the minimum number of distinct eigenvalues of a graph ... · unique shortest paths consider the...

The Minimum Number of Distinct Eigenvaluesof a Graph

Representing Saskatchewan: hard to spell, easy to draw

University of Regina’s Discrete Math Group: BahmanAhmadi, Fatemeh Alinaghipour, Robert Bailey, Michael

Cavers, Shaun Fallat, Karen Meagher, Shahla Nasserasr

University of Regina, Saskatchewan

CanaDAM, June 2013

The Discrete Math Group

Our Website is:http://www.math.uregina.ca/~kmeagher/DMRG.html

Publications:

1. Minimum number of distinct eigenvalues of graphs.2. The minimum rank of universal adjacency matrices.3. Generalized covering designs and clique coverings.

http://www.math.uregina.ca/~kmeagher/DMRG.html

The Problem

Let G be a graph,

the vertices of G are {1, 2, ..., n}.

The set of matrices compatible with G is

S(G) = {A ∈ Mn : A = AT , for i 6= j , aij 6= 0 iff {i , j} ∈ E(G)}.

Let q(A) denote the number of distinct eigenvalues of A, and

q(G) = min{q(A) : A ∈ S(G)}.

1. For a graph G, find q(G).2. Which graphs G have q(G) small or large?3. How does changing G change q(G)?

The Problem

Let G be a graph, the vertices of G are {1, 2, ..., n}.




q(G) = min{q(A) : A ∈ S(G)}.


The Problem



S(G) = {A ∈ Mn :

A = AT , for i 6= j , aij 6= 0 iff {i , j} ∈ E(G)}.


q(G) = min{q(A) : A ∈ S(G)}.


The Problem



S(G) = {A ∈ Mn : A = AT ,

for i 6= j , aij 6= 0 iff {i , j} ∈ E(G)}.


q(G) = min{q(A) : A ∈ S(G)}.


The Problem





q(G) = min{q(A) : A ∈ S(G)}.


The Problem





q(G) = min{q(A) : A ∈ S(G)}.

1. For a graph G, find q(G).

2. Which graphs G have q(G) small or large?3. How does changing G change q(G)?

The Problem





q(G) = min{q(A) : A ∈ S(G)}.

1. For a graph G, find q(G).2. Which graphs G have q(G) small

or large?3. How does changing G change q(G)?

The Problem





q(G) = min{q(A) : A ∈ S(G)}.

1. For a graph G, find q(G).2. Which graphs G have q(G) small or large?

3. How does changing G change q(G)?

The Problem





q(G) = min{q(A) : A ∈ S(G)}.


Minimum rank

Compatible matrices for a graph are well-studied.

The minimum rank of a graph G is

mr(G) = min{rank(A) : A ∈ S(G)}.

For a graph G

q(G) ≤ mr(G) + 1

Simple Results

For a graph G on n vertices

1. 1 ≤ q(G) ≤ n.2. q(G) = 1 if and only if G has no edges.3. For any integer n ≥ 2, we have q(Kn) = 2.4. For any integers n1, n2, . . . nk

(with at least one ni greater than 2)

q(k⋃

i=1

Kni ) = 2.

5. q(Cn) =⌈n

2

⌉.

Simple Results

For a graph G on n vertices1. 1 ≤ q(G) ≤ n.

2. q(G) = 1 if and only if G has no edges.3. For any integer n ≥ 2, we have q(Kn) = 2.4. For any integers n1, n2, . . . nk


q(k⋃

i=1

Kni ) = 2.

5. q(Cn) =⌈n

2

⌉.

Simple Results

For a graph G on n vertices1. 1 ≤ q(G) ≤ n.2. q(G) = 1 if and only if G has no edges.

3. For any integer n ≥ 2, we have q(Kn) = 2.4. For any integers n1, n2, . . . nk


q(k⋃

i=1

Kni ) = 2.

5. q(Cn) =⌈n

2

⌉.

Simple Results

For a graph G on n vertices1. 1 ≤ q(G) ≤ n.2. q(G) = 1 if and only if G has no edges.3. For any integer n ≥ 2, we have q(Kn) = 2.

4. For any integers n1, n2, . . . nk(with at least one ni greater than 2)

q(k⋃

i=1

Kni ) = 2.

5. q(Cn) =⌈n

2

⌉.

Simple Results

For a graph G on n vertices1. 1 ≤ q(G) ≤ n.2. q(G) = 1 if and only if G has no edges.3. For any integer n ≥ 2, we have q(Kn) = 2.4. For any integers n1, n2, . . . nk


q(k⋃

i=1

Kni ) = 2.

5. q(Cn) =⌈n

2

⌉.

Unique Shortest Paths

Consider the following graph G:

v1 v2 v3 v4 v5v6

v7

1. There is one unique path of length 5 from v6 to v1.2. Let A ∈ S(G), the entry in the (v1, v6)-position of Ai will be

0 if i < 5.3. The (v1, v6)-position of A5 will be the product of terms from

A,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7

1. There is one unique path of length 5 from v6 to v1.

2. Let A ∈ S(G), the entry in the (v1, v6)-position of Ai will be0 if i < 5.

3. The (v1, v6)-position of A5 will be the product of terms fromA,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.

4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7

1. There is one unique path of length 5 from v6 to v1.2. Let A ∈ S(G),

the entry in the (v1, v6)-position of Ai will be0 if i < 5.





v1 v2 v3 v4 v5v6

v7


0 if i < 5.





v1 v2 v3 v4 v5v6

v7



A,

(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7



A,(namely a1,2a2,3a3,4a4,5a5,6).

Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7



A,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.




v1 v2 v3 v4 v5v6

v7



A,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.

5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7



A,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.

6. Thus q(A) ≥ 6.



v1 v2 v3 v4 v5v6

v7



A,(namely a1,2a2,3a3,4a4,5a5,6). Thus it will be non-zero.4. So A0, A1, . . . , A5 are linearly independent.5. The minimal polynomial of A has degree at least 6.6. Thus q(A) ≥ 6.

A Lower Bound

TheoremIf there are vertices u, v in a connected graph G at distance d

and the path of length d from u to v is unique, then

q(G) ≥ d + 1.

1. If Pn is a path on n vertices than q(Pn) = n.2. If T is a tree, then q(T ) ≥ diam(T ) + 1.

A Lower Bound

TheoremIf there are vertices u, v in a connected graph G at distance dand the path of length d from u to v is unique,

then

q(G) ≥ d + 1.


A Lower Bound

TheoremIf there are vertices u, v in a connected graph G at distance dand the path of length d from u to v is unique, then

q(G) ≥ d + 1.


A Lower Bound


q(G) ≥ d + 1.

1. If Pn is a path on n vertices than q(Pn) = n.

2. If T is a tree, then q(T ) ≥ diam(T ) + 1.

A Lower Bound


q(G) ≥ d + 1.


Another Example

v4 u4 v3 u3 v1

u1v2 u5 v5 u6 v6

u2

This graph has 6 distinct eigenvalues.

v4 u4 v3 u3

v1

u1v2 u5 v5 u6 v6

u2

This graph has at least 10 distinct eigenvalues.

Graphs with two distinct eigenvalues

Suppose G is a connected graph on n vertices with n ≥ 3 andq(G) = 2 then

I G has no pendant vertex.I There is no cut edge in the graph G.I Any two non-adjacent vertices of G must have at least two

common neighbors.

For any n > 1I q(Kn) = 2I q(Kn,n) = 2I q(Kn\e) = 2 where n > 3 and e is any edge.


Suppose G is a connected graph on n vertices with n ≥ 3

andq(G) = 2 then


common neighbors.





common neighbors.




I G has no pendant vertex.

I There is no cut edge in the graph G.I Any two non-adjacent vertices of G must have at least two

common neighbors.




I G has no pendant vertex.I There is no cut edge in the graph G.

I Any two non-adjacent vertices of G must have at least twocommon neighbors.





common neighbors.





common neighbors.

For any n > 1

I q(Kn) = 2I q(Kn,n) = 2I q(Kn\e) = 2 where n > 3 and e is any edge.




common neighbors.

For any n > 1I q(Kn) = 2

I q(Kn,n) = 2I q(Kn\e) = 2 where n > 3 and e is any edge.




common neighbors.

For any n > 1I q(Kn) = 2I q(Kn,n) = 2

I q(Kn\e) = 2 where n > 3 and e is any edge.




common neighbors.


Join of two graphs

Let G and H be graphs, then the join of G and H

is the graphwith

I vertex set V (G) ∪ V (H)I and edge set

E(G) ∪ E(H) ∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

K3,3 = K3 ∨ K3.

P3 ∨ P3

TheoremLet G be a connected graph, then q(G ∨G) = 2.

Join of two graphs

Let G and H be graphs, then the join of G and H is the graphwith

I vertex set V (G) ∪ V (H)

I and edge set

E(G) ∪ E(H) ∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

K3,3 = K3 ∨ K3.

P3 ∨ P3


Join of two graphs



E(G)

∪ E(H) ∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

K3,3 = K3 ∨ K3.

P3 ∨ P3


Join of two graphs



E(G) ∪ E(H)

∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

K3,3 = K3 ∨ K3.

P3 ∨ P3


Join of two graphs



E(G) ∪ E(H) ∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

K3,3 = K3 ∨ K3.

P3 ∨ P3


Join of two graphs



E(G) ∪ E(H) ∪ {{g, h} | g ∈ V (G), h ∈ V (H)}.

P3 ∨ P3


Outline of Proof

The goal of this proof is to construct a matrix P such that

Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√

I − P has all non-zero entries,3.√

P and√

I − P commute.

ThenI Q ∈ S(G ∨G) andI Q2 = I, which implies that Q has eigenvalues 1 and −1.

Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),

2.√


P and√

I − P commute.


Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√

I − P has all non-zero entries,

3.√

P and√

I − P commute.


Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√


P and√

I − P commute.


Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√


P and√

I − P commute.

ThenI Q ∈ S(G ∨G) and

I Q2 = I, which implies that Q has eigenvalues 1 and −1.

Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√


P and√

I − P commute.

ThenI Q ∈ S(G ∨G) andI Q2 = I,

which implies that Q has eigenvalues 1 and −1.

Outline of Proof


Q =

[ √P

√I − P√

I − P −√

P

]with:

1.√

P in S(G),2.√


P and√

I − P commute.


Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.

2. Set R = I − P then R is an invertible “M-matrix”, so it has asquare root. (old and non-trivial result).

3. This square root has the form I − Y ∗, where Y ∗ is the limitof the sequence generated by

Yi+1 =12(P + Y 2

i ), Y0 = 0.

4. Every Yi is a symmetric matric which commutes with√

P,so Y ∗ does too.

5. Since G is connected and A(G) is positive, so the entriesof Y ∗ are positive.

6. Every eigenvalue of Y ∗ is smaller than 1n , so I − Y ∗ is

non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.

2. Set R = I − P

then R is an invertible “M-matrix”, so it has asquare root. (old and non-trivial result).


Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.

2. Set R = I − P then R is an invertible “M-matrix”, so it has asquare root.

(old and non-trivial result).3. This square root has the form I − Y ∗, where Y ∗ is the limit

of the sequence generated by

Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.


3. This square root has the form I − Y ∗,

where Y ∗ is the limitof the sequence generated by

Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.


P,

so Y ∗ does too.5. Since G is connected and A(G) is positive, so the entries

of Y ∗ are positive.6. Every eigenvalue of Y ∗ is smaller than 1

n , so I − Y ∗ isnon-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.




6. Every eigenvalue of Y ∗ is smaller than 1n ,

so I − Y ∗ isnon-zero.

Outline of Proof

1. Set

P =2n − 1

4n2

(1n

A(G) + I)2

.



Yi+1 =12(P + Y 2

i ), Y0 = 0.





non-zero.

Graphs with a Large Number of Distinct Eigenvalues

For which graphs does q(G) = |V (G)| − 1?

I The maximum multiplicity for G must be 2.I Graphs with maximum multiplicity 2 have been

characterized,I they are only the “parallel paths”.



I The maximum multiplicity for G must be 2.

I Graphs with maximum multiplicity 2 have beencharacterized,

I they are only the “parallel paths”.




characterized,

I they are only the “parallel paths”.




characterized,I they are only the “parallel paths”.

Two Graphs with a Large Number of DistinctEigenvalues

v1 v2 v3 v4 v5

v6

v7

v1 v2 v3 v4 v5 v6

v7

We suspect that these are all the graphs withq(G) = |V (G)| − 1!

the minimum number of distinct eigenvalues of a graph ... · unique shortest paths consider the...

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