Combinatorial Matrix Theory...Combinatorial Matrix Theory Fusion of Graph Theory and Matrix Theory Background in Graph Theory. 6 vertices 9 edges Cycles in a graph Wayne Barrett (BYU

Combinatorial Matrix Theory

Wayne Barrett

August 30, 2013

Brigham Young University

Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 1 / 55


Fusion of Graph Theory and Matrix Theory

Background in Graph Theory.

6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph





6 vertices 9 edges

Cycles in a graph


Hamiltonian cycles

Definition

A Hamiltonian cycle is one that passes through all the vertices. A graph isHamiltonian if it contains a Hamiltonian cycle.

is Hamiltonian is not

A simple necessary condition for a graph to be Hamiltonian is that everyvertex have degree at least 2.


Hamiltonian cycles

Definition





Hamiltonian cycles

Definition


is Hamiltonian

is not



Hamiltonian cycles

Definition





Hamiltonian cycles

Definition





Hamiltonian cycles

Not sufficient K2,3


Hamiltonian cycles

Not sufficient K2,3

no cycle of odd lengthno 5-cycleno Hamiltonian cycle

No known necessary and sufficient condition(s) for a graph to have aHamiltonian cycle (other than the definition).

NP hard.


Hamiltonian cycles

Not sufficient K2,3

no cycle of odd length

no 5-cycleno Hamiltonian cycle


NP hard.


Hamiltonian cycles

Not sufficient K2,3

no cycle of odd lengthno 5-cycle

no Hamiltonian cycle


NP hard.


Hamiltonian cycles

Not sufficient K2,3



NP hard.


Hamiltonian cycles

Not sufficient K2,3



NP hard.


Hamiltonian cycles

Not sufficient K2,3



NP hard.


Two renowned 3-regular graphs

5-prism Petersen graph

















Spectral Graph Theory

Adjacency matrix of a graph G

1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.

Definition

If G is a graph on n vertices, its adjacency matrix A(G ) is the n × nsymmetric (0, 1)-matrix defined by

aij =

{1 if {i , j} is an edge of G

0 otherwise.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.

Definition


aij =


0 otherwise.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.

Definition


aij =


0 otherwise.




1

3 4

2

A(G ) =

0 1 1 01 0 1 01 1 0 10 0 1 0

.

Definition


aij =


0 otherwise.



Goal of spectral graph theory:

Determine properties of G from theeigenvalues of A(G ) or the eigenvalues of other matrices associated withG .

Example:Eigenvalues of Cn.

n

1

6 4

2

5

3 A(Cn) =

0 1 0 0 11 0 1 0 0

0 1 0 1. . .

0 1 0. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . 1

1 0 0 1 0

.



Goal of spectral graph theory: Determine properties of G from theeigenvalues of A(G ) or the eigenvalues of other matrices associated withG .


n

1

6 4

2

5

3 A(Cn) =

0 1 0 0 11 0 1 0 0

0 1 0 1. . .

0 1 0. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . 1

1 0 0 1 0

.





n

1

6 4

2

5

3 A(Cn) =

0 1 0 0 11 0 1 0 0

0 1 0 1. . .

0 1 0. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . 1

1 0 0 1 0

.





n

1

6 4

2

5

3

A(Cn) =

0 1 0 0 11 0 1 0 0

0 1 0 1. . .

0 1 0. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . 1

1 0 0 1 0

.





n

1

6 4

2

5

3 A(Cn) =

0 1 0 0 11 0 1 0 0

0 1 0 1. . .

0 1 0. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . 1

1 0 0 1 0

.


Cn

A(Cn) = Qn + Q−1n Qn =

0 1 0 0

0 0 1. . .

0 0. . . 0

0. . .

. . . 11 0 0 0

.

Qn is orthogonal so all eigenvalues have absolute value 1.

Char poly PQn(t) = tn − 1.

Eigenvalues of Qn are n roots of unity λj = e2πijn , j = 0, 1, 2, . . . , n − 1.

Qn is normal so has an orthonormal basis {u1, . . . , un} of eigenvectors.

Qnuj = λjuj .


Cn


0 1 0 0

0 0 1. . .

0 0. . . 0

0. . .

. . . 11 0 0 0

.





Qnuj = λjuj .


Cn


0 1 0 0

0 0 1. . .

0 0. . . 0

0. . .

. . . 11 0 0 0

.





Qnuj = λjuj .


Cn


0 1 0 0

0 0 1. . .

0 0. . . 0

0. . .

. . . 11 0 0 0

.





Qnuj = λjuj .


Cn


0 1 0 0

0 0 1. . .

0 0. . . 0

0. . .

. . . 11 0 0 0

.





Qnuj = λjuj .


Cn

The uj are also eigenvectors of Q−1n

Q−1n uj =

(1

λj

)uj .

So the uj are also eigenvectors of A(Cn).

A(Cn)uj = (Qn + Q−1n )uj =

(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .

FACT The eigenvalues of A(Cn) are 2 cos2πj

n, j = 0, . . . n − 1.


Cn

The uj are also eigenvectors of Q−1n Q−1

n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .


A(Cn)uj

= (Qn + Q−1n )uj =

(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .


A(Cn)uj = (Qn + Q−1n )uj

=(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj

= 2 cos 2πjn uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Cn


n uj =

(1

λj

)uj .



(λj + 1

λj

)uj

=(

e2πijn + e−

2πijn

)uj = 2 cos 2πj

n uj .


n, j = 0, . . . n − 1.


Laplacian matrix of a graph

1

2

3 4 L(G ) =

2 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

.

Definition

If G is a graph on n vertices, L(G ) is the n × n symmetric integer matrixdefined by

ìj =

deg(i) if i = j

−1 if i 6= j and {i , j} is an edge of G .

0 otherwise.



1

2

3 4

L(G ) =

2 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

.

Definition


ìj =

deg(i) if i = j


0 otherwise.



1

2

3 4 L(G ) =

2 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

.

Definition


ìj =

deg(i) if i = j


0 otherwise.



1

2

3 4 L(G ) =

2 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

.

Definition


ìj =

deg(i) if i = j


0 otherwise.



1

2

3 4 L(G ) =

2 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

.

Definition


ìj =

deg(i) if i = j


0 otherwise.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).

Eigenvalues: L(Cn)uj = (2In − A(Cn))uj =

(2− 2 cos

2πj

n

)uj .

FACT: Eigenvalues of L(Cn) are 2− 2 cos2πj

n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).


(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).


(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).

Eigenvalues:

L(Cn)uj = (2In − A(Cn))uj =

(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).

Eigenvalues: L(Cn)uj = (2In − A(Cn))uj

=

(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).


(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


L(Cn)

L(Cn) =

2 −1 0 0 −1−1 2 −1 0 0

0 −1 2 −1. . .

0 −1 2. . .

. . .. . .

. . .. . .

. . . 0

0. . .

. . .. . . −1

−1 0 0 −1 2

= 2In − A(Cn).


(2− 2 cos

2πj

n

)uj .


n, j = 0, . . . n − 1.


Weyl’s inequalities

If A, B are real symmetric n × n matrices, what is the relationship of theeigenvalues of A + B to those of A and B?

Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)

be the eigenvalues of the real symmetric n × n matrices A, B. Then

λj(A) + λ1(B) ≥ λj(A + B) ≥ λj(A) + λn(B).

Corollary

If B is positive semidefinite, λj(A + B) ≥ λj(A).




Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)



Corollary





Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)

be the eigenvalues of the real symmetric n × n matrices A, B.

Then


Corollary





Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)



Corollary





Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)



Corollary

If B is positive semidefinite,

λj(A + B) ≥ λj(A).




Theorem

[Weyl] Let

λ1(A) ≥ λ2(A) ≥ · · · ≥ λn(A)

λ1(B) ≥ λ2(B) ≥ · · · ≥ λn(B)



Corollary



Second Way to Define L(G)

Example

1

2

3 4



Example

1

2

3 4

E{1, 2} =

1 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

,



Example

1

2

3 4

E{1, 2} =

1 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

, E{1, 3} =

1 0 −1 00 0 0 0−1 0 1 0

0 0 0 0



Example

1

2

3 4

E{1, 2} =

1 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

, E{1, 3} =

1 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

E{2, 3} =

0 0 0 00 1 −1 00 −1 1 00 0 0 0

,Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 23 / 55


Example

1

2

3 4

E{1, 2} =

1 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

, E{1, 3} =

1 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

E{2, 3} =

0 0 0 00 1 −1 00 −1 1 00 0 0 0

, E{3, 4} =

0 0 0 00 0 0 00 0 1 −10 0 −1 1



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).

