Locating the extremal entries of the Fiedler vector for rose trees

In this note we locate the extremal entries of the Fiedler vector for the class of rose trees, which consists of paths with a star graph attached to it. We derive directly from the eigenvector equation conditions to characterize in which cases the extremal values are located either at the end points of the path or on the pendant vertices of the star graph.

Location of the extremal entries. It is a well known result due to Fiedler [5] that the extremal entries of the Fiedler vector are located at the pendant vertices of the tree, see also [6,Lemma 8]. This result was recently extended to quantum graphs in [8]. If a(G) = λ 2 (G) < 1 then one can see directly from the eigenvector equation L(G)x = a(G)x, x ∈ R n , that the absolute value at the pendant vertex v is strictly larger then the value at the unique vertex w with vw ∈ E(G). However it remains unclear at which pendant vertices these entries are located.
It was conjectured that the extremal entries are located at the most distant vertices in the tree. A counter example to this conjecture are rose trees studied in [1,3,7]. A rose tree is given by two paths P l and P t with l and t vertices, respectively, and a star graph S r with r vertices where we select a pendant vertex in each of the graphs and glue them together at a vertex v B which satisfies then deg(v B ) = 3. Roughly speaking, it was shown in [7] that for fixed l and t the number r can be made large enough, such that one extremal entry moves from the end point of the paths P l or P t to the pendant vertices of the star graph S r .
Before we characterize the locations of the extremal entries for rose trees, we state the following auxiliary result. Lemma 1 Let G be a connected graph with Laplacian L(G), eigenvalue λ > 0 and eigenvector x = (x i ) n i=1 then n i=1 x i = 0 and the following holds. (a) Let P be a permutation matrix, i.e. P 2 = I n with P L(G)P = L(G) then P x is also an eigenvector of L(G) for λ.
(b) Let P n be a path with n vertices v 1 , . . . , v n then x n = 2 implies that x is orthogonal to this vector and therefore Therefore P x = 0 is also an eigenvector of λ. In particular if λ is simple then there exists α = 0 with P x = αx. The proof of (b) is based on deriving a linear second-order difference equation from the eigenvector equation and can be found e.g. in [6,Lemma 15] and (c) follows directly from the eigenvector equation L(G)x = λx.
In the following we denote the entries at the pendant vertices in the rose tree by x P l , x Pt and x Sr . Below we state the main result on the location of extremal entries of rose trees and provide an elementary proof. For arbitrary trees one can characterize the extremal entries of the Fiedler vector using the Schur reduction method from [6], which is a technique to obtain a smaller (weighted) graph in such a way that the original eigenvector is preserved.

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Section 23: Applied operator theory Theorem 2 Let T be a rose tree with parameters l, t, r and 1 ≤ l ≤ t and r ≥ 2 then the following holds for λ = a(T ).

(b)
The entries x Sr , x P l and x Pt of the Fiedler vector are related by If l < t then x P l and x B are nonzero and have the same sign.
(c) If r/2 − r 2 /4 − 1 < 2 − 2 cos( π 2t−1 ) then the entries x P l , x B , x Pt are non-zero and have the same sign. Therefore the extremal entries are x Sr and x Pt .
(d) For l, r fixed and t sufficiently large there is one extremal entry x Pt and a second extremal entry of opposite sign x Sr , if (e) If 1−(2r) −1 ≤ cos( π l+t−1 ) and −r(2−2 cos( π t+l+r−1 ))+1 then the extremal entries are x P l and x Pt .
P r o o f. The upper bound in (a) follows from [2, p. 187] and the lower bound can be found in [4]. The simplicity can be obtained from a(T ) < 1 and that we can resolve the eigenvector equations hence there is up to scaling at most one solution of the eigenvector equation. We see by a direct calculation from the eigenvector equation L(T )x = λx and the previous Lemma 1 that where x 0 is the center of the star graph. If l < t and x P l > 0 then the upper bound for a(T ) in (a) implies x l > 0 which proves (b). Next, we prove (c). For this we use the eigenvector equation L(T )x = λx at the row that corresponds to x B and obtain where x 0 is again the center of the star graph and x t−1 and x l−1 are the vertices on the paths connected to x B . We assume that λ < 2 − 2 cos(π/(2t − 1)) then we have x B = 0 and we show that there is a positive solution to the above equation. One of these possible solutions is then a(T ). Since the Fiedler vector is unique up to scaling, we can assume without restriction that x B = 1 and therefore we obtain with the formulas in (b) and Lemma 1 (b) in (2) that Recall that ζ = arccos(1 − λ/2), hence λ = 0 is a solution to the above equation. Furthermore, we have poles for λ ∈ {r/2 ± r 2 /4 − 1, 2 − 2 cos(π/(2l − 1)), 2 − 2 cos(π/(2t − 1))}.
Observe that there is no solution of (3) for λ ∈ (0, r/2− r 2 /4 − 1) because then all eigenvector entries would be positive. This contradicts Lemma 1, where we have shown that the sum of all eigenvector entries is zero. Since the function tends to ∞ for λ ↓ r/2 − r 2 /4 − 1 and to −∞ for λ ↑ 2 − 2 cos(π/(2t − 1)) there is at least one solution λ = 0 in this interval. For all this solutions the corresponding eigenvector has the property that x Sr = 1 λ 2 −rλ+1 < 0 and x Pt > x P l > x B > 0. Hence the Fiedler vector also has this property.
We continue with the proof of (d). The upper bound for a(T ) in (a) implies that a(T ) → 0 as t → ∞. The coefficients in (b) can be viewed as functions of λ and they have the same value at λ = 0. Therefore it remains to compare the derivatives of these functions, to decide which entry x Sr or x P l is larger. Indeed we find the derivative of these functions at λ = 0 to be −r and −l(l + 1)/2. This proves (d). For the proof of (e) we refer to [6, p. 111-112].
Note that (a) and (c) for l = t, i.e. perfect rose trees, were already obtained in [7].