Solutions for the problems of problem set 1
1. Let the central vertex be v1 , its neighbors v2 , v3 , v4 , and
√ their neighbors
v5 , v6 , v7 , v8 , v9 , v10 . Then the eigenvector corresponding to 5 is
√ √ √
(3, 5, 5, 5, 1, 1, 1, 1, 1, 1).
√
Similarly, the eigenvector belonging to − 5 is
√
√
√
(3, − 5, − 5, − 5, 1, 1, 1, 1, 1, 1).
√
The eigenspace for eigenvalue 2 is
√ √ √
V√2 = {(0, 2a, 2b, 2c, a, a, b, b, c, c) | a + b + c = 0}.
√
The eigenspace belonging to − 2 is
√
√
√
V−√2 = {(0, − 2a, − 2b, − 2c, a, a, b, b, c, c) | a + b + c = 0}.
Finally, the eigenspace corresponding to 0 is
V0 = {(d, 0, 0, 0, a, −a − d, b, −b − d, c, −c − d)}.
(Sorry for not writing the vectors to pictures as I required in the problem, but
I’m not really good at drawing pictures.)
2. We show that the eigenvalues of Kn,n,n are 2n, −n(2) , 0(3n−3) . Indeed, the
constant 1 is an eigenvector belonging to 2n. If we have constant a, b and c
on the 3 clusters such that a + b + c = 0, then it is a 2 dimensional eigenspace
belonging to −n. If we have a vector x such that the sum of the entries on
each cluster separately is 0, then it is an eigenspace belonging to 0, and its
dimension is 3(n − 1).
3. (a) Let µ(G) = µ1 ≥ · · · ≥ µn be the eigenvalues of G. Then we have
n
∑
2
µ(G) ≤
µ2i (G) = 2e(G).
Hence µ(G) ≤
(b) We have
2
µ(G) =
√
i=1
2e(G).
(
−
n
∑
)2
µi
≤ (n − 1)
µ2i = (n − 1)(2e(G) − µ(G)2 ).
i=2
i=2
Hence
n
∑
(
)
1
µ(G) ≤ 2e(G) 1 −
.
n
2
1
2
√
4. By part (a) of Problem 3 we know that µ(G) ≤ 2e(G). Hence
√( )
√
√
√
√
n
µ(G) + µ(G) ≤ 2e(G) + 2e(G) ≤ 2 e(G) + e(G) = 2
≤ 2n.
2
5. First of all, let us do practice problem 1. On the figure you can find the
eigenvectors belonging to eigenvalue 2. We see that these eigenvectors all have
only positive eigenvalues so 2 is indeed the largest eigenvalue of these graphs.
1
2
1
2
2
2
2
1
2
3
4
3
2
1
5
4
3
2
1
1
1
1
2
2
3
3
2
1
2
4
6
1
Note that a cycle has also largest eigenvalue 2 since it is a 2–regular graph.
Note that the first graph can be generalized as follows, take a path of length
n − 4 and put 2-2 pendant edges to both ends then the obtained graph Fn has
largest eigenvalue 2 too. The first graph depicted on the figure is F8 .
So by the monotonicity of the largest eigenvalue any graph containing one
of these graphs has largest eigenvalue at least 2.
Note that we only have to describe the connected graphs with largest eigenvalue less than 2 as the disconnected ones will be the unions of some connected
graphs with largest eigenvalue less than 2.
If G is a connected graph with largest eigenvalue less than 2 then it cannot
contain a cycle, so it must be a tree. If G has a degree 4 vertex then it contains
K1,4 and its largest eigenvalue is again at least 2. So G is a tree with largest
degree 3. Note that if it has at least two vertices of degree 3 then it contains
an Fk for some k which has largest eigenvalue 2. So our graph G has at most
1 degree 3 vertex. If there is no degree 3 vertex, then G is simply the path
Pn . It is a proper subgraph of Fn+4 so its largest eigenvalue is indeed less
than 2. Now assume that there is a degree 3 vertex. Let v be the degree 3
vertex and let the distances of v from the leaves (degree 1 vertices) a, b and c. If
min(a, b, c) ≥ 2 then our graph contains the third graph on the picture so it has
largest eigenvalue at least 2 again. So we can assume that a = min(a, b, c) = 1.
