Let A denote the adjacency matrix of Erdös-Rényi random graph G with parameters
n, p. Let λ1 ≥ λ2 ≥ · · · ≥ λn be the eigenvalues of A. Let davg , dmin , and dmax denote the
average, min, and max degree of G respectively.
Proposition: If np >> log n then λ1 = np(1 + o(1)).
Proof. Let v = (v1 , ..., vn ) denote the eigenvector corresponding to λ1 . Since p >> log n/n
we know A is connected and hence irreducible. By Perron-Frobenius theorem we know that:
1) λ1 is unique, and
2) vi > 0 for all i.
Let kuk1 =
Pn
i=1 |ui |
be the 1-norm of a vector. Consider the sum
k(Av)k1 =
n X
n
X
i=1 j=1
n
X
= λ1 kvk1 =
aij vj =
λ 1 vj ≤
j=1
n
X
dj vj
j=1
n
X
dmax vj
j=1
Hence λ1 ≤ dmax . For an upper bound note that kAuk1 /kuk1 ≤ λ1 for all vectors u.
Plugging in the all-ones vector gives davg ≤ λ1 .
Let Si denote the sum of the entries in the ith row. Each entry in the ith row is
independent from the other entries in that row, so we can use the Chernoff Bound to get
P[Si ≥ (1 + δ)ESi ] =P[Si ≥ (1 + δ)np]
≤e−np((1+δ)log(1+δ)−δ)
2
≤ 2e−npδ .
We can choose δ = O(1/n1/3 ) such that P[Si − np ≥ δnp] = o(1). Using the union bound
2
P[Si − np ≥ δnp for all i] ≤ 2ne−npδ = o(1).
Hence dmax = np(1 + o(1))
The law of large numbers says davg ≥ (1 + o(1)). Therefore
λ1 = np(1 + o(1)).
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