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Ma/CS 6b Class 23: Eigenvalues in Regular Graphs
By Adam Sheffer
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
𝑠𝑝𝑒𝑐 𝐺 =𝜆1, … , 𝜆𝑡𝑚1, … ,𝑚𝑡
,
where 𝜆1, … , 𝜆𝑡 are the eigenvalues of 𝐴 and 𝑚𝑖 is the multiplicity of 𝜆𝑖.
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Example: Spectrum
𝐴 =
0 1 0 11 0 1 001100 11 0
det 𝜆𝐼 − 𝐴 = det
𝜆 −1 0 −1−1 𝜆 −1 00−1−10𝜆 −1−1 𝜆
= 𝜆2 𝜆 − 2 𝜆 + 2 .
𝑠𝑝𝑒𝑐 𝐶4 =0 2 −22 1 1
𝑣1 𝑣2
𝑣3 𝑣4
Slight Change of Notation
Instead of multiplicities, let 𝜆1, … , 𝜆𝑛 be the not necessarily distinct eigenvalues of 𝑛.
For example, if the spectrum is 2 −12 2
,
we write 𝜆1 = 𝜆2 = 2 and 𝜆3 = 𝜆4 = −1 (instead of 𝜆1 = 2, 𝑚1 = 2, 𝜆2 = −1,𝑚2= 2).
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Recall: 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 = 𝜆 − 𝜆𝑖𝑛𝑖=1 .
◦ The multiplicity of an eigenvalue 𝜆 is 𝑛 − 𝑟𝑎𝑛𝑘 𝜆𝐼 − 𝐴 .
More Examples
We already derived the following:
◦ 𝑠𝑝𝑒𝑐 𝐾𝑛 =−1 𝑛 − 1𝑛 − 1 1
◦ 𝑠𝑝𝑒𝑐 𝐾𝑛,𝑚 =0 𝑚𝑛 − 𝑚𝑛
𝑚 + 𝑛 − 2 1 1
Our next goal is to study regular graphs.
◦ Can you come up with an eigenvector of any regular graph with 𝑛 vertices?
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Eigenvalues of Regular graphs
If 𝐴 is the adjacency matrix of a 𝑑-regular graph, then any row of 𝐴 contains exactly 𝑑 1’s.
◦ Thus, the vector 1𝑛 = 1,1,… , 1 is an eigenvector of 𝐴 with eigenvalue 𝑑.
Theorem. Let 𝐺 be a connected graph. The eigenvalue of 𝐺 of largest absolute value is the maximum degree if and only if 𝐺 is regular.
Proof
𝐴 – 𝑛 × 𝑛 adjacency matrix of a graph 𝐺.
Δ 𝐺 – the maximum degree of 𝐺.
𝑥 = 𝑥1, … , 𝑥𝑛 – eigenvector of eigenvalue 𝜆 of largest absolute value.
𝑥𝑗 = max𝑖|𝑥𝑖| .
𝜆 𝑥𝑗 = 𝐴𝑥 𝑗 = 𝑥𝑖𝑣𝑖∈𝑁 𝑣𝑗
≤ deg 𝑣𝑗 𝑥𝑗 ≤ Δ 𝐺 𝑥𝑗 ,
So 𝜆 ≤ Δ 𝐺 .
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Δ 𝐺 – the maximum degree of 𝐺.
We proved that the absolute value of any eigenvalue of 𝐴 is at most Δ 𝐺 , using
𝜆 𝑥𝑗 = 𝐴𝑥 𝑗 = 𝑥𝑖𝑣𝑖∈𝑁 𝑣𝑗
≤ deg 𝑣𝑗 𝑥𝑗 ≤ Δ 𝐺 𝑥𝑗 .
For equality to hold, we need
◦ deg 𝑣𝑗 = Δ 𝐺 .
◦ 𝑥𝑖 = 𝑥𝑗 for each 𝑣𝑖 ∈ 𝑁 𝑣𝑗 .
