The star is the tree with greatest greatest Laplacian eigenvalue

62
The number of first neighbors of a vertex v is the degree of this vertex and is
denoted by d(v) . Note that if v i is a vertex of the graph G , then d(v i ) is equal to the
sum of the i-th row of the adjacency matrix A(G) .
Let D(G) be a square matrix whose diagonal entries are d(v 1 ), d(v 2 ), . . . , d(v n ) and
the off-diagonal elements are zero.
Then L(G) = D(G) − A(G) is the Laplacian matrix of G .
The eigenvalues µ 1 , µ 2 , . . . , µ n of L(G) are called the Laplacian eigenvalues of the
graph G and the respective eigenvectors C ⃗ 1 , C ⃗ 2 , . . . , C ⃗ n the Laplacian eigenvectors of
the graph G . Thus the equality L(G) C ⃗ i = µ i Ci ⃗ is obeyed for all i = 1, 2, . . . , n . We
label the Laplacian eigenvalues so that µ 1 ≥ µ 2 ≥ · · · ≥ µ n .
Details of the theory of Laplacian spectra of graphs can be found in some of the
numerous reviews published on this topic [2, 3, 4]. Here we only mention that it is
always µ n = 0 .
SOME AUXILIARY RESULTS
Lemma 1. The vector ⃗j is a Laplacian eigenvector of any n-vertex graph, corresponding
to the eigenvalue µ n = 0 .
Proof. Because the sum of any row of L(G) is equal to zero, L(G)⃗j = ⃗0 = 0 ·⃗j .
In view of Lemma 1 we may choose C ⃗ n = ⃗j . In what follows we assume that the
Laplacian eigenvectors C ⃗ i , i = 1, 2, . . . , n − 1 , are orthogonal to C ⃗ n . If µ n−1 ≠ 0
then this orthogonality condition is automatically satisfied. If however, µ n−k = 0 for
some k ≥ 1 , then the requirement that the eigenvectors C ⃗ n−1 , . . . , C ⃗ n−k are chosen
to be orthogonal to C ⃗ n must be additionally stipulated. Recall that µ n−1 ≠ 0 if and
only if the graph G is connected [2, 3, 4].
From the fact that for any vector C ⃗ the scalar product ⃗j t • C ⃗ is equal to σ( C) ⃗ , it
follows:
Lemma 2. For any graph G and any 1 ≤ i ≤ n − 1 , σ( C ⃗ i ) = 0 .