On the Fiedler vectors of graphs that arise from trees by Schur complementation of the Laplacian

Abstract

The utility of Fiedler vectors in interrogating the structure of graphs has generated intense interest and motivated the pursuit of further theoretical results. This paper focuses on how the Fiedler vectors of one graph reveal structure in a second graph that is related to the first. Specifically, we consider a point of articulation r in the graph G whose Laplacian matrix is L and derive a related graph G_{r} whose Laplacian is the matrix obtained by taking the Schur complement with respect to r in L. We show how Fiedler vectors of G_{r} relate to the structure of G and we provide bounds for the algebraic connectivity of G_{r} in terms of the connected components at r in G. In the case where G is a tree with points of articulation r ∈ R, we further consider the graph G_R derived from G by taking the Schur complement with respect to R in L. We show that Fiedler vectors of G_R valuate the pendent vertices of G in a manner consistent with the structure of the tree.

1. INTRODUCTION

Let G be a connected weighted graph with vertex set V and edge set E such that each edge is associated with a positive weight. For such a weighted graph, let A (G) = (a_ij) denote the adjacency matrix of G given by

$$a_{\mathit{\text{ij}}}=\{\begin{array}{cc}\omega \hfill & \text{ if }(i,j)\in E\text{ and the weight of the edge is }\omega \hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$$

Let D (G) = (d_ij) be the diagonal matrix with d_ii = ∑_j≠i a_ij. We use L (G) = D (G) − A (G) to denote the Laplacian matrix of G and suppress the G throughout when the context is clear. We follow the convention of using e to denote a conformal vector of ones, and it is clear that Le = 0. It is well known that zero is the smallest eigenvalue of L, and when G is connected, as is assumed here, that smallest eigenvalue is simple [1]. The second smallest eigenvalue of L is referred to as the algebraic connectivity of the graph G and often denoted μ. In this context, vectors Y satisfying the eigenvalue-eigenvector relationship LY = μY are known as Fiedler vectors in honor of the mathematician Miroslav Fiedler [2].

Fiedler first elucidated the properties of the vectors that bear his name, and of primary interest is how such vectors relate back to the graph from which they are derived. Each entry of Y corresponds to a vertex in the graph and Fiedler showed, among other things, that the sets {υ ∈ V : Y (υ) ≥ 0} and {υ ∈ V : Y (υ) ≤ 0} induce subgraphs of G that are connected. Thus, the sign pattern of a Fiedler vector Y reveals important structure in the graph G, and the body of literature exploiting that fact is extensive. Similarly, theoretical studies of Fiedler vectors and algebraic connectivity abound, including a number of recent works that consider how μ is affected by graph perturbation (e.g. [3]). Much rarer are perturbation results about the Fiedler vectors themselves, possibly because most interesting perturbations change the vertex set of the graph. We elucidate the properties of one such perturbation here.

Our particular interest is Schur complementation. Throughout the text we adopt the standard Schur complement notation of $B/B_{11}=B_{22}-B_{21}{B}_{11}^{-1}{B}_{12}$ for a block matrix

$$B=\left[\begin{array}{cc}\hfill B_{11}\hfill & \hfill B_{12}\hfill \\ \hfill B_{21}\hfill & \hfill B_{22}\hfill \end{array}\right].$$

For any proper subset S of V, we use L_S to denote the principal submatrix of L whose rows and columns are indexed by elements in S. Thus, for a vertex υ ∈ V corresponding to the m^th row and column of L, L_{V −{υ}} = L (m|m) and L_{υ} = L_m,m. The matrix

$$L/L_{\{\upsilon \}}=L_{V-\{\upsilon \}}-\frac{L_{V-\{\upsilon \}}{\mathit{\text{ee}}}^T{L}_{V-\{\upsilon \}}}{e^T{L}_{V-\{\upsilon \}}e}$$ (1.1)

is the Laplacian of a graph that we will denote G_{υ}, and it is this perturbation of G that forms the basis of our study. In Section 3 we build upon our knowledge of G_{υ} to consider the graph G_S whose Laplacian L/L_S is obtained by removing the vertices in S from G by Schur complementation. To that end, we embark on a study of the graph G_{υ} whose Laplacian L/L_{υ} is given in (1.1).