If a = 1 and b, c ≥ 3 then it contains the second graph on the picture so it
has largest eigenvalue at least 2 again. So we can assume that a = 1 and
3
b = min(b, c) ≤ 2. If b = 1 then our graph is a subgraph of Fn+2 so our graph
Rn has indeed largest eigenvalue less than 2. Finally if a = 1, b = 2 then c ≤ 4
since for c ≥ 5 it would contain the fourth graph on the picture. On the other
hand, for a = 1, b = 2 and c ≤ 4 it is the subgraph of this fourth graph so it
has indeed largest eigenvalue less than 2. Hence the connected components of
our graph are one of the graphs below, where the graphs on the left hand side
represents infinite families.
E6
Pn
E7
Rn
E8
6. Note that A(G) + A(G) = J − I. Let u1 , . . . , un be orthogonal eigenvectors
of A(G) such that u1 = √1n 1, and A(G)ui = µi ui (we can choose them this
way). Note that A(G)1 = (n − d − 1)1 as G is n − d − 1 regular. If i > 1, then
ui is orthogonal to 1 and so we have
A(G)ui = (J − I − A(G))ui = 0 − ui − µi ui = (−1 − µi )ui .
Hence the eigenvalues of G are the following: n−d−1, −1−µ2 , −1−µ3 . . . , −1−
µn .
7. We will show that the spectrum of G[k] is kµ1 + k − 1, . . . , kµn + k − 1
together with −1 of multiplicity n(k − 1).
Assume that x is an eigenvector of G belonging to some eigenvalue µ. Let
x[k] be the vector whose entries on the cluster of vertex i are constant xi . Then
for a vertex v in cluster of i is
∑
∑
x[k]
=
k
xj + (k − 1)xi = kµxi + (k − 1)xi = (kµ + (k − 1))xi .
u
u∈NG[k] (v)
j∈NG (i)
Next let us consider the vectorspace V consisting of the vectors y for which
the following holds true: the sum of the entries on each cluster is 0. Then
∑
∑
yu =
yv − yu = 0 − yu = −yu .
u∈NG[k] (v)
v∈Cu
Hence each such vector is an eigenvector belonging to −1.
4
We need to prove that we have found all eigenvalues. Note that if x1 , x2 , . . . , xn
is an orthonormal eigenbasis of G belonging to eigenvalues µ1 , . . . , µn then
[k]
[k]
[k]
[k]
x1 , x2 , . . . , xn are also pairwise orthogonal. Moreover each xi is also orthogonal to V . Note that dim V = n(k − 1). This means that we have found
nk pairwise orthogonal eigenvectors.
8. Let the eigenvalues of G1 be µ1 ≥ · · · ≥ µn , and let the eigenvalues of G2
be ρ1 ≥ · · · ≥ ρm . Let A(G1 )xi = µi xi and A(G2 )y j = ρj y j , where x1 , . . . , xn
and y 1 , . . . , y m are pairwise orthogonal system of vectors. We will show that
µi ρj + µi + ρj are the mn eigenvalues of G1 G2 by constructing mn pairwise
orthogonal eigenvectors belonging to them.
For x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , ym ) let us consider the vector
x ⊗ y = (x1 y1 , x1 y2 , . . . , x1 ym , x2 y1 , . . . , x2 ym , . . . , xn ym ). Then
∑
∑
∑
ajk (G2 )xi yk +
ati (G1 )xt yj +
ajk ati yk xt =
(A(G1 G2 )x ⊗ y)ij =
k
= xi
∑
k
aik (G2 )yk + yj
∑
(
atj (G1 )xt +
k
t
∑
k
)(
aik (G2 )yk
j,t
∑
atj (G1 )xt
)
=
k
= xi ρj yj + yj µi xi + (ρj yj )(µi xi ) = (µi + ρj + µi ρj )xi yj .
(This step can be understood much easier if you write the numbers xi yj on the
vertices to the graph G1 G2 , and you sum these numbers for the neighbors
of (i, j) in the row and the column and the "diagonal" entries separately.) On
the other hand it is easy to see that xi ⊗ y j and xr ⊗ y s are orthogonal if
(i, j) ̸= (r, s). Hence this is indeed mn orthogonal vectors.