FACT. Let G be a graph on n vertices.

For each edge {r , s} ∈ G , let E{r , s} be the n × n matrix defined by

E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775

= L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.



E{1, 2}+ E{1, 3}+ E{2, 3}+ E{3, 4}

=

26641 −1 0 0−1 1 0 0

0 0 0 00 0 0 0

3775 +

26641 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

3775 +

26640 0 0 00 1 −1 00 −1 1 00 0 0 0

3775 +

26640 0 0 00 0 0 00 0 1 −10 0 −1 1

3775

=

26642 −1 −1 0−1 2 −1 0−1 −1 3 −1

0 0 −1 1

3775 = L(G).



E{r , s}ij =

8><>:1 if i = j = r or i = j = s

−1 if i = r , j = s or i = s, j = r

0 otherwise.

Then L(G) =X

{r,s}∈E(G)

E{r , s}.


Edge Deletion

Definition

Given a graph G and an edge e = {i , j} of G , G − e is the graph obtainedfrom G by deleting e.

Observation

1. L(G ) = L(G − e) + E{i , j} 2. E{i , j} is positive semidefinite.

Corollary

Let G be a graph, let e be an edge of G, and let

λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G )

λ1(G − e) ≥ λ2(G − e) ≥ · · · ≥ λn(G − e)

be the eigenvalues of L(G ) and L(G − e).

Then λj(G ) ≥ λj(G − e), j = 1, . . . , n.


Edge Deletion

Definition


Observation

1. L(G ) = L(G − e) + E{i , j}

2. E{i , j} is positive semidefinite.

Corollary


λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G )

λ1(G − e) ≥ λ2(G − e) ≥ · · · ≥ λn(G − e)


Then λj(G ) ≥ λj(G − e), j = 1, . . . , n.


Edge Deletion

Definition


Observation


Corollary


λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G )

λ1(G − e) ≥ λ2(G − e) ≥ · · · ≥ λn(G − e)


Then λj(G ) ≥ λj(G − e), j = 1, . . . , n.


Edge Deletion

Definition


Observation


Corollary


λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G )

λ1(G − e) ≥ λ2(G − e) ≥ · · · ≥ λn(G − e)


Then λj(G ) ≥ λj(G − e), j = 1, . . . , n.


Edge Deletion

Definition


Observation


Corollary


λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G )

λ1(G − e) ≥ λ2(G − e) ≥ · · · ≥ λn(G − e)


Then λj(G ) ≥ λj(G − e), j = 1, . . . , n.


Hamiltonian Graphs

Observation

If G is a Hamiltonian graph on n vertices, there exist edges e1, e2, . . . , ek

of G such that G − e1 − e2 − · · · − ek = Cn.

Corollary

Let G be a Hamiltonian graph on n vertices, let

λ1(G ) ≥ λ2(G ) ≥ · · · ≥ λn(G ) be the eigenvalues of L(G )

and λ1(Cn) ≥ λ2(Cn) ≥ · · · ≥ λn(Cn) be the eigenvalues of L(Cn).

Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


Hamiltonian Graphs

Observation



Corollary




Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


Hamiltonian Graphs

Observation



Corollary

Let G be a Hamiltonian graph on n vertices,

let



Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


Hamiltonian Graphs

Observation



Corollary




Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


Hamiltonian Graphs

Observation



Corollary




Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


Hamiltonian Graphs

Observation



Corollary




Then λj(G ) ≥ λj(Cn), j = 1, . . . , n − 1.


The Petersen Graph

1

2

3

9

5

4

106

7

8

L(P) =

3 −1 0 0 −1 −1 0 0 0 0−1 3 −1 0 0 0 −1 0 0 0

0 −1 3 −1 0 0 0 −1 0 00 0 −1 3 −1 0 0 0 −1 0−1 0 0 −1 3 0 0 0 0 −1−1 0 0 0 0 3 0 −1 −1 0

0 −1 0 0 0 0 3 0 −1 −10 0 −1 0 0 −1 0 3 0 −10 0 0 −1 0 −1 −1 0 3 00 0 0 0 −1 0 −1 −1 0 3

Eigenvalues: 5, 5, 5, 5, 2, 2, 2, 2, 2, 0.