2. PRELIMINARY RESULTS

Recall that a point of articulation (or cutpoint) in G is a vertex r ∈ V whose deletion induces a subgraph G − r with two or more connected components. Let C₀,…, C_k be the connected components at r in G and let L_Ci (i = 0,…, k) denote the principal submatrix of L whose rows and columns correspond to the vertices in C_i. It is well known that $L_{C_i}^{-1}$ is an entrywise positive matrix whose maximal eigenvalue (henceforth Perron value) $\rho \left(L_{C_i}^{-1}\right)$ is simple. The component C_i at r for which $\rho \left(L_{C_i}^{-1}\right)$ is maximal is called a Perron component at r, and we remark that there may be more than one such component [4, 5, 6]. Recent studies have described an intimate relationship between Fiedler vectors and Perron components, and we build upon that line of research below.

The results in this section are general and pertain to arbitrary graphs G. We consider a point of articulation r ∈ V and study the graph G_{r} whose Laplacian is L/L_{r}. That L/L_{r} is a Laplacian is well known [7], and the structure of its graph is intuitive. Specifically, the vertex set of G_{r} is V − {r}, and any edges in G that are not incident to r persist in G_{r}. The effect of Schur complementation is to create edges in G_{r} between any pairs of vertices that are adjacent to r in G. Thus, the structure of G_{r} is related to the structure of G, and in what follows we describe how Fiedler vectors of G_{r} relate to the structure of G as well. We begin with a useful lemma that describes how the algebraic connectivity of G_{r} depends on the connected components at r in G.

Lemma 2.1. Let G be a connected graph with vertex set V and let r ∈ V be a point of articulation in G such that the connected components at r are labeled C₀, C₁,…, C_k. Let L be the Laplacian of G and let G_{r} be the graph whose Laplacian is the Schur complement L/L_{r}. Let L_Ci denote the principal submatrix of L whose rows and columns correspond to the vertices in C_i, let ${\lambda }_i=\rho \left(L_{C_i}^{-1}\right)$ be the Perron value of $L_{C_i}^{-1}$, and suppose that the C_i are labeled so that the sequence λ_i, i = 0,…, k is nonincreasing. Let Y₀ and Y₁ be Perron vectors (i.e. eigenvectors of unit norm) corresponding to λ₀ and λ₁. Then the algebraic connectivity μ of G_{r} satisfies

$${(\frac{{(e^T{Y}_0)}^2}{{(e^T{Y}_0)}^2+{(e^T{Y}_1)}^2}{\lambda }_1^{-1}+\frac{{(e^T{Y}_1)}^2}{{(e^T{Y}_0)}^2+{(e^T{Y}_1)}^2}{\lambda }_0^{-1})}^{-1}\le {\mu }^{-1}\le {\lambda }_0$$

Proof. The Laplacian L can be written as

$$L=\left[\begin{array}{ccccc}\hfill L_{C_0}\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill -L_{C_0}e\hfill \\ \hfill 0\hfill & \hfill L_{C_1}\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill & \hfill -L_{C_1}e\hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill 0\hfill & \hfill ⋮\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill L_{C_k}\hfill & \hfill -L_{C_k}e\hfill \\ \hfill -e^T{L}_{C_0}\hfill & \hfill -e^T{L}_{C_1}\hfill & \hfill \cdots \hfill & \hfill -e^T{L}_{C_k}\hfill & \hfill {\displaystyle {\sum }_i}{e}^T{L}_{C_i}e\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill B\hfill & \hfill -\mathit{\text{Be}}\hfill \\ \hfill -e^{T}B\hfill & \hfill e^T\mathit{\text{Be}}\hfill \end{array}\right]$$

where B represents the upper 4 × 4 block. Each of L_Ci is an M-matrix and has an inverse that is positive; hence the Perron vectors Y₀ and Y₁ in the statement of the theorem exist. Moreover, by Perron’s Theorem, the entries of Y₀ and Y₁ are positive, so that e^TY₀ and e^TY₁ are nonzero. Consider the vectors $Y_0^{*}\text{ and }{Y}_1^{*}$ obtained from Y₀ and Y₁ by appending zero entries to each conformally with B:

$$Y_0^{*}=\left[\begin{array}{c}\hfill Y_0\hfill \\ \hfill 0\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\text{ and }{Y}_1^{*}=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill Y_1\hfill \\ \hfill ⋮\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right].$$

It is clear that $B^{-1}{Y}_0^{*}={\lambda }_0{Y}_0^{*}\text{ and }{B}^{-1}{Y}_1^{*}={\lambda }_1{Y}_1^{*}$, and therefore ${\mathit{\text{BY}}}_0^{*}={\lambda }_0^{-1}{Y}_0^{*}\text{ and }{\mathit{\text{BY}}}_1^{*}={\lambda }_1^{-1}{Y}_1^{*}$ as well. We introduce the vector Z defined as

$$Z=\frac{(e^T{Y}_0^{*})Y_1^{*}-(e^T{Y}_1^{*})Y_0^{*}}{\sqrt{{(e^T{Y}_0^{*})}^2+{(e^T{Y}_1^{*})}^2}}$$

and consider the quadratic form Z^TBZ. Notice that by construction $Y_0^{*}\text{ and }{Y}_1^{*}$ are orthonormal and that $Y_0^{*T}{\mathit{\text{BY}}}_1^{*}=Y_1^{*T}{\mathit{\text{BY}}}_0^{*}=0$. Moreover, from the eigenvalue-eigenvector relationship we have that $Y_0^{*T}{\mathit{\text{BY}}}_0^{*}={\lambda }_0^{-1}\text{ and }{Y}_1^{*T}{\mathit{\text{BY}}}_1^{*}={\lambda }_1^{-1}$. It follows that

$$Z^T\mathit{\text{BZ}}=\frac{{\left(\right(e^T{Y}_0^{*})Y_1^{*}-(e^T{Y}_1^{*}\left)Y_0^{*}\right)}^{T}B\left(\right(e^T{Y}_0^{*})Y_1^{*}-(e^T{Y}_1^{*}\left)Y_0^{*}\right)}{{\left(e^T{Y}_0^{*}\right)}^2+{\left(e^T{Y}_1^{*}\right)}^2}=\frac{({\left(e^T{Y}_1^{*}\right)}^2{Y}_0^{*T}{\mathit{\text{BY}}}_0^{*}+{\left(e^T{Y}_0^{*}\right)}^2{Y}_1^{*T}{\mathit{\text{BY}}}_1^{*})}{{\left(e^T{Y}_0^{*}\right)}^2+{\left(e^T{Y}_1^{*}\right)}^2}=\frac{({\left(e^T{Y}_1^{*}\right)}^2{\lambda }_0^{-1}+{\left(e^T{Y}_0^{*}\right)}^2{\lambda }_1^{-1})}{{\left(e^T{Y}_0^{*}\right)}^2+{\left(e^T{Y}_1^{*}\right)}^2}=\frac{{\left(e^T{Y}_0\right)}^2}{{\left(e^T{Y}_0\right)}^2+{\left(e^T{Y}_1\right)}^2}{\lambda }_1^{-1}+\frac{{\left(e^T{Y}_1\right)}^2}{{\left(e^T{Y}_0\right)}^2+{\left(e^T{Y}_1\right)}^2}{\lambda }_0^{-1}$$

and note that Z^TZ = 1 and e^TZ = 0 by design. We seek to bound the algebraic connectivity μ of L/L_{r}, and the Courant-Fischer minimax principle tells us that

$$\mu =\underset{X:e^{T}X=0}{\text{min}}\frac{X^T(L/L_{\{r\}})X}{X^{T}X}.$$

From this representation, it follows immediately that

$$\mu \le \frac{Z^T(L/L_{\{r\}})Z}{Z^{T}Z}=Z^T(B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}})Z=Z^T\mathit{\text{BZ}}-\frac{\left(Z^T\mathit{\text{Be}}\right)\left(e^T\mathit{\text{BZ}}\right)}{e^T\mathit{\text{Be}}}\le Z^T\mathit{\text{BZ}}$$

where the final inequality holds because e^TBe = ∑_i e^TL_Cie > 0. Thus we have

$$\mu \le \frac{{\left(e^T{Y}_0\right)}^2}{{\left(e^T{Y}_0\right)}^2+{(e^T{Y}_1)}^{{}^2}}{\lambda }_1^{-1}+\frac{{\left(e^T{Y}_1\right)}^2}{{\left(e^T{Y}_0\right)}^2+{\left(e^T{Y}_1\right)}^2}{\lambda }_0^{-1}$$

and since μ > 0 the left-hand inequality of the lemma is established. The right-hand inequality is an application of Weyl’s inequality relating the eigenvalues of B to those of the rank-one perturbation $B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}}$ (see for example Theorem 1.1 of [8]). By assumption, μ is the second smallest eigenvalue of $B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}}$ (the smallest, 0, is simple), whereas ${\lambda }_0^{-1}$ is the smallest eigenvalue of B. These are guaranteed to interlace so that $0\le {\lambda }_0^{-1}\le \mu$. Neither are zero, and so μ⁻¹ ≤ λ₀ as desired.

Previous results from Fallat and Kirkland [9] and from Bapat and Pati [10] established deep relationships between the algebraic connectivity of a weighted graph G and the connected components at any of its points of articulation. Our purpose in establishing Lemma 2.1, as revealed in the following series of corollaries, is to show how the connected components at a point of articulation r in G can be used to bound the algebraic connectivity of the derived graph G_{r}.

Corollary 2.1. Under the conditions of Lemma 2.1, if there is a unique Perron component at r in G, then

$${\lambda }_1<{\mu }^{-1}<{\lambda }_0$$

Proof. That there is a unique Perron component means exactly that λ₁ < λ₀. Both λ₁ and λ₀ are positive, and recall from the proof of Lemma 2.1 that e^TY₀ > 0 and e^TY₁ > 0. Therefore, we have that

$$0<\frac{{\left(e^T{Y}_0\right)}^2}{{\left(e^T{Y}_0\right)}^2+{\left(e^T{Y}_1\right)}^2}<1.$$

Now, for any c strictly between 0 and 1,

$${(c{\lambda }_1^{-1}+(1-c){\lambda }_0^{-1})}^{-1}=\frac{{\lambda }_0{\lambda }_1}{c{\lambda }_0+(1-c){\lambda }_1}$$

is strictly between λ₁ and λ₀. Thus, λ₁ is strictly less then the lower bound in Lemma 2.1, and we conclude that λ₁ < μ⁻¹. To see that μ⁻¹ < λ₀, suppose μ⁻¹ = λ₀ by way of contradiction. Then there exists an eigenvector Y of L/L_{r} with unit norm such that (L/L_{r}) Y = μY. Writing

$$L/L_{\{r\}}=B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}}$$

as in the proof of Lemma 2.1, we have that

$$(B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}})Y=\mu Y.$$

Noting that B is invertible, this can be rearranged as

$$B^{-1}Y={\mu }^{-1}[Y-(\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\left)e\right]$$

from which it follows that

$$Y^T{B}^{-1}Y={\mu }^{-1}[1-(\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\left)Y^{T}e\right].$$

On the other hand, because Y is an eigenvector of L/L_{r} with eigenvalue μ > 0, it is orthogonal to eigenvectors with eigenvalue 0, namely e. Thus Y^Te = 0 and we have that Y^TB⁻¹Y = μ⁻¹ = λ₀. Recall that we have assumed that there is exactly one Perron component at r in G. This implies that the largest eigenvalue of B⁻¹, i.e. λ₀, is simple. Defining $Y_0^{*}$ as in the proof of Lemma 2.1, it follows that $Y_0^{*}$ spans the eigenspace of λ₀, and thus our vector Y is a scalar times $Y_0^{*}$. But whereas e^TY = 0, we have seen that $e^T{Y}_0^{*}$ is nonzero. This contradiction completes the proof.

Corollary 2.2. Under the conditions of Lemma 2.1, if there is not a unique Perron component at r in G, then

$${\lambda }_1={\mu }^{-1}={\lambda }_0$$

Proof. The assumption that there are at least two Perron components means exactly that λ₀ = λ₁. Substituting this into the bounds of Lemma 2.1 yields the desired result.

Lemma 2.1 and its corollaries bound the algebraic connectivity of G_{r} in terms of the connected components at r in G and characterize the conditions under which these inequalities are strict. When there is not a unique Perron component at r in G, so that λ₁ = μ⁻¹ = λ₀, the number of Perron components relates to the multiplicity of μ as an eigenvalue of L/L_{r}. To establish this relationship requires use of the following lemma.

Lemma 2.2. Let B be a symmetric positive definite real matrix with eigenvalues ordered λ₀ (B) ≥ λ₁ (B) ≥ ⋯ λ_n (B) > 0. Let $M=B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}}$ and order its eigenvalues λ₀ (M) ≥ λ₁ (M) ≥ ⋯ λ_n (M) = 0.. Suppose that λ_n (B) = λ_n−1 (M), and let Y be an eigenvector of M with eigenvalue λ_n−1 (M). Then Y is an eigenvector of B as well.

Proof. Let Y be as described in the statement of the lemma and suppose without loss of generality that Y^TY = 1. We have that MY = λ_n (B) Y and so

$$(B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}})Y={\lambda }_n(B)Y$$ (2.1)

as well. Noting that by assumption B is invertible, (2.1) can be rearranged as

$$B^{-1}Y={\lambda }_n{(B)}^{-1}[Y-(\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\left)e\right]$$

from which it follows that

$$Y^T{B}^{-1}Y={\lambda }_n{(B)}^{-1}[1-(\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\left)Y^{T}e\right].$$ (2.2)

Because Y is an eigenvector of M with eigenvalue λ_n−1 (M), the Courant-Fischer minimax principle guarantees that Y is orthogonal to eigenvectors of M with eigenvalue λ_n (M), namely e. Thus Y^Te = 0 and (2.2) becomes Y^TB⁻¹Y = λ_n (B)⁻¹. Now λ_n (B)⁻¹ is the largest eigenvalue of B⁻¹, and so Y^TB⁻¹Y = λ_n (B)⁻¹ implies that B⁻¹Y = λ_n (B)⁻¹ Y as well. Hence BY = λ_n (B) Y, completing the proof of the lemma.

In the context of Lemma 2.1, we have now established that in the case where there are two or more Perron components at r in G, the Fiedler vectors of L/L_{r} are also eigenvectors corresponding to eigenvalues of L_V−{r} of maximum modulus. This has interesting implications. Suppose that there are m ≥ 2 Perron components at r in G. We know from Corollary 2.2 that λ₁ = μ⁻¹ = λ₀, and it is clear that this implies λ₀ = λ₁ = ⋯ = λ_m−1 = μ⁻¹. Let Y₀, Y₁,…, Y_m−1 denote the corresponding Perron vectors, and define $Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*}$ as in the proof of Lemma 2.1. The vectors $Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*}$ span the eigenspace of λ₀ in L_{r} and thus span $(\{Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*})\}$ contains the eigenspace of μ in L/L_{r}. On the other hand, it easy to verify that the eigenspace of μ in L/L_{r} contains all linear combinations of $Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*}$ that are orthogonal to the vector e. From this we deduce that the eigenspace of μ in L/L_{r} is exactly $\{\upsilon :(L/L_{\{r\}})\upsilon =\mu \upsilon \}=\{\upsilon ={\displaystyle {\sum }_{i=0}^{m-1}}{a}_i{Y}_i^{*}:a_0,\dots ,a_{m-1}\in ℝ\text{ and }{e}^T\upsilon =0\}$ ${\displaystyle {\sum }_{i=0}^{m-1}}{a}_i{Y}_i^{*}$. The dimension of this space is evidently m − 1, one less than the number of Perron components. We collect these results in a proposition for future reference.

Proposition 2.1. Suppose that there are m ≥ 2 Perron components at r in G, and let $Y_0^{*},Y_1^{*},\dots ,Y_m^{*}$ be as defined in the proof of Lemma 2.1. Then the Fiedler vectors Y of L/L_{r} satisfy e^TY = 0 and can be written as ${\displaystyle {\sum }_{i=0}^{m-1}}{a}_i{Y}_i^{*}$ for some scalars a₀,…, a_m−1 ∈ ℝ. Conversely, for any scalars a₀,…, a_m−1 ∈ ℝ, if $Y={\displaystyle {\sum }_{i=0}^{m-1}}{a}_i{Y}_i^{*}$ satisfies e^TY = 0 then it is a Fiedler vector of L/L_{r}.

3. MAIN RESULTS

In the previous section we described the relationship between Fiedler vectors of G_{r} and the connected components at r in G. In this section we make the additional assumptions that r is adjacent to every other vertex in V and that the connected components at r in G are complete. As the following theorem shows, these further restrictions guarantee that the sign pattern of a Fiedler vector Y of G_{r} can be used to cut the graph G. The proof indicates where the assumptions can be weakened without invalidating its result.

Theorem 3.1. Let G be a connected graph with vertex set V and let r ∈ V be a point of articulation in G adjacent to every other vertex in V and such that the connected components at r are complete. Let L be the Laplacian of G and let G_{r} be the graph whose Laplacian is the Schur complement L/L_{r}. For any Fiedler vector Y of L/L_{r} corresponding to the algebraic connectivity μ of G_{r}, let V₊ = {υ ∈ V − {r} : Y (υ) > 0} and let V₋ be its complement in V − {r}. Then there exists a Fiedler vector Y such that (1) both V₊ and V₋ ∪ {r} induce connected subgraphs in G or (2) both V₋ and V₊ ∪ {r} induce connected subgraphs in G.

Proof. The proof hinges on the structure of the graph G. As in the proof of Lemma 2.1, enumerate the connected components at r as C₀, C₁,…, C_k and permute the Laplacian of G so that it can be written as

$$L=\left[\begin{array}{ccccc}\hfill L_{C_0}\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill -L_{C_0}e\hfill \\ \hfill 0\hfill & \hfill L_{C_1}\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill & \hfill -L_{C_1}e\hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill 0\hfill & \hfill ⋮\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill L_{C_k}\hfill & \hfill -L_{C_k}e\hfill \\ \hfill -e^T{L}_{C_0}\hfill & \hfill -e^T{L}_{C_1}\hfill & \hfill \cdots \hfill & \hfill -e^T{L}_{C_k}\hfill & \hfill {\sum }_i{e}^T{L}_{C_i}e\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill B\hfill & \hfill -\mathit{\text{Be}}\hfill \\ \hfill -e^{T}B\hfill & \hfill e^T\mathit{\text{Be}}\hfill \end{array}\right]$$

where B represents the upper 4 × 4 block. Let λ_i once again denote the Perron value $\rho \left(L_{C_i}^{-1}\right)$ of the entrywise positive matrix $L_{C_i}^{-1}$, and without loss of generality assume the sequence λ_i, i = 0,…, k is nonincreasing. We are interested in the matrix L/L_{r}, which in terms of B can be written as

$$L/L_{\{r\}}=B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}}$$

Now consider the Fiedler vector Y of L/L_{r} and partition it conformally as

$$Y=\left[\begin{array}{c}\hfill Y_0\hfill \\ \hfill Y_1\hfill \\ \hfill ⋮\hfill \\ \hfill Y_n\hfill \end{array}\right].$$

The eigenvector-eigenvalue relationship gives (L/L_{r}) Y = μY, or in other words

$$(B-\frac{{\mathit{\text{Bee}}}^{T}B}{e^T\mathit{\text{Be}}})Y=\mu Y$$

which can be rewritten as

$$(B^{-1}-{\mu }^{-1}I)Y=\left({\mu }^{-1}\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\right)e.$$

The right-hand side of this equation is a scalar times e, while the left-hand can be expanded as

$$(B^{-1}-{\mu }^{-1}I)Y=\left[\begin{array}{c}\hfill (L_{C_0}^{-1}-{\mu }^{-1}I)Y_0\hfill \\ \hfill (L_{C_1}^{-1}-{\mu }^{-1}I)Y_1\hfill \\ \hfill ⋮\hfill \\ \hfill (L_{C_k}^{-1}-{\mu }^{-1}I)Y_k\hfill \end{array}\right].$$

Thus we have for each i = 0,…, k that $(L_{C_i}^{-1}-{\mu }^{-1}I)Y_i=\left({\mu }^{-1}\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\right)e$.

First consider the case where there is a unique Perron component at r in G_{r}. Corollary 2.1 guarantees that λ₁ < μ⁻¹, and since λ₁ is the largest of the eigenvalues of $L_{C_i}^{-1}$ it is clear that μ⁻¹ cannot be an eigenvalue of $L_{C_i}^{-1}$ for any i ≥ 1. This implies that $L_{C_i}^{-1}-{\mu }^{-1}I$ is invertible, which allows us to solve for Y_i, i ≥ 1 as

$$Y_i={(L_{C_i}^{-1}-{\mu }^{-1}I)}^{-1}\left({\mu }^{-1}\frac{e^T\mathit{\text{BY}}}{e^T\mathit{\text{Be}}}\right)e.$$

Next, note that ${\mu }^{-1}I-L_{C_i}^{-1}$ is by definition an M-matrix since $L_{C_i}^{-1}$ is a positive matrix whose spectral radius (Perron value) is strictly less than μ⁻¹ (see, for example [11]). Thus, the matrix ${\mu }^{-1}I-L_{C_i}^{-1}$ is inverse positive, meaning that the entries of ${({\mu }^{-1}I-L_{C_i}^{-1})}^{-1}$ are strictly greater than zero. This implies that the entries of ${(L_{C_i}^{-1}-{\mu }^{-1}I)}^{-1}$ are negative for each i ≥ 1. Thus, the entries of ${(L_{C_i}^{-1}-{\mu }^{-1}I)}^{-1}$ e are negative, which proves that each of the entries of Y_i is the opposite sign of e^TBY (since μ⁻¹ and e^TBe are positive). First suppose that e^TBY is positive, so that none of the vertices of C_i, i ≥ 1 are contained in V₊. Then V₊ contains only vertices in C₀, and since C₀ is complete, V₊ induces a connected subgraph in G. Because r has an edge incident to every vertex, V₋ ∪ {r} must be connected as well. Conversely, if e^TBY is negative, it is V₋ that contains none of the vertices of C_i, i ≥ 1. Again, because C₀ is complete, V₋ induces a connected subgraph in G. That V₊ ∪ {r} is connected follows as before.

Thus, the theorem holds in the case where there is a unique Perron component, and we are left to consider the alternative. Suppose that there are m ≥ 2 Perron components at r in G_{r} labelled C₀, C₁,…, C_m−1 as before. Let Y₀, Y₁,…, Y_m−1 be Perron vectors corresponding respectively to the $L_{C_i}^{-1}$, and define $Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*}$ as in the proof of Lemma 2.1. Thus, as before, for 0 ≤ i ≤ m − 1, $Y_i^{*}$ valuates the vertices of C_i as positive and valuates all other vertices as zero. Proposition 2.1 characterizes the Fiedler vectors of L/L_{r} as vectors $Y={\displaystyle {\sum }_{i=0}^{m-1}}{a}_i{Y}_i^{*}$, a₀,…, a_m−1 ∈ ℝ that satisfy e^TY = 0. Therefore, in light of how the $Y_0^{*},Y_1^{*},\dots ,Y_{m-1}^{*}$ were constructed, any vertices $\upsilon \in V-{\cup }_{i=0}^{m-1}{C}_i$ (i.e. those that do not belong to one of the m Perron components at r in G) are valuated as zero by all Fiedler vectors of L/L_{r}. The remaining vertices can be valuated as either positive or negative depending on the choice of Fiedler vector; however, vertices from the same connected component in G are always valuated with the same sign. To complete the proof, we simply choose a Fiedler vector whose positive valuations are restricted to exactly one Perron component, say C₀. Consider, for example, the vector

$$Y\triangleq \frac{\left({\displaystyle {\sum }_{i=1}^{m-1}}{e}^T{Y}_i^{*}\right)Y_0^{*}-(e^T{Y}_0^{*}){\displaystyle {\sum }_{i=1}^{m-1}}{Y}_i^{*}}{\sqrt{{\left({\displaystyle {\sum }_{i=1}^{m-1}}{e}^T{Y}_i^{*}\right)}^2+{(e^T{Y}_0^{*})}^2}}$$

which is clearly a Fiedler vector of L/L_{r} by Proposition 2.1. Note that ${\displaystyle {\sum }_{i=1}^{m-1}}{e}^T{Y}_i^{*}>0$ so that $\left({\displaystyle {\sum }_{i=1}^{m-1}}{e}^T{Y}_i^{*}\right)Y_0^{*}$ valuates the vertices of C₀ as positive and valuates all remaining vertices as zero. In addition, $e^T{Y}_0^{*}>0$ and the vector ${\displaystyle {\sum }_{i=1}^{m-1}}{Y}_i^{*}$ valuates the vertices of C_i, 1 ≤ i ≤ m − 1 as positive and valuates all remaining vertices as zero. Taken together, the vector Y is positive on C₀, negative on C_i, 1 ≤ i ≤ m − 1, and zero elsewhere. We conclude that V₊ contains only vertices in C₀, which as above implies that both V₊ and V₋ ∪ {r} are connected.

We now specialize to the case where the graph G is a tree. As such, the vertex set V can be partitioned into points of articulation R and pendent vertices P. Recall that for any S ⊂ V, L_S denotes the principal submatrix of L that corresponds to the vertices in S. If S is a proper subset of the vertex set V of G, the matrix L/L_S can also be viewed as a Laplacian, and we call its graph G_S. In light of the discussion in Section 2, we can make the following claim:

Claim 3.2. Let G be a tree and let V be its vertex set. Let S be a proper subset of V and let G_S be the graph whose Laplacian matrix is L/L_S. Then there is an edge in G_S connecting i,j ∈ V − S if and only if either (1) i and j are adjacent in G or (2) there is a path from i to j in G that only traverses vertices in S.

When R ⊂ S so that S includes all of the points of articulation of the tree G, it follows that G_S is complete. We focus on the case R = S, as the graph G_R holds special interest. In particular, we can obtain the graph G_R from the graph G_R−{r} by Schur complementation for any point of articulation r ∈ R in G, and G_R−{r} retains some structure of the tree G at the vertex r. Specifically, the graph G_R−{r} contains exactly one point of articulation, namely r. Moreover, the connected components at r in G_R−{r} are complete, and r is adjacent to each of the vertices in V − R. Thus, G_R−{r} is of exactly the structure required to invoke Theorem 3.1. Therefore, if Y is a Fiedler vector corresponding to the graph G_R that valuates the vertices in V − R, it follows that either (1) both V₊ and V₋ ∪ {r} induce connected subgraphs in G_R−{r} or (2) both V₋ and V₊ ∪ {r} induce connected subgraphs in G_R−{r}. Crucially, this is true for every r ∈ R, and as the following theorem shows, a consequence is that meaningful structure in G can be recovered.

Theorem 3.3. Let G be a tree with vertex set V and let R ⊂ V be the set of points of articulation in G. Let L be the Laplacian of G and let G_R be the graph whose Laplacian is the Schur complement L/L_R. Let Y be a Fiedler vector of L/L_R corresponding to the algebraic connectivity μ of G_R. Let V₊ = {υ ∈ V − R : Y (υ) > 0} and let V₋ be its complement in V − R. Then there exists a subset S of R such that both V₊ ∪ S and V₋ ∪ (R − S) induce connected subgraphs in G.

Proof. Our proof is constructive. Partition the vertex set V of G into two sets R and P containing points of articulation and pendent vertices, respectively, and note that the vertex set of G_R is V − R = P. Index the vertices in R as r_k arbitrarily and consider the graphs G_R−{rk}. We remark for the sake of clarity that the vertices in V₊ and V₋ are pendent in G, that the vertex set of G_R−{rk} is P ∪ {r_k}, and that V₊ ∪ V₋ = P. Now from Theorem 3.1, we know that either (1) V₊ and V₋ ∪ {r_k} induce connected subgraphs in G_R−{rk} or (2) V₋ and V₊ ∪ {r_k} induce connected subgraphs in G_R−{rk}. Let S = {r_k : V₋ and V₊ ∪ {r_k} are connected in G_R−{rk}}. We claim that V₊ ∪ S and V₋ ∪ (R − S) induce connected subgraphs in G

Let a, b ∈ V₊ ∪ S. Because G is a tree, there exists a unique path from a to b in G. To show that V₊ ∪ S is connected in G, we must prove that each vertex along that path belongs to V₊ ∪ S as well. If a and b are adjacent in G, we are done. Otherwise, let υ be any vertex distinct from a and b that lies on the path between them. By definition, the vertex υ is not pendent, and so we may assume υ ∈ R. Since V₊ ⊂ V − R = P, υ ∉ V₊, and to prove that υ ∈ V₊ ∪ S requires us to show that υ ∈ S. We assume by way of contradiction that υ ∈ R − S.

We proceed in cases, considering first the case where a, b ∈ V₊. Because υ ∈ R − S, V₊ is connected in G_R−{υ}, which by Claim 3.2 is possible only if V₊ is contained in exactly one connected component at υ in G. But for υ to lie on the path between a and b in G means that a and b fall in distinct connected components, contradicting our assumption that υ ∈ R − S. We conclude that if a, b ∈ V₊, then the path between a and b in G is comprised exclusively of vertices in S.

Next, suppose that a ∈ V₊ and b ∈ S. Again, because υ ∈ R − S, V₊ is connected in G_R−{υ}, and so V₊ must be contained in exactly one connected component at υ in G, in this case the one that includes the vertex a. The vertex b ∈ S is in a distinct connected component, and all of the pendent vertices in G that belong to this connected component must therefore belong to V₋. Now because b is a point of articulation, there must be at least one pendent vertex x ∈ V₋ in this connected component whose path in G from x to υ contains b. On the other hand, because υ ∈ R − S, V₋ is not connected in G_R−{υ}, and so there must exist y ∈ V₋ in a connected component at υ in G that is distinct from the one in which x resides. In particular, by construction the path from x to y in G contains b; however, because b ∈ S it must be that V₋ is connected in G_R−{b}. These two statements are contradictory, and so it cannot be that υ ∈ R − S.

Finally, suppose that a, b ