rounded to two decimal places, is given by
[0.71, 0.29, −0.06, −0.21, 0.04, −0.23, −0.56]0.
Thus, vertices 1, 2 and 5 are positive and they induce a connected subgraph. Ver-tices 3, 4, 6 and 7 are negative and they induce a connected subgraph as well.
7.2 Classification of trees
We now consider the case of trees in greater detail. Let T be a tree with V (T ) = {1, . . . , n} and the edge set E(T ) = {e1, . . . , en−1}. Let L be the Laplacian of T and µ be the algebraic connectivity. Let x be an eigenvector of L corresponding to µ . We refer to x as a Fiedler vector of L. First, suppose that x has no zero coordinate.
Then
V+= {i : xi> 0}, V−= {i : xi< 0}
give a partition of V (T ). By Theorem 7.2, the subgraphs induced by V+and V− must be connected and then, clearly, they must both be trees. Recall that a vertex i is positive or negative according as xi> 0 or xi< 0, respectively. Then there must be precisely one edge such that one of its end-vertices is positive and the other negative.
Such an edge is called a characteristic edge (with respect to x). Any other edge has either both its end-vertices positive or both negative.
We turn to the case where a Fiedler vector has a zero coordinate. This case re-quires a closer analysis by means of some subtle properties of interlacing of eigen-values. Note that Lx = µx implies that
∑
j∼ixj= (di− µ)xi, (7.5)
where diis the degree of i. If xi= 0 then (7.5) implies that either xj= 0 for all j adjacent to i or i is adjacent to a positive vertex as well as a negative vertex. A zero vertex is called a characteristic vertex (with respect to x) if it is adjacent to a positive vertex and a negative vertex. It is evident from (7.5) that a pendant vertex cannot be a characteristic vertex. Our goal is to prove the interesting fact that corresponding to any Fiedler vector a tree has at most one characteristic vertex.
We first develop some preliminary results. If A is an n × n symmetric matrix, then p+(A), p−(A) and p0(A) will denote, respectively, the number of positive, neg-ative and zero eigenvalues of A. Thus, p+(A) + p−(A) + p0(A) = n. The 3-tuple (p+(A), p−(A), p0(A)) is called the inertia of A.
Lemma 7.4. Let B be a symmetric n × n matrix and let c be a vector of order n × 1.
Suppose there exists a vector u such that Bu= 0 and c0u6= 0. Let A= B c
c0d
,
where d is a real number. Then
p+(A) = p+(B) + 1, p−(A) = p−(B) + 1 and p0(A) = p0(B) − 1.
Proof. First note that u 6= 0 since c0u6= 0. Then Bu = 0 implies that B is singular and 0 is an eigenvalue of B. If c were in the column space of B, then c would be equal to By for some vector y. Then u0c= u0By= 0, since Bu = 0. This is a contradiction since c0u6= 0. Therefore, c is not in the column space of B. Thus,
rank (A) = rank [B, c] + 1 = rank (B) + 2. (7.6) Since the rank of an m × m symmetric matrix is m minus the multiplicity of the zero eigenvalue, it follows from (7.6) that p0(A) = p0(B) − 1. By the interlacing theorem, p+(B) ≤ p+(A) ≤ p+(B) + 1 and p−(B) ≤ p−(A) ≤ p−(B) + 1. These conditions together imply that p+(A) = p+(B) + 1 and p−(A) = p−(B) + 1. That completes
the proof. ut
Corollary 7.5. Let A be a symmetric matrix partitioned as A= A11A12
A21A22
,
where A11and A22are square. Let u be a vector such that A11u= 0 and A21u6= 0.
Then p−(A) ≥ p−(A11) + 1.
Proof. Since A21u6= 0, there exists a column c of A12such that c0u6= 0. Let d be the diagonal entry of A22corresponding to the column c of A12. Consider the matrix
X= A11 c c0 d
.
By Lemma 7.4, p−(X ) = p−(A11) + 1. Also, since X is a principal submatrix of A, by the interlacing theorem, p−(A) ≥ p−(X ). It follows that p−(A) ≥ p−(A11) + 1.
u t In Theorem 6.2, Chapter 6, we proved the main aspects of Perron–Frobenius theory confining ourselves to adjacency matrices. The theorem can be proved, es-sentially by the same method, for any nonnegative, “irreducible” matrix. Here we do not yet need the theorem in its full generality, however we do need it for a small modification of the adjacency matrix. The result is stated next. The proof is along the lines of that of Theorem 6.2.
Theorem 7.6. Let G be a connected graph with n ≥ 2 vertices, and let A be the ad-jacency matrix of G. Let E ≥ 0 be a diagonal matrix. Then the following assertions hold:
(i) E+ A has an eigenvalue λ > 0 and an associated eigenvector x > 0.
(ii) For any eigenvalue µ 6= λ of E + A, −λ ≤ µ < λ .
(iii) If u is an eigenvector of E+ A for the eigenvalue λ , then u = αx for some α.
7.2 Classification of trees 85 We will refer to the eigenvalue λ in (i) of Theorem 7.6 as the Perron eigenvalue of E + A.
Corollary 7.7. Let G be a connected graph with n vertices and let A be the adja-cency matrix of G. Let E be a diagonal matrix of order n and let τ1≤ τ2· · · ≤ τn
be the eigenvalues of E− A. Then the algebraic multiplicity of τ1is1 and there is a positive eigenvector of E− A corresponding to τ1.
Proof. Let B = kI − (E − A), where k > 0 is sufficiently large so that kI − E ≥ 0.
The eigenvalues of B are k − τ1≥ k − τ2· · · ≥ k − τn. Since B = (kI − E) + A, by Theorem 7.6 k − τ1, which is the Perron eigenvalue of B, has algebraic multiplicity 1 and there is a positive eigenvector corresponding to the same. It follows that τ1, as an eigenvalue of E − A, has algebraic multiplicity 1 with an associated positive
eigenvector. ut
For a symmetric matrix B, let τ(B) denote the least eigenvalue of B.
Theorem 7.8. Let T be a tree with V (T ) = {1, . . . , n}. Let L be the Laplacian of T and µ the algebraic connectivity. Let x be a Fiedler vector and suppose n is a characteristic vertex. Let T1, . . . , Tkbe the components of T\ {n}. Then for any
j= 1, . . . , k, the vertices of V (Tj) are either all positive, all negative or all zero.
Proof. Recall that since n is a characteristic vertex, xn= 0 and n is adjacent to a positive as well as a negative vertex. As observed earlier, n cannot be a pendant vertex and hence k ≥ 2. Partition L and x conformally so that Lx = µx is expressed
as obtained by deleting the row and the column n. By Corollary 7.7 there exists a vector u > 0 such that L2u= τ(L2)u. Augment u suitably by zeros to get the vector
Now suppose τ(L1) = τ(L2) ≤ µ. Then L1 0 0 L2
has at least two eigenvalues not exceeding µ. By the interlacing theorem, L must have two eigenvalues not exceed-ing τ(L1). It follows that τ(L1) = τ(L2) = µ. By Corollary 7.7 it follows that xj> 0 or xj< 0 for j = 1, 2. A similar argument shows that for j = 3, . . . , k, if xj6= 0 then either xj> 0 or xj< 0. That completes the proof. ut Corollary 7.9. Let T be a tree with V (T ) = {1, . . . , n}. Let L be the Laplacian of T and µ the algebraic connectivity. Let x be a Fiedler vector. Then T has at most one characteristic vertex with respect to x.
Proof. Suppose i 6= j are both characteristic vertices with respect to x. Then xi= xj= 0. By Theorem 7.7 all vertices of the component of T \ {i} that contains j are zero vertices. Then j cannot be adjacent to a nonzero vertex and thus it cannot be a
characteristic vertex. ut
Proof. We prove the result by induction on n. The proof is easy when n = 2. Assume the result to be true for matrices of order n − 1. We assume, without loss of general-ity, that vertex n is pendant and is adjacent to n−1. Let z be a vector such that Az = 0.
Then the nth equation gives an−1,nzn−1+ annzn= 0. Since an−1,n= −ann6= 0, it fol-lows that zn−1= zn. As usual, let A(n, n) be the submatrix obtained by deleting row and column n of A. Also, let z(n) be the vector obtained by deleting the last coordi-nate of z. The first n − 1 equations from Az = 0 give that GB is the tree T \ {n} and B1 = 0. By, induction assumption it follows that rank (B) = n − 2 and therefore z(n) must be a scalar multiple of 1. It follows that z is a scalar multiple of 1 and hence rank (A) = n − 1. ut Theorem 7.11. Let G be a tree with V (G) = {1, . . . , n}. Let L be the Laplacian of G and µ the algebraic connectivity. Suppose there exists a Fiedler vector with no zero coordinate. Then µ has algebraic multiplicity 1.