The Petersen Graph

1

2

3

9

5

4

106

7

8

L(P) =

3 −1 0 0 −1 −1 0 0 0 0−1 3 −1 0 0 0 −1 0 0 0

0 −1 3 −1 0 0 0 −1 0 00 0 −1 3 −1 0 0 0 −1 0−1 0 0 −1 3 0 0 0 0 −1−1 0 0 0 0 3 0 −1 −1 0

0 −1 0 0 0 0 3 0 −1 −10 0 −1 0 0 −1 0 3 0 −10 0 0 −1 0 −1 −1 0 3 00 0 0 0 −1 0 −1 −1 0 3

Eigenvalues: 5, 5, 5, 5, 2, 2, 2, 2, 2, 0.


The Petersen Graph

1

2

3

9

5

4

106

7

8

L(P) =

3 −1 0 0 −1 −1 0 0 0 0−1 3 −1 0 0 0 −1 0 0 0

0 −1 3 −1 0 0 0 −1 0 00 0 −1 3 −1 0 0 0 −1 0−1 0 0 −1 3 0 0 0 0 −1−1 0 0 0 0 3 0 −1 −1 0

0 −1 0 0 0 0 3 0 −1 −10 0 −1 0 0 −1 0 3 0 −10 0 0 −1 0 −1 −1 0 3 00 0 0 0 −1 0 −1 −1 0 3

Eigenvalues: 5, 5, 5, 5, 2, 2, 2, 2, 2, 0.


L(C10)

Eigenvalues: 2− 2 cos2πj

10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.

If the Petersen graph is Hamiltonianλ5(P) ≥ λ5(C10)

⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π

5< 0 The Petersen graph is not Hamiltonian.


L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π



L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π



L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5

⇔cos 3π

5 ≥ 0

3π5

cos3π



L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π



L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π



L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π

5< 0

The Petersen graph is not Hamiltonian.


L(C10)


10, j = 0, . . . , 9.

Ordered:

4, 2− 2 cos4π

5(2), 2− 2 cos

3π

5(2), 2− 2 cos

2π

5(2), 2− 2 cos

π

5(2), 0.


⇔2 ≥ 2− 2 cos 3π

5⇔

cos 3π5 ≥ 0

3π5

cos3π



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√

5 > 1. The Petersen graph is not Hamiltonian


√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)

⇔2 ≥ 2− 2 cos 3π

5⇔

2 ≥ 2− 1−√

5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2

⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√

5 > 1.

The Petersen graph is not Hamiltonian


√5 > 1

2 cos3π

5=

1−√

5

2

λ5(P) ≥ λ5(C10)⇔

2 ≥ 2− 2 cos 3π5

⇔2 ≥ 2− 1−

√5

2⇔1−√

5

2≥ 0

⇔1 ≥√

5

But√



Spectral Graph Theory and Combinatorial Matrix Theory


Matrix information (eigenvalues) −→ Graph information.

Combinatorial Matrix Theory is the Reverse direction

Graph information −→ Matrix information

Symmetric matrices that arise in applications often have a specific pattern.

A graph is the best way to indicate the pattern of zeros and nonzeros insuch a matrix.










































Symmetric Matrix associated with a Graph

Sn - set of all n × n real symmetric matrices

Given A ∈ Sn, let G (A) be the graph with

vertex set V = {1, 2, . . . , n} and

edge set E = {{i , j}|aij 6= 0}

For any graph G , let S(G ) = {A ∈ Sn |G (A) = G}

1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n}

and



1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n} and



1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n} and



1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n} and



1

2

3 4b

a

c

dG

A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n} and



1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

paw





vertex set V = {1, 2, . . . , n} and



1

2

3 4b

a

c

dG A =

d1 a b 0a d2 c 0b c d3 d0 0 d d4

∈ S(G )

pawWayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 32 / 55

Questions

What possible eigenvalues can a matrix in S(G ) have?

What possible ranks can a matrix in S(G ) have?

