2/26/2015
Ma/CS 6b
Class 22: Spectral Graph Theory 2
𝑣1
𝑣2
𝑣4
𝑣3
Eigenvalues: -2,0,0,2.
𝑣1
𝑣2
𝑣3
Eigenvalues: − 2, 0, 2.
By Adam Sheffer
Chromatic number

Given a graph 𝐺, the chromatic number
𝜒(𝐺) is the minimum number of colors
required to color the vertices of 𝐺.
𝜒 𝐺 =3
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Coloring Graphs with Bounded
Degrees

Claim. Let 𝐺 = (𝑉, 𝐸) be a graph with
maximum degree 𝑘, then 𝜒 𝐺 ≤ 𝑘 + 1.

Proof.
◦ At each step choose an arbitrary uncolored
vertex 𝑣.
◦ Since 𝑣 has at most 𝑘 neighbors, one of the
𝑘 + 1 colors must be OK for 𝑣.
Example: 𝑘 + 1-coloring
1
2
3
4
5
6
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Sometimes We Cannot Do Better



𝐾𝑛 - complete graph
with 𝑛 vertices.
Max degree: 𝑛 − 1.
𝜒 𝐾𝑛 = 𝑛.



𝐶𝑛 - cycle of odd
length 𝑛.
Max degree: 2.
𝜒 𝐾𝑛 = 3.
Today’s Goal

The naïve coloring bound implies that the
simple graph below requires three colors.
◦ Using eigenvalues of graph, we can obtain a
better bound.
𝑣1
𝑣2
𝑣3
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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 distinct eigenvalues of
𝐴 and 𝑚𝑖 is the multiplicity of 𝜆𝑖 .
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
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Slight Change of Notation
Instead of multiplicities, let 𝜆1 , … , 𝜆𝑛 be
the not necessarily distinct eigenvalues of
𝐺.
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).

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 .
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Matrix Polynomials

𝐴 – 𝑛 × 𝑛 adjacency matrix of a graph 𝐺.

Consider the matrices 𝐼, 𝐴, 𝐴2 , … , 𝐴𝑛 .
Since each matrix has 𝑛2 coordinates, we can
2
consider each matrix as a point in ℝ𝑛 .
Since this is a space of dimension 𝑛2 and we
have 𝑛2 + 1 matrices, they cannot be
independent.
That is, there exist constants 𝑐0 , … , 𝑐𝑛2 such
2
that not all are zero and 𝑐0 𝐼 + 𝑛𝑖=1 𝑐𝑖 𝐴𝑖 = 0.
In other words, the polynomial
2
𝑓 𝑀 = 𝑐0 𝐼 + 𝑛𝑖=1 𝑐𝑖 𝑀𝑖 vanishes on 𝐴.




2
Minimum Polynomials



𝐴 – 𝑛 × 𝑛 adjacency matrix of a graph 𝐺.
The minimum polynomial of 𝐴 (or of 𝐺) is the
smallest degree polynomial with leading
coefficient 1 that vanishes on 𝐴.
Is the minimum polynomial always unique?
◦ Yes. If there are two polynomials of the
minimum degree, by subtracting them we
obtain a smaller degree polynomial.
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Minimum Polynomial Property

Claim. The minimum polynomial 𝑓 of 𝐴
must divide every other polynomial that
vanishes on 𝐴.
◦ Assume, for contradiction, that there exists a
polynomial 𝑔 that vanishes on 𝐴 such that 𝑓
does not divide 𝑔.
◦ The remainder of dividing 𝑔 by 𝑓 is a smaller
degree polynomial that vanishes on 𝐴,
contradicting the minimality of 𝑓.
◦ (𝑔 = 𝑓𝑞 + 𝑟)
The Minimum Polynomial
Theorem (without proof). The minimum
𝑡
polynomial of 𝐴 is Π𝑖=1
𝑀 − 𝜆𝑖 ⋅ 𝐼 , where
𝜆1 , … , 𝜆𝑡 are the distinct eigenvalues of 𝐴.
0 1 0 1
1 0 1 0 .
 Example. 𝐴 =
0 1 0 1
1 0 1 0 𝑣1
𝑣2
0 2 −2
𝑠𝑝𝑒𝑐 𝐶4 =
.
2 1 1

The minimum polynomial is
𝑀 𝑀 − 2𝐼 𝑀 + 2𝐼 .
𝑣4
𝑣3
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Recall: Diameter

