CME 305: Discrete Mathematics and Algorithms
Instructor: Professor Amin Saberi ([email protected])
HW#4 – Due 03/18/11
1. In this problem, we use the
reflection principle
from the study of partial differential
equations to find the eigenvalues and eigenvectors of the path graph, i.e. two vertices
have degree one and all other vertices have degree two.
(a) Write the Laplacian matrix for the path graph on
N
vertices, and for the ring
graph (a simple cycle) on 2
N
vertices.
(b) Show that if
v
∈
R
N
is an eigenvector for the path graph, then
w
= (
v
1
, v
2
, . . . , v
n
, v
n
, v
n

1
, . . . , v
1
)
>
is an eigenvector for the ring graph on 2
N
vertices. Likewise, show that if
w
is
an eigenvector of the ring graph obeying
w
i
=
w
2
N
+1

i
for
i
= 1
, . . . , N
, then
v
= (
w
1
, . . . , w
n
)
>
is an eigenvector for the path graph.
(c) Recall the geometric argument from class showing that the eigenvectors of the
ring graph on
N
vertices are
(
x
k
)
i
= cos
2
πki
N
and
(
y
k
)
i
= sin
2
πki
N
Use this to find the eigenvalues and eigenvectors of the path graph on
N
vertices.
Hint:
Both
x
k
and
y
k
have the same eigenvalue.
So any vector in the space
spanned by these vectors is an eigenvector with the same eigenvalue.
(d) This trick is called the reflection principle. If a neighbor in a mirror is treated
exactly as a real neighbor, show how to set up mirrors next to a path graph in
order to make the connection between the ring and path graphs geometrically
obvious.
Solution:
(a) The Laplacian for the path graph is
L
=
1

1

1
2
.
.
.

1
.
.
.

1
.
.
.
2

1

1
1
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For the ring graph, the only difference is that the first and last vertices are
neighbors, so
L
=
2

1

1

1
2
.
.
.

1
.
.
.

1
.
.
.
2

1

1

1
2
(b) If
v
∈
R
N
is an eigenvector for the path graph, then
L
p
v
=
λv
If
w
= (
v
1
, v
2
, . . . , v
n
, v
n
, v
n

1
, . . . , v
1
)
>
, then (
L
r
v
)
i
=
λv
i
for
i
=
{
2
, . . . , N

1
, N
+ 2
, . . . ,
2
N

1
}
since the ring graph Laplacian is the same on these
entries as the path graph Laplacian, and the vectors are the same (up to a
reversal of order on the last half of the vector, which doesn’t matter). On the
first coordinate, we know that since
v
is an eigenvector of
L
p
,
v
1

v
2
=
λv
1
, so
(
L
r
w
)
1
= 2
w
1

w
2
N

w
2
= 2
v
1

v
2

v
1
=
v
1

v
2
=
λv
1
The argument is similar for the other endpoints, so
w
is an eigenvector of
L
r
.
Likewise, if
w
is an eigenvector of the ring graph obeying
w
i
=
w
2
N
+1

i
for
i
= 1
, . . . , N
, then we can repeat the same argument in reverse to show that
v
= (
w
1
, . . . , w
n
)
>
is an eigenvector for the path graph.
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 Spring '09
 Eigenvalue, eigenvector and eigenspace, vertices, Discrete Mathematics and Algorithms, path graph

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