In order to determine all possible ranks, it suffices to determine theminimum possible rank.

Every rank larger than the minimum rank is attainable.

Determining the minimum possible rank is called the minimum rankproblem.


Questions







Questions







Questions







Questions







Minimum Rank Problem

Let mr(G ) be the minimum rank over all matrices in S(G ).

Let M(G ) to be the maximum nullity over all matrices in S(G ).

Since mr(G ) + M(G ) = n, computing the minimum rank and themaximum nullity are equivalent problems.

Computing M(G ) or mr(G ) for a general graph is hard.

Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1







Easy for n < 6.

mr(paw) = 2

d1 a b 0a d2 c 0b c d3 d0 0 d d4

1 1 1 01 1 1 01 1 2 10 0 1 1


Extreme examples

1. Complete graph Kn

K K K K1 2 3 4

Jn =

1 1 . . . 11 1 . . . 1...

.... . .

...1 1 . . . 1

∈ S(Kn) =⇒ mr(G ) = 1, n ≥ 2

Kn is the only connected graph with minimum rank 1.


Extreme examples


K K K K1 2 3 4

Jn =

1 1 . . . 11 1 . . . 1...

.... . .

...1 1 . . . 1

∈ S(Kn) =⇒ mr(G ) = 1, n ≥ 2



Extreme examples


K K K K1 2 3 4

Jn =

1 1 . . . 11 1 . . . 1...

.... . .

...1 1 . . . 1

∈ S(Kn)

=⇒ mr(G ) = 1, n ≥ 2



Extreme examples


K K K K1 2 3 4

Jn =

1 1 . . . 11 1 . . . 1...

.... . .

...1 1 . . . 1

∈ S(Kn) =⇒ mr(G ) = 1, n ≥ 2



Extreme examples


K K K K1 2 3 4

Jn =

1 1 . . . 11 1 . . . 1...

.... . .

...1 1 . . . 1

∈ S(Kn) =⇒ mr(G ) = 1, n ≥ 2



Extreme examples

2. Pn . . .

Any A ∈ S(Pn) has the form A =

a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0

Deleting the first column and last row gives an invertible lower triangularmatrix =⇒ rank A ≥ n − 1 =⇒ mr(Pn) ≥ n − 1

L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)

Rows sum to 0, so L is singular =⇒ rank L = n − 1 =⇒ mr(Pn) = n − 1


Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0

Deleting the first column and last row gives an invertible lower triangularmatrix

=⇒ rank A ≥ n − 1 =⇒ mr(Pn) ≥ n − 1

L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0

Deleting the first column and last row gives an invertible lower triangularmatrix =⇒ rank A ≥ n − 1

=⇒ mr(Pn) ≥ n − 1

L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)

Rows sum to 0, so L is singular

=⇒ rank L = n − 1 =⇒ mr(Pn) = n − 1


Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)

Rows sum to 0, so L is singular =⇒ rank L = n − 1

=⇒ mr(Pn) = n − 1


Extreme examples

2. Pn . . .


a1 b1

b1 a2 b2

b2 a3. . .

. . .. . . bn−1

bn−1 an

, bi 6= 0


L(Pn) =

1 −1−1 2 −1

−1 2. . .

. . .. . . −1−1 1

∈ S(Pn)



Induced subgraphs

G v G − v P4

Definition

An induced subgraph H of a graph G is a graph that can be obtained fromG by successively deleting vertices.

Example

−→ −→


Induced subgraphs

G v

G − v P4

Definition


Example

−→ −→


Induced subgraphs

G v G − v

P4

Definition


Example

−→ −→


Induced subgraphs

G v G − v P4

Definition


Example

−→ −→


Induced subgraphs

G v G − v P4

Definition


Example

−→ −→


Induced subgraphs

G v G − v P4

Definition


Example

−→ −→


Induced subgraphs

G v G − v P4

Definition


Example

−→

−→


Induced subgraphs

G v G − v P4

Definition


Example

−→ −→


Trees

Definition

A graph T is a tree if

T is connected

T contains no cycle


Trees

Definition

A graph T is a tree if

T is connected

T contains no cycle


Path Cover Number

Definition

A path cover of a graph G is a set of disjoint, induced paths in G thatcover all the vertices.

The path cover number P(G ) is the minimum number of paths in a pathcover of G .

Theorem (Duarte-Johnson)

If T is a tree, then M(T ) = P(T ).


Path Cover Number

Definition






Path Cover Number

Definition






Path Cover Number

Definition






Unicyclic Graphs

Definition

A connected graph is unicyclic if it contains exactly one cycle.

1

3 4

2

Theorem (Barioli,Fallat,Hogben)

If G is unicyclic, then either M(G ) = P(G ) or else M(G ) = P(G )− 1.

Furthermore, the circumstances in which each occur are characterized.

Best reference: p. 49 of John Sinkovic’s Ph.D. dissertation


Unicyclic Graphs

Definition


1

3 4

2






Unicyclic Graphs

Definition


1

3 4

2






Unicyclic Graphs

Definition


1

3 4

2






Unicyclic Graphs

Definition


1

3 4

2






Outerplanar Graphs

Definition

A graph is outerplanar if it has an embedding in the plane with everyvertex adjacent to the unbounded face.

Example

Theorem (John Sinkovic)

If G is outerplanar, then M(G ) ≤ P(G ).


Outerplanar Graphs

Definition


Example




Outerplanar Graphs

Definition


Example




Induced Subgraphs and Minimum Rank

1

2 3

4 5

1

2

4 5

house house delete vertex 3 (P4)

A =

266664d1 u v 0 0u d2 w x 0v w d3 0 y0 x 0 d4 z0 0 y z d5

377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)

rank A ≥ rank A(3)⇒ mr(house) ≥ mr(P4) = 3



1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house)

A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)




1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)




1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)




1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)

rank A ≥ rank A(3)

⇒ mr(house) ≥ mr(P4) = 3



1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)

rank A ≥ rank A(3)⇒ mr(house) ≥ mr(P4)

= 3



1

2 3

4 5

1

2

4 5


A =


377775 ∈ S(house) A(3) =


377775

=

2664d1 u 0 0u d2 x 00 x d4 z0 0 z d5

3775 ∈ S(P4)




Observation

Let G be a graph and v a vertex of G . Then mr(G ) ≥ mr(G − v).

Observation

If H is an induced subgraph of G , then mr(G ) ≥ mr(H).



Observation

Let G be a graph and v a vertex of G . Then mr(G ) ≥ mr(G − v).

Observation

If H is an induced subgraph of G , then mr(G ) ≥ mr(H).


Minimum rank 2 graphs

Problem

Which graphs G have minimum rank ≤ 2?

Work with Hein van der Holst and Raphael Loewy.

Necessary condition. If mr(H) ≥ 3, then H cannot be induced in G .

Definition

If mr(H) ≥ 3, but mr(H − v) ≤ 2 for any vertex v of H, then H is aminimal forbidden subgraph for the minimum rank 2 problem.

Example: P4 mr(P4) = 3

P4 − v 2 possibilities up to isomorphism

P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition


Example: P4

mr(P4) = 3


P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1



Problem




Definition




P3 P2 ∪ K1

mr(P3) = 2 mr(P2 ∪ K1) = 1Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 44 / 55

Minimal Forbidden Subgraphs

1 2

3

4 5

A =

d1 a b 0 0a d2 c 0 0b c d3 x y0 0 x d4 00 0 y 0 d5

folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0

which is always invertible (det = −axy)

=⇒ rank B = 3 =⇒ rank A ≥ 3 =⇒ mr(folding stool) ≥ 3



1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0





1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0


=⇒ rank B = 3

=⇒ rank A ≥ 3 =⇒ mr(folding stool) ≥ 3



1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0


=⇒ rank B = 3 =⇒ rank A ≥ 3

=⇒ mr(folding stool) ≥ 3



1 2

3

4 5

A =


folding stool

B = A({2, 5}, {1, 4}) =


=

a b 0c d3 y0 x 0


=⇒ rank B = 3 =⇒ rank A ≥ 3 =⇒ mr(folding stool) ≥ 3Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 45 / 55


1 2

3

4 5

mr(folding stool) ≥ 3

Also mr(folding stool − v) ≤ 2 for each vertex v .

folding stool is a minimal forbidden subgraph.

A similar argument shows that

is a minimal forbidden subgraph.

dart

For a month or more in 2001 Raphi Loewy and I thought these were all.



1 2

3

4 5






dart




1 2

3

4 5






dart




1 2

3

4 5






dart




1 2

3

4 5






dart

For a month or more in 2001 Raphi Loewy and I thought these were all.Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 46 / 55

K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3

a4 b4 c4 x1 y1 z1 d7 0 0a5 b5 c5 x2 y2 z2 0 d8 0a6 b6 c6 x3 y3 z3 0 0 d9

∈ S(K3,3,3)

What is the smallest possible rank of A?

Case 1a d1 d2 d3 6= 0 rank A ≥ 31b d4 d5 d6 6= 0 rank A ≥ 31c d7 d8 d9 6= 0 rank A ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)




K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)




K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)


Case 1a d1 d2 d3 6= 0

rank A ≥ 31b d4 d5 d6 6= 0 rank A ≥ 31c d7 d8 d9 6= 0 rank A ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)


Case 1a d1 d2 d3 6= 0 rank A ≥ 3

1b d4 d5 d6 6= 0 rank A ≥ 31c d7 d8 d9 6= 0 rank A ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)


Case 1a d1 d2 d3 6= 0 rank A ≥ 31b d4 d5 d6 6= 0 rank A ≥ 3

1c d7 d8 d9 6= 0 rank A ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)




K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)

Case 2

For some integers i , j , k ,

1 ≤ i ≤ 3, 4 ≤ j ≤ 6, 7 ≤ k ≤ 9,

di = dj = dk = 0.

All cases are alike. So say d3 = d6 = d8 = 0


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)

Case 2 For some integers i , j , k ,

1 ≤ i ≤ 3, 4 ≤ j ≤ 6, 7 ≤ k ≤ 9,

di = dj = dk = 0.



K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)


1 ≤ i ≤ 3, 4 ≤ j ≤ 6, 7 ≤ k ≤ 9,

di = dj = dk = 0.

All cases are alike.

So say d3 = d6 = d8 = 0


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 d3 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 d6 z1 z2 z3


∈ S(K3,3,3)


1 ≤ i ≤ 3, 4 ≤ j ≤ 6, 7 ≤ k ≤ 9,

di = dj = dk = 0.

All cases are alike. So say d3 = d6 = d8 = 0Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 48 / 55

K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3

a4 b4 c4 x1 y1 z1 d7 0 0a5 b5 c5 x2 y2 z2 0 0 0a6 b6 c6 x3 y3 z3 0 0 d9

∈ S(K3,3,3)


1 ≤ i ≤ 3, 4 ≤ j ≤ 6, 7 ≤ k ≤ 9,

di = dj = dk = 0.



K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0

det B = 2c3c5z2 6= 0 =⇒ rank B = 3 =⇒ rank A ≥ 3

In either case, rank A ≥ 3 =⇒ mr(K3,3,3) ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0




K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0

det B = 2c3c5z2 6= 0

=⇒ rank B = 3 =⇒ rank A ≥ 3



K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0

det B = 2c3c5z2 6= 0 =⇒ rank B = 3

=⇒ rank A ≥ 3



K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0




K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0


In either case, rank A ≥ 3

=⇒ mr(K3,3,3) ≥ 3


K3,3,3

A =

d1 0 0 a1 a2 a3 a4 a5 a6

0 d2 0 b1 b2 b3 b4 b5 b6

0 0 0 c1 c2 c3 c4 c5 c6

a1 b1 c1 d4 0 0 x1 x2 x3

a2 b2 c2 0 d5 0 y1 y2 y3

a3 b3 c3 0 0 0 z1 z2 z3


∈ S(K3,3,3)

Let B =

0 c3 c5

c3 0 z2

c5 z2 0




K3,3,3 − v

K3,3,3 − v ∼= K3,3,2 for every v

A =

0 0 0 1 1 1 1 10 0 0 1 1 1 1 10 0 0 1 1 1 1 11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 1 1 1 2 0−1 −1 −1 1 1 1 0 2

∈ S(K3,3,2)

rank A = 2 =⇒ mr(K3,3,2) ≤ 2

So K3,3,3 is also a minimal forbidden subgraph for the minimum rank 2problem.