The diameter of 𝐺 is the maximum
distance between two vertices of 𝐺
◦ That is, max 𝑑 𝑢, 𝑣 .
𝑢,𝑣

What is the diameter of the Petersen
graph? 2
Diameter and Eigenvalues


Theorem. The diameter of a connected graph 𝐺
is less than the number of distinct eigenvalues
of 𝐺.
Proof.
◦ 𝐴 – the adjacency matrix of 𝐺.
◦ 𝑘 – diameter of 𝐺.
◦ To prove the theorem, it suffices to prove that
𝐴0 , 𝐴1 , … , 𝐴𝑘 are linearly independent.
◦ The degree of the minimum polynomial 𝑓 of
𝐴 is the number of distinct eigenvalues.
◦ If 𝑓 is of degree 𝑟, then 𝐴0 , 𝐴1 , … , 𝐴𝑟 must
satisfy a linear relation.
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Linear Independence






𝐴 – the adjacency matrix of 𝐺.
𝑘 – diameter of 𝐺.
It remains to prove that 𝐴0 , 𝐴1 , … , 𝐴𝑘 are
linearly independent.
For that, we show that 𝐴𝑖 is not a linear
combination of 𝐴0 , 𝐴1 , … , 𝐴𝑖−1 , for any 𝑖 ≤ 𝑘.
Recall that 𝐴𝑖 contains the number of paths of
length 𝑖 between any two vertices of 𝐺.
Since 𝑖 ≤ 𝑘, there exist vertices 𝑣𝑥 , 𝑣𝑦 such that
𝑑 𝑣𝑥 , 𝑣𝑦 = 𝑖. The claim holds since 𝐴𝑖 𝑥𝑦 > 0
and 𝐴 𝑗
𝑥𝑦
= 0, for every 0 ≤ 𝑗 < 𝑖.
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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Eigenvalues and Induced Subgraphs

Theorem.
◦ Let 𝐺 be a graph with eigenvalues 𝜆min = 𝜆1
≤ 𝜆2 ≤ ⋯ ≤ 𝜆𝑛 = 𝜆max .
◦ Let 𝐺 ′ be an induced subgraph of 𝐺 with
eigenvalues 𝜆′min = 𝜆1′ ≤ 𝜆′2 ≤ ⋯ ≤ 𝜆′𝑡
= 𝜆′max .
◦ Then 𝜆min ≤ 𝜆′min ≤ 𝜆′max ≤ 𝜆max .

We now prove 𝜆′max ≤ 𝜆max . The min part is
proved symmetrically.

𝐴, 𝐴′ – the adjacency matrices of 𝐺, 𝐺 ′ .
We reorder the vertices of 𝐺 such that we first
have the vertices that are also in 𝐺 ′ (this does
not change the eigenvalues of 𝐴).
◦ 𝐴′ is obtained by removing the 𝑛 − 𝑡 last rows
and columns from 𝐴.
𝑧 ′ – an eigenvector such that 𝐴′ 𝑧 ′ = 𝜆′max .
𝑧 – the vector obtained by appending 𝑡 zeros at
the end of 𝑧 ′ .
By the previous lemma, for every vector 𝑥 ∈ ℝ𝑛
we have 𝜆min ≤ 𝑅 𝐴, 𝑥 ≤ 𝜆max .
𝑇
𝑧 ′ 𝐴′ 𝑧 ′ 𝑧 𝑇 𝐴𝑧
′
′ ′
𝜆max = 𝑅 𝐴 , 𝑧 = ′ 𝑇 ′ = 𝑇 ≤ 𝜆max .
𝑧 𝑧
𝑧 𝑧




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

0 1 0 1
𝐴= 1 0 1 0 .
0 1 0 1
1 0 1 0
◦ Eigenvalues: -2,0,0,2.
0 1 0
𝐴′ = 1 0 1
0 1 0
◦ Eigenvalues: − 2, 0, 2.
′
𝑇
𝑧 ′ 𝐴′ 𝑧 ′

𝑧 = 1, 2, 1 ,

𝑧 = 1, 2, 1,0 ,
𝑇
𝑧′ 𝑧′
𝑧 𝑇 𝐴𝑧
𝑧𝑇𝑧
=
𝑧′
𝑇
2𝑧 ′
𝑇
𝑧′ 𝑧′
𝑣1
𝑣2
𝑣4
𝑣3
𝑣1
𝑣2
= 2.
𝑣3
= 2.
Degrees and 𝜆max

Lemma. Let 𝐺 = 𝑉, 𝐸 be a graph. Than
𝐺’s maximum eigenvalue 𝜆max is
◦ at least the average degree
2𝐸
𝑉
, and
◦ at most the maximum degree of 𝐺.
0 1
 Example. 𝐴′ = 1 0
0 1
◦ Eigenvalues: − 2, 0,
0
1
0
2.
𝑣1
𝑣2
𝑣3
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Proof
𝐴 – the adjacency matrix of 𝐺.
 𝜆 – an eigenvalue of 𝐴 with eigenvector
𝑥 = 𝑥1 , … , 𝑥𝑛 . 𝑥𝑗 = max 𝑥𝑖 .