9. First proof (only sketch): since µ(Pn ) ≤ µ(Cn ) = 2 we need to consider
only those trees Tn fro which µ(Tn ) ≤ 2. By problem 5 we know exactly the
list of these trees. For n ≤ 8 one can check them one by one. For n ≥ 9 we
know that the only possible tree for which µ(Tn ) < µ(Pn ) is the tree Rn . On
the other hand, one can show that µ(Rn ) = µ(P2n−3 ) > µ(Pn ).
Remark 0.1. If you want to see something interesting then copy the code below to http://sagecell.sagemath.org/ and push the evaluate button (the even
and odd n cases behave in a slightly different way):
n=6
g=graphs.PathGraph(2*n-1)
h=graphs.PathGraph(n)
h.add_edges([(1,n)])
5
g.show()
h.show()
print g.spectrum()
print h.spectrum()
Second proof (this is an exotic proof ): Let us consider the following transformation on a tree T , it is called generalized tree shift. If Pu,v is a path
connecting the vertices u and v such that every inner vertex of Pu,v has degree 2, then we can consider the tree T ′ = GT S(T, Pu,v ) obtained as follows:
delete all edges between v and its neighbors except the edge (v, w), where w
is the unique neighbor of v on the path Pu,v , and add all edges (u, s), where
s ∈ NT (v) \ {w}. One important property of this transformation is that if we
change the role of u and v we get the very same tree, so only the path Pu,v
matters. (In the picture u,v and w are x, y and z, sorry.)
0
x
0
1 . . . k−1 k
1
x
z
y
k−1
y
A
B
k
A
B
Figure 1. The generalized tree shift.
The key observation is that µ(T ′ ) ≥ µ(T ). Indeed, let x be the non-negative
eigenvector of unit length corresponding to the largest eigenvalue of the tree
T . We can assume that xu ≥ xv since the roles of u and v are exchangable.
Furthermore, let A(T ) and A(T ′ ) be the adjacency matrices of the tree T and
T ′ . Then
∑
xs ≤ xT A(T ′ )x ≤
µ(T ) = xT A(T )x = xT A(T ′ )x − 2(xu − xv )
s∈NT (v)\{w}
≤ max y T A(T ′ )y = µ(T ′ ).
||y||=1
The last step of the proof is to show that if T is a tree on n vertices different
from Pn then there exists a tree T1 and a path P1 such that T = GT S(T1 , P1 ),
so µ(T ) ≥ µ(T1 ). So we actually apply the transformation backwards. Indeed,
such a tree T1 exists, we can do it in 4 small steps: (1) let v be a vertex of
degree 1 in T (it is called a leaf), (2) let u be the unique vertex of degree at
least 3 which is closest to v (such a vertex exists since T is not Pn ), (3) then
6
decompose the set of neighbors of u into two non-empty subsets A and B such
that the neighbor of u lying on the path Pu,v is in A, (4) next delete the edges
between u and the set B and add all edges between v and B. Let the obtained
tree be T1 , it is clear that T = GT S(T1 , Pu,v ). Note that T1 has one less leaf
than T . So as long as T is not Pn we can apply this transformation backwards
obtaining trees with fewer leaves and a non-increasing largest eigenvalue.
Remark 0.2. Note that this transformation determines a partially ordered
set on the set of trees with exactly n vertices. Now we have seen that the
minimal element of this poset is the path, and it’s very easy to show that the
maximal element is the star. In particular, we get that µ(Tn ) ≤ µ(Sn ) for any
tree Tn , where Sn is the star on n vertices.
Figure 2. The induced poset of the generalized tree shift on
trees of 6 vertices.
10. Let µs , . . . , µn be the set of non-positive eigenvalues. Then
0 = 6 · number of triangles =
n
∑
i=1
Hence
n
∑
i=s
(−µi )3 ≥ µ31 .
µ3i ≥ µ31 +
n
∑
i=s
µ3i .
7
On the other hand
n
n
∑
∑
(−µi )3 ≤ (−µn )
(−µi )2 ≤ (−µn )(2e(G) − µ21 ) ≤ (−µn )(nµ1 − µ21 ).
i=s
i=s
Hence
µ31 ≤ (−µn )(nµ1 − µ21 ).
Thus
µ21 ≤ (−µn )(n − µ1 ),
or in other words,
−nµn
.
µ1 − µn
Now let us solve the constrained maximization problem:
{
}
µ1 + µn
−nµn
max
| µ1 ≤
.
n
µ1 − µn
µ1 ≤
Let a = µ1 , b = −µn then we have
a≤
nb
a−b
which is equivalent to
a2
≤ b.
n−a
Hence
a−b
1
≤
n
n
(
a−
a2
n−a
)
=
an − 2a2
.
n(n − a)
So with the notation α = a/n we need to maximize
f (α) :=
Its derivative is
α − 2α2
.
1−α
1 − 4α + 2α2
(1 − α)2
√
which is 0 at α = 1 ± 1/ 2. Note that µ1 ≤ ∆ ≤ n − 1, where ∆ is√the largest
degree, so 0 ≤ α ≤ 1. So we only need to consider α = 1 − 1/ 2 and the
extreme points
of the interval, √
α = 0 and 1, to see that f (α) is indeed maximal
√
at 1 − 1/ 2 and f (α) = 3√− 2 2.
Hence µ1 + µn ≤ (3 − 2 2)n.
8
11. (a) Let x = (x1 , . . . , xn ) be a non-negative eigenvector belonging to µ(G).
Let us consider the following inequalities: let (i, j) ∈ E(G) such that j is
oriented towards i:
x2
2x x
x2
2x x
√ i j ≤ √ i j ≤ +i + −j ,
−
di
dj
D+ D−
d+
i dj
−
where d+
i is the indegree of vertex i and dj is the outdegree of vertex j. Now
let us add these inequalities for all (oriented) edges. Then
(
)
2
2
∑
∑
∑
x
x
1
j
i
2
√
x2j = 2||x||2 .
x
+
xT Ax ≤
+
=
i
−
d+
d
D+ D−
i
j
i
j
(i,j)∈E(G)
√
Since xT Ax = µ(G)||x||2 we have µ(G) ≤ 2 D+ D− .
Second solution for part (a):
µ(G) = 2
∑
xi xj = 2
i=1
(i,j)∈E(G)
n
∑
√
−
≤2 D
xi
i=1
√
= 2 D− · 1 ·
(
n
∑
∑
)1/2
x2j
j:i→j
(
n
∑
j=1
xi
∑
)
xj
≤2
j:i→j
n
∑
i=1
√
xi
(
d−
i
∑
)1/2
x2j
≤
j:i→j
( n
)1/2 ( n
)1/2
∑
∑∑
√
≤ 2 D−
x2i
x2j
=
i=1
)1/2
2
d+
j xj
(
i=1 j:i→j
( n
)1/2
∑
√ √
√ √
≤ 2 D− D+
x2j
= 2 D− D+ .
j=1
(b) Let us pick a vertex v of degree 1 and orient every edges towards this
vertex. (Note that in a tree there is exactly one path between any two vertices
so it makes sense to say such a thing that "orient every edges towards this
vertex".) Then the outdegree of every vertex is exactly 1 except the chosen
vertex v which is 0, consequently every indegree is at most D − 1 except the
indegree of v which is 1. Note that since n ≥ 3 we have D ≥ 2, so D − 1 ≥ 1.
By part (a) we have
√
√
µ(G) ≤ 2 D+ D− = 2 D − 1.
12. This is a half algebraic, half probabilistic solution. It is enough to check
the statement for connected graphs since if the connected components of G
9
are G1 , G2 , . . . , Gr then
r
r
∑
∑
e(Gi )
e(Gi )
e(G)
≥
=
.
2
2
2
µ(G
)
µ(G)
µ(G)
i
i=1
i=1
From now on assume that G is a connected graph and x be an eigenvector
belonging to the eigenvalue µ = µ(G). We can assume that all entries of x are
positive.
At each vertex i let us choose a neighbor j of i randomly with probability
x
pij = µxji . Note that it is indeed a probability distribution:
∑ xj
∑
pij =
= 1.
µxi
j∈N (i)
j∈N (i)
Let us say that the vertices i and j form a pair if they choose each other. Let
X be the number of pairs, this is a random variable. The expected number of
pairs is:
∑
∑
xj xi
e(G)
pij pji =
EX =
= 2 .
µxi µxj
µ
(i,j)∈E(G)
(i,j)∈E(G)
So with positive probability we have at least as many pairs. But, of course,
pairs form a matching, so we have a matching of size at least e(G)
.
µ2
Remark 0.3. The length of the solution of this problem is very deceiving, it
is very poor, for most
is not a simple problem. Surprisingly the bound e(G)
µ2
graph it is nowhere near tight, but one particular example where it is tight is
the star K1,n .
Second solution: (Joint work with Timothy Ngotiaoco). Let ν(G) denote the
size of the largest matching. Again it is enough to check the statement for
connected graphs since if the connected components of G are G1 , G2 , . . . , Gr
then
r
r
∑
∑
e(Gi )
e(Gi )
e(G)
≥
=
.
2
2
µ(Gi )
µ(G)
µ(G)2
i=1
i=1
We prove the claim by induction on the number of edges. Since the claim
doesn’t make sense if e(G) = 0 and consequently µ(G) = 0, we start with case
e(G) = 1 then G = K2 and µ(G) = 1, and the largest matching has size 1 so
the claim holds true. Let e = (u, v) be an edge which we choose later. Let H
be the graph obtained from G by deleting the vertices u and v. By induction
we have
e(H)
e(G) − du − dv + 1
e(H)
≥1+
=1+
.
ν(G) ≥ ν(H) + 1 ≥ 1 +
2
2
µ(H)
µ(G)
µ(G)2
10
We used the inequality µ(G) ≥ µ(H). Note that if H is empty then
doesn’t make sense, but the inequality
e(H)
µ(H)2
e(H)
=1
µ(G)2
ν(G) ≥ 1 +
makes sense even in this case. So we have
e(G) + µ(G)2 − du − dv + 1
ν(G) ≥
.
µ(G)2
Note that
∑
(du + dv ) =
(u,v)∈E(G)
∑
d2u .
u∈V (G)
So there must be an edge e = (u, v) for which we have
1 ∑ 2
di .
du + dv ≤
e(G)
i∈V (G)
Let’s choose this edge in our argument:
1
e(G) + µ(G)2 − e(G)
e(G) + µ(G)2 − du − dv + 1
ν(G) ≥
≥
µ(G)2
µ(G)2
∑
i∈V (G)
d2i + 1
.
We would be able to finish the proof if we can show that
1 ∑ 2
µ(G)2 −
di + 1 ≥ 0.
e(G)
i∈V (G)
We will show that it is indeed true. Let’s introduce a few notations:
√
√
∑
D
1 ∑ 2
2e(G)
1 ∑
2
di , s2 :=
di , s1 :=
di .
D :=
=
=
n
n
n
n
i∈V (G)
i∈V (G)
i∈V (G)
With these notations we need to prove that
µ(G)2 + 1 ≥
D
2s2
= 2.
e(G)
s1
It is known that
µ(G) ≥ s2 ≥ s1 .
Indeed, the second inequality comes the quadratic-arithmetic mean, the first
one comes from
D
µ(G)2 = max xT A2 x ≥ v1 A2 v1 = ,
||x||=1
n
11
where v1 =
√1 (1, 1, . . . , 1).
n
In particular, if s1 ≥ 2 then
2s22
2s2
≥ 2,
2
s1
and we are done even without the extra term +1. If s1 < 2 then e(G) < n.
Since G is connected we must have e(G) = n − 1, and G is a tree. Hence
. If we can show that s22 ≤ n − 1 then we are done since we have
s1 = 2(n−1)
n
µ(G)2 ≥ s22 ≥
s22
n 2 2s22
=
s =
.
n−1
n−1 2
s1
Finally, to prove s22 ≤ n − 1 we use that n − 1 ≥ µ(G)2 ≥ s22 for trees:
1
µ(G)2 ≤ (µ21 + · · · + µ2n ) = e(G) = n − 1,
2
where we used the fact that trees are bipartite graphs so µn = −µ1 . So in all
cases we proved that
2s2
µ(G)2 + 1 ≥ 2
s1
for connected graphs which finishes our proof.
µ(G)2 + 1 ≥ s22 + 1 ≥ s22 +