K3,3,3 − v

K3,3,3 − v ∼= K3,3,2 for every v

A =

0 0 0 1 1 1 1 10 0 0 1 1 1 1 10 0 0 1 1 1 1 11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 1 1 1 2 0−1 −1 −1 1 1 1 0 2

∈ S(K3,3,2)

rank A = 2 =⇒ mr(K3,3,2) ≤ 2



K3,3,3 − v

K3,3,3 − v ∼= K3,3,2 for every v

A =

0 0 0 1 1 1 1 10 0 0 1 1 1 1 10 0 0 1 1 1 1 11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 1 1 1 2 0−1 −1 −1 1 1 1 0 2

∈ S(K3,3,2)

rank A = 2

=⇒ mr(K3,3,2) ≤ 2



K3,3,3 − v

K3,3,3 − v ∼= K3,3,2 for every v

A =

0 0 0 1 1 1 1 10 0 0 1 1 1 1 10 0 0 1 1 1 1 11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 1 1 1 2 0−1 −1 −1 1 1 1 0 2

∈ S(K3,3,2)

rank A = 2 =⇒ mr(K3,3,2) ≤ 2



K3,3,3 − v

K3,3,3 − v ∼= K3,3,2 for every v

A =

0 0 0 1 1 1 1 10 0 0 1 1 1 1 10 0 0 1 1 1 1 11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 0 0 0 1 −11 1 1 1 1 1 2 0−1 −1 −1 1 1 1 0 2

∈ S(K3,3,2)

rank A = 2 =⇒ mr(K3,3,2) ≤ 2



Minimum Rank 2 Theorem

These are all the minimal forbidden subgraphs.

Theorem

Let G be a connected graph. Then mr(G ) ≤ 2 if and only if none of the 4graphs

P4 dart folding stool K3,3,3

is induced in G .

Reverse implication is much longer and depends on a characterization ofthe complements of the class of graphs that are {P4, dart, folding stool,K3,3,3}−free




Theorem



is induced in G .





Theorem



is induced in G .



Other formulations

Let G be a connected graph.

Then M(G ) ≥ n − 2 if and only if none of the 4 graphs P4, dart, foldingstool, K3,3,3 is induced in G .

M(G ) ≤ n − 3 if and only if one of the 4 graphs P4, dart, folding stool,K3,3,3 is induced in G .

Eigenvalue formulation: A matrix in S(G ) can have an eigenvalue ofmaximum multiplicity at most n − 3 if and only if one of the graphs P4,dart, folding stool, K3,3,3 is induced in G .


Other formulations






Other formulations






Other formulations






Minimum Rank 2 Theorem over any field

For any field F , one can define S(F ,G ) to be the class of all symmetricmatrices corresponding to G with entries in F .

For each such field there is a characterization of the graphs such thatmr(F ,G ) ≤ 2 in terms of forbidden subgraphs.

For any infinite field with char F 6= 2, the result is the same.

For Hermitian matrices, the result is the same except K3,3,3 drops off thelist.

For disconnected graphs, the result is the same except that one has toinclude two disconnected graphs

P3 ∪ K2 3K2

in the list of forbidden subgraphs.








P3 ∪ K2 3K2









P3 ∪ K2 3K2









P3 ∪ K2 3K2









P3 ∪ K2 3K2









P3 ∪ K2 3K2

in the list of forbidden subgraphs.Wayne Barrett (BYU ) Combinatorial Matrix Theory August 30, 2013 54 / 55

The minimum rank 3 Problem

What graphs G have mr(G ) ≤ 3?

What are the forbidden subgraphs?

Major question of my Fulbright proposal for leave in Israel in 2007.

Tracy Hall demolished any hope of solving this problem a week or twobefore I left by showing that the list of forbidden subgraphs is infinite.

Picture description



What graphs G have mr(G ) ≤ 3? What are the forbidden subgraphs?



Picture description






Picture description





Tracy Hall demolished any hope of solving this problem a week or twobefore I left

by showing that the list of forbidden subgraphs is infinite.

Picture description






Picture description






Picture description


Combinatorial Matrix Theory...Combinatorial Matrix Theory Fusion of Graph Theory and Matrix Theory Background in Graph Theory. 6 vertices 9 edges Cycles in a graph Wayne Barrett (BYU

Documents