𝑖
𝑁 𝑣𝑗 – the set of neighbors of 𝑣𝑗 .
 We have

𝜆𝑥𝑗 = 𝐴𝑥
𝑗
=
𝑥𝑖 ≤ deg 𝑣𝑗 𝑥𝑗 ,
𝑣𝑖 ∈𝑁 𝑣𝑗
so 𝜆 is at most the max degree of 𝐺.
Proof (cont.)
𝐴 – the adjacency matrix of 𝐺 = 𝑉, 𝐸 .
 𝜆max – the max eigenvalue of 𝐴.

𝑥 𝑇 𝐴𝑥
We have 𝜆max ≥ 𝑅 𝑥 = 𝑇 for every
𝑥 𝑥
𝑛
𝑥∈ℝ .
 By taking 𝑥 = 1𝑛 = 1,1, … , 1 , we have

𝜆max
1𝑇𝑛 𝐴1𝑛 1
≥ 𝑇
=
𝑛
1𝑛 1𝑛
𝐴𝑖𝑗 =
𝑖,𝑗
2𝐸
.
𝑉
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Back to Graph Coloring

Theorem. Let 𝐺 = 𝑉, 𝐸 be a graph with
maximum eigenvalue 𝜆max . Then
𝜒 𝐺 ≤ 1 + 𝜆max .
0 1 0
 Example. 𝐴 = 1 0 1
0 1 0
𝑣1
◦ Eigenvalues: − 2, 0, 2.
 Coloring by max degree implies 3 colors.
 Coloring by 𝜆max implies 2 colors.
𝑣2
𝑣3
Proof

𝐺 = 𝑉, 𝐸 – a graph with 𝜒 𝐺 = 𝑘.
◦ We repeatedly remove vertices from 𝐺, if the
removal does not decrease 𝜒 𝐺 .
◦ We obtain an induced subgraph 𝐺 ′ = 𝑉 ′ , 𝐸′
such that 𝜒 𝐺 ′ = 𝑘, and for every 𝑣 ∈ 𝑉 ′ ,
we have 𝜒 𝐺 ′ − 𝑣 < 𝑘.
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The Minimum Degree of 𝐺 ′

We obtain an induced subgraph 𝐺 ′ = 𝑉 ′ , 𝐸′
such that 𝜒 𝐺 ′ = 𝑘, and for every 𝑣 ∈ 𝑉 ′ , we
have 𝜒 𝐺 ′ − 𝑣 < 𝑘.
𝛿(𝐺 ′ ) – the minimum degree of 𝐺 ′ .

Claim. 𝛿 𝐺 ′ ≥ 𝑘 − 1.

◦ Assume, for contradiction, that there exists
𝑣 ∈ 𝑉 ′ such that deg 𝑣 < 𝑘 − 1.
◦ We color 𝐺 ′ − 𝑣 using at most 𝑘 − 1 colors.
We can then use one of these 𝑘 − 1 colors to
also color 𝑣, contradicting 𝜒 𝐺 ′ = 𝑘.
Completing the Proof

𝐺 = 𝑉, 𝐸 – a graph with 𝜒 𝐺 = 𝑘.
◦ We repeatedly remove vertices from 𝑣, if the
removal does not decrease 𝜒 𝐺 .
◦ We obtain an induced subgraph 𝐺 ′ = 𝑉 ′ , 𝐸′
such that 𝜒 𝐺 ′ = 𝑘, and for every 𝑣 ∈ 𝑉 ′ ,
we have 𝜒 𝐺 ′ − 𝑣 < 𝑘.
2𝑉
𝑘 − 1 ≤ 𝛿 𝐺′ ≤
≤ 𝜆max .
𝐸
◦ Equivalently,
𝜒 𝐺 ′ = 𝑘 ≤ 1 + 𝜆max .
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The End
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