have
1𝑆𝑇 𝐴1𝑆 = 𝑖,𝑗∈S 𝐴𝑖𝑗 = 0.
Since 𝐺 is 𝑑-regular, 1𝑆𝑇 𝐴1𝑛 = 1𝑆𝑇 ⋅ 𝑑1𝑛 = 𝑑 𝑆 ,
and also 1𝑇𝑛 𝐴1𝑛 = 1𝑇𝑛 ⋅ 𝑑1𝑛 = 𝑑𝑛.
 Combining the above, we have
𝑦 𝑇 𝐴𝑦 = 0 − 2 𝑆 𝑛 ⋅ 𝑑 𝑆 + 𝑆 2 ⋅ 𝑑𝑛 = − 𝑆 2 𝑑𝑛.
𝑦 𝑇 𝑦 = 𝑛2 1𝑆𝑇 1𝑆 − 2 𝑆 𝑛1𝑇𝑆 1𝑛 + 𝑆 2 1𝑇𝑛 1𝑛
= 𝑛2 𝑆 − 2 𝑆 2 𝑛 + 𝑆 2 𝑛 = 𝑆 𝑛 𝑛 − 𝑆 .

Completing the Proof

By the lemma, we have
𝑦 𝑇 𝐴𝑦
≥ 𝜆𝑛 .
𝑦𝑇 𝑦

We have

Thus
𝑦 𝑇 𝐴𝑦 = − 𝑆 2 𝑑𝑛.
𝑦𝑇 𝑦 = 𝑆 𝑛 𝑛 − 𝑆 .
− 𝑆 2 𝑑𝑛
𝜆𝑛 ≤
𝑆 𝑛 𝑛− 𝑆
𝜆𝑛 𝑛 − 𝑆
≤ −𝑑 𝑆
→
=
−𝑑 𝑆
.
𝑛− 𝑆
𝑆 ≤
𝑛
1 − 𝑑/𝜆𝑛
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The End
17