That is, 𝑣𝑗 and all of its neighbors are of degree
Δ 𝐺 . Repeating the same argument for a neighbor 𝑣𝑖 implies that 𝑣𝑖’s neighbors are also of degree Δ 𝐺 . We continue to repeat the argument to obtain that the graph is regular.
Multiplicity
The previous proof also shows that any eigenvector 𝑥1, … , 𝑥𝑛 of the eigenvalue 𝑑 satisfies 𝑥1 = 𝑥2 = ⋯ = 𝑥𝑛.
Thus, the space of eigenvectors of the eigenvalue 𝑑 is of dimension 1 (that is, 𝑑 has multiplicity 1).
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The Spectrum of the Petersen Graph The Petersen graph 𝐺 = 𝑉, 𝐸 is a 3-
regular graph with 10 vertices.
◦ We know that it has eigenvalue 3 with eigenvector 110.
To find the other eigenvalues, we notice some useful properties:
◦ If 𝑢, 𝑣 ∈ 𝐸 then 𝑢 and 𝑣 have no common neighbors.
◦ If 𝑢, 𝑣 ∉ 𝐸 then 𝑢 and 𝑣 have exactly one common neighbor.
The Adjacency Matrix
The number of neighbors shared by 𝑣𝑖 and 𝑣𝑗 is 𝐴2 𝑖𝑗 . That is
𝐴2 𝑖𝑗 =
3, if 𝑖 = 𝑗,
0, if 𝑖 ≠ 𝑗 and 𝑣𝑖, 𝑣𝑗 ∈ 𝐸,
1, if 𝑖 ≠ 𝑗 and 𝑣𝑖, 𝑣𝑗 ∉ 𝐸.
That is, 𝐴2 + 𝐴 − 2𝐼 = 1𝑛×𝑛.
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The Additional Eigenvalues
We have 𝐴2 + 𝐴 − 2𝐼 = 1𝑛×𝑛.
Since 3 is an eigenvalue with eigenvector 1𝑛, the other eigenvectors are orthogonal to 1𝑛.
Thus, for an eigenvector 𝑣 of eigenvalue 𝜆 ≠ 3: 1𝑛×𝑛𝑣 = 0𝑛.
That is, 𝐴2 + 𝐴 − 2𝐼 𝑣 = 0𝑛.
If 𝑣 is an eigenvector of 𝜆 then we have 𝜆2𝑣 + 𝜆𝑣 − 2𝑣 = 0𝑛.
Thus, the additional eigenvalues of 𝐴 are 1,−2.
𝑠𝑝𝑒𝑐 𝐺 =3 1 −21 𝑚2 𝑚3
.
The Multiplicities
𝑠𝑝𝑒𝑐 𝐺 =3 1 −21 𝑚2 𝑚3
.
Recall that 𝜆𝑖10𝑖=1 = 𝑡𝑟𝑎𝑐𝑒 𝐴 = 0.
That is, 3 +𝑚2 − 2𝑚3 = 0.
Combining this with 𝑚2 +𝑚3 = 9 and 𝑚2, 𝑚3 ≥ 0, we obtain the unique solution 𝑚2 = 5,𝑚3 = 4.
𝑠𝑝𝑒𝑐 𝐺 =3 1 −21 5 4
.
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Petersen Graph and 𝐾10
Problem. Can we partition the edges of 𝐾10 into three disjoint sets, such that each set forms a Petersen graph?
◦ 𝐾10 has 102= 45 edges, and the Petersen
graph has 15.
◦ In 𝐾10 every vertex is of degree 9, and in the Petersen graph 3.
Disproof
Assume, for contradiction, that the partition exists, and let 𝐴, 𝐵, 𝐶 be the adjacency matrices of the three copies of the Petersen graph.
The adjacency matrix of 𝐾10 is 110×10 − 𝐼.
That is, 𝐴 + 𝐵 + 𝐶 = 110×10 − 𝐼.
𝑉𝐴, 𝑉𝐵 – the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in 𝐴 and 𝐵.
We know that dim𝑉𝐴 = dim𝑉𝐵 = 5.
Since both 𝑉𝐴 and 𝑉𝐵 are orthogonal to 110, they are not disjoint (otherwise we would have a set of 11 orthogonal vectors in ℝ10).
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Disproof (cont.) 𝐴, 𝐵, 𝐶 – the adjacency matrices of the three
copies of the Petersen graph in 𝐾10.
𝐴 + 𝐵 + 𝐶 = 1𝑛×𝑛 − 𝐼.
𝑉𝐴, 𝑉𝐵 – the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in 𝐴 and 𝐵.
𝑉𝐴 and 𝑉𝐵 are not disjoint.
Let 𝑧 ∈ 𝑉𝐴 ∩ 𝑉𝐵. Since every vector in 𝑉𝐴 and 𝑉𝐵 is orthogonal to 1𝑛, so is 𝑧.
We have 𝐶𝑧 = 1𝑛×𝑛 − 𝐼 − 𝐴 − 𝐵 𝑧 = 0 − 𝑧 − 𝐴𝑧 − 𝐵𝑧= −3𝑧.
Contradiction since -3 is not an eigenvalue of 𝐶.
Four Things You Did Not Know About The Petersen Graph It has 15 edges and 2000 spanning trees.
It is the smallest 3-regular graph of girth 5 (this is called a 3,5 -cage graph).
It likes gardening, ballet, and building airplane models.
It has gotten divorced three times.
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Moore Graphs
A Moore graph is a graph that is 𝑑-regular, of diameter 𝑘, and whose number of vertices is
1 + 𝑑 𝑑 − 1 𝑖𝑘−1
𝑖=0
.
As can be easily checked, this is the minimum possible number of vertices of any graph of diameter 𝑘 and minimum degree 𝑑.
Examples of Moore Graphs
A Moore graph is a graph that is 𝑑-regular, of diameter 𝑘, and whose number of vertices is
1 + 𝑑 𝑑 − 1 𝑖𝑘−1
𝑖=0
.
What Moore graphs do we know?
◦ The Petersen graph is 3-regular, of diameter
2, and contains 1 + 3 3 − 1 𝑖1𝑖=0 = 10
vertices.
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Examples of Moore Graphs
A Moore graph is a graph that is 𝑑-regular, of diameter 𝑘, and whose number of vertices is
1 + 𝑑 𝑑 − 1 𝑖𝑘−1
𝑖=0
.
The Petersen graph is 3-regular, of diameter 2,
and contains 1 + 3 3 − 1 𝑖1𝑖=0 = 10 vertices.
Is there a Moore graph that is 2-regular and of diameter 2? 𝐶5
Moore Graphs of Diameter 2 and Girth 5 Recall that the girth of a graph is the
length of the shortest cycle in it.
Theorem. There exist 𝑑-regular Moore graphs with diameter 2 and girth 5 only for 𝑑 = 2,3,7, and possibly 57.
◦ The case of 𝑑 = 57 is an open problem.
◦ If it exists, it has 3250 vertices, and 92,625 edges.
The case of 𝑑 = 7
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𝐺 = 𝑉, 𝐸 – a 𝑑-regular graph of diameter 2, girth 5, and with
𝑉 = 1 + 𝑑 𝑑 − 1 𝑖1
𝑖=0
= 1 + 𝑑2.
Since the girth is five, if 𝑣𝑖 , 𝑣𝑗 ∈ 𝐸 then 𝑣𝑖 and
𝑣𝑗 have no common neighbors.
Since the diameter is two, if 𝑣𝑖, 𝑣𝑗 ∉ 𝐸 then 𝑣𝑖
and 𝑣𝑗 have exactly one common neighbor.
Thus, the adjacency matrix 𝐴 satisfies
𝐴2 𝑖𝑗 =
𝑑, if 𝑖 = 𝑗,
0, if 𝑖 ≠ 𝑗 and 𝑣𝑖, 𝑣𝑗 ∈ 𝐸,
1, if 𝑖 ≠ 𝑗 and 𝑣𝑖 , 𝑣𝑗 ∉ 𝐸.
That is, we have 𝐴2 + 𝐴 − 𝑑 − 1 𝐼 = 1𝑛×𝑛.
Proof (cont.)
𝐺 = 𝑉, 𝐸 – a 𝑑-regular graph of diameter 2, girth 5, and with 𝑉 = 1 + 𝑑2.
𝐴2 + 𝐴 − 𝑑 − 1 𝐼 = 1𝑛×𝑛.
𝜆 ≠ 𝑑 – an eigenvalue of 𝐴 with eigenvector 𝑣. Since 𝜆 ≠ 𝑑, 𝑣 is orthogonal to 1𝑛. Thus 𝐴2 + 𝐴 − 𝑑 − 1 𝐼 𝑣 = 0𝑛, or 𝜆2𝑣 + 𝜆𝑣 − 𝑑 − 1 𝑣 = 0.
This implies that 𝜆2 + 𝜆 − 𝑑 + 1 = 0, so
𝜆2,3 = −1 ± 1 + 4 𝑑 − 1
2= −1 ± 4𝑑 − 3
2.
We thus have 𝑠𝑝𝑒𝑐 𝐺 =𝑑 𝜆2 𝜆31 𝑚2 𝑚3
.
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Finding the Multiplicities
We have 𝜆2 = −1+ 4𝑑−3
2, 𝜆3 = −
1− 4𝑑−3
2,
𝑠𝑝𝑒𝑐 𝐺 =𝑑 𝜆2 𝜆31 𝑚2 𝑚3
.
0 = 𝑡𝑟𝑎𝑐𝑒 𝐴 = 𝑑 + 𝜆2𝑚2 + 𝜆3𝑚3
= 𝑑 −𝑚2 +𝑚32+𝑚3 −𝑚22
4𝑑 − 3.
Since 𝑚2 +𝑚3 = 𝑛 − 1 = 𝑑2, we have
𝑑2 − 2𝑑 = 𝑚3 −𝑚2 4𝑑 − 3.
This can happen if either 𝑚2 = 𝑚3 or 4𝑑 − 3 = 𝑠2 for some integer 𝑠.
If 𝑚2 = 𝑚3 than 𝑑2 − 2𝑑 = 0, implying 𝑑 = 2.
The Case of 4𝑑 − 3 = 𝑠2
𝑑2 − 2𝑑 = 𝑚3 −𝑚2 4𝑑 − 3.
Assume that 4𝑑 − 3 = 𝑠2 for some integer 𝑠. That is, 𝑑 = 𝑠2 + 3 /4.
Substituting into the above equation: 𝑠2 + 3 2
16−2𝑠2 + 6
4= 𝑚3 −𝑚2 𝑠.
Setting 𝑚3 −𝑚2 = 2𝑚3 − 𝑑2, we have
𝑠4 + 6𝑠2 + 9 − 8𝑠2 − 24 = 32𝑚3𝑠 − 𝑠5 − 6𝑠3 − 9𝑠.
𝑠5 + 𝑠4 + 6𝑠3 − 2𝑠2 + 9 − 32𝑚3 𝑠 = 15.
So 𝑠 must divide 15, and we get 𝑠 ∈ ± 1,3,5,15 , which implies 𝑑 ∈ 1,3,7,57 .
The case 𝑑 = 1 leads to 𝐾2. Not a Moore graph.
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Recall: Independent Sets
Consider a graph 𝐺 = 𝑉, 𝐸 . An independent set in 𝐺 is a subset 𝑉′ ⊂ 𝑉 such that there is no edge between any two vertices of 𝑉′.
Finding a maximum independent set in a graph is a major problem in theoretical computer science.
◦ No polynomial-time algorithm is known.
Past Bounds
Let 𝐺 = (𝑉, 𝐸) be a graph.
Already proved:
◦ 𝐺 has an independent set of size at least
1
1+ deg 𝑣𝑣∈𝑉
.
◦ If 𝐸 = 𝑉 ⋅𝑑
2, then 𝐺 has an independent
set of size at least 𝑉 /2𝑑.
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An Upper Bound
Theorem. For a 𝑑-regular graph 𝐺 = 𝑉, 𝐸 with smallest (most negative) eigenvalue 𝜆𝑛, the size of the largest
independent set of 𝐺 is at most 𝑛
1−𝑑/𝜆𝑛.
Example.
◦ 𝑠𝑝𝑒𝑐 𝐶4 =0 2 −22 1 1
.
◦ So at most 4
1−2
−2
= 2.
𝑣1 𝑣2
𝑣3 𝑣4
Recall: The Rayleigh Quotient The Rayleigh quotient is 𝑅 𝐴, 𝑥 = 𝑅(𝑥)
=𝑥𝑇𝐴𝑥
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?
◦𝑥𝑇 𝐴𝑥
𝑥𝑇𝑥=𝑥𝑇𝜆𝑥
𝑥𝑇𝑥= 𝜆.
◦ Thus, the min and max values of 𝑅 𝑥 are the min and max eigenvalues of 𝐴.
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𝜆𝑛 – the most negative eigenvalue of 𝐺.
𝑆 – a largest independent set of 𝐺.
1𝑆 = 𝑥1, … , 𝑥𝑛 – a vector with 𝑥𝑖 = 1 if 𝑣𝑖 ∈ 𝑆 (otherwise 𝑥𝑖 = 0).
𝑦 = 𝑛1𝑆 − 1𝑛 ⋅ 𝑆 .
𝑦𝑇𝐴𝑦 = 𝑛2 ⋅ 1𝑆𝑇𝐴1𝑆 − 2 𝑆 𝑛 ⋅ 1𝑆
𝑇𝐴1𝑛 + 𝑆2 ⋅ 1𝑛𝑇𝐴1𝑛.
Since 𝑆 is an independent set, we have
1𝑆𝑇𝐴1𝑆 = 𝐴𝑖𝑗𝑖,𝑗∈S = 0.
Since 𝐺 is 𝑑-regular, 1𝑆𝑇𝐴1𝑛 = 1𝑆
𝑇 ⋅ 𝑑1𝑛 = 𝑑 𝑆 , and also 1𝑛
𝑇𝐴1𝑛 = 1𝑛𝑇 ⋅ 𝑑1𝑛 = 𝑑𝑛.
Combining the above, we have 𝑦𝑇𝐴𝑦 = 0 − 2 𝑆 𝑛 ⋅ 𝑑 𝑆 + 𝑆 2 ⋅ 𝑑𝑛 = − 𝑆 2𝑑𝑛. 𝑦𝑇𝑦 = 𝑛21𝑆
𝑇1𝑆 − 2 𝑆 𝑛1𝑆𝑇1𝑛 + 𝑆
21𝑛𝑇1𝑛
= 𝑛2 𝑆 − 2 𝑆 2𝑛 + 𝑆 2𝑛 = 𝑆 𝑛 𝑛 − 𝑆 .
Completing the Proof
By the lemma, we have 𝑦𝑇𝐴𝑦
𝑦𝑇𝑦≥ 𝜆𝑛.
We have 𝑦𝑇𝐴𝑦 = − 𝑆 2𝑑𝑛. 𝑦𝑇𝑦 = 𝑆 𝑛 𝑛 − 𝑆 .
Thus
𝜆𝑛 ≤− 𝑆 2𝑑𝑛
𝑆 𝑛 𝑛 − 𝑆=−𝑑 𝑆
𝑛 − 𝑆.
𝜆𝑛 𝑛 − 𝑆 ≤ −𝑑 𝑆 → 𝑆 ≤𝑛
1 − 𝑑/𝜆𝑛
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The End