Proof. Let Ly = µy where yi6= 0, i = 1, . . . , n. Let E = diag(y1, . . . , yn) and let C = E(L − µI)E. Then GCis a tree and C1 = 0. It follows by Lemma 7.9 that rank (C) = n− 1. Then rank (L − µI) is n − 1 as well, and hence µ has algebraic multiplicity 1.
u t
7.2 Classification of trees 87 Let T be a tree with V (T ) = {1, . . . , n}. Let L be the Laplacian of T and µ the al-gebraic connectivity. Let x be a Fiedler vector and suppose x has no zero coordinate.
Then by Theorem 7.11, µ has algebraic multiplicity 1, and hence any other Fiedler vector must be a scalar multiple of x. Thus, in this case there is an edge of T that is the characteristic edge with respect to every Fiedler vector. An analogous result holds for a characteristic vertex as well, as seen in the next result.
Theorem 7.12. Let T be a tree with V (T ) = {1, . . . , n}. Let L be the Laplacian of T and µ the algebraic connectivity. Let x and y be Fiedler vectors. Then a vertex i is a characteristic vertex with respect to x if and only if it is a characteristic vertex with respect to y.
Proof. At the outset we note a consequence of Theorem 7.8, which will be used. If xis a Fiedler vector of a tree and has a zero coordinate, then for any vertices i, j and kof the tree such that j is on the i − k path, if xi= xk= 0 then xj= 0.
We turn to the proof. If µ has algebraic multiplicity 1 then x is a scalar multiple of y and the result is obvious. So, suppose µ has algebraic multiplicity greater than 1, and let
V0= { j ∈ V (T ) : zj= 0, for any Fiedler vector z}.
If V0= φ then for each vertex j we can find a Fiedler vector zj such that the jth coordinate of zjis nonzero. Then there must be a vector z with no zero entry that is a linear combination of zj, j = 1, . . . , n. Note that z is a Fiedler vector, contradicting Theorem 7.11. Therefore, V06= φ .
There must be a vertex k ∈ V0that is adjacent to a vertex not in V0. Suppose there are two vertices k1, k2∈ V0adjacent to vertices not in V0. Specifically, suppose k1
is adjacent to `16∈ V0and k2is adjacent to `26∈ V0. Then there are Fiedler vectors w1and w2such that the ki-coordinate of wiis zero, while the `i-coordinate of wiis nonzero, i = 1, 2. We may take a linear combination w of w1 and w2, which then is a Fiedler vector, with respect to which both k1and k2are characteristic vertices, contradicting Corollary 7.10. Hence, k ∈ V0is the unique vertex adjacent to a vertex not in V0.
We claim that k is the characteristic vertex with respect to any Fiedler vector.
Suppose i 6= k is the characteristic vertex with respect to the Fiedler vector x. There must be a vertex j adjacent to i such that xj6= 0. Thus, i 6∈ V0and there is a Fiedler vector y such that yi6= 0. Since xi= xk= 0 and xj6= 0, by the structure implied by Theorem 7.8, i is on the j − k path. It follows by the observation in the beginning that yj6= 0.
We may take a linear combination z of x and y satisfying zj= 0 and zi6= 0.
However, zk= 0, which again contradicts the observation in the beginning since i is on the j − k path. We conclude that k is the characteristic vertex with respect to any
Fiedler vector. ut
We are now in a position to describe a classification of trees. Let T be a tree with V(T ) = {1, . . . , n}. We say that T is of Type I if it has a characteristic vertex with respect to any Fiedler vector, while T is said to be of Type II if it has a characteristic
edge with respect to any Fiedler vector. As discussed earlier, neither the character-istic vertex nor the charactercharacter-istic edge depend on the particular Fiedler vector. Note that every tree must be of one of the two types. A tree cannot be both Type I and Type II. Indeed, in that case µ must have algebraic multiplicity at least 2 and then, by Theorem 7.11, it cannot have a Fiedler vector with all coordinates nonzero, a contradiction.
It must be remarked that if µ has algebraic multiplicity greater than 1 then T is necessarily of Type I. However, the converse is not true. If T is the path on 3 vertices then the central vertex is a characteristic vertex, although the algebraic multiplicity of µ = 1 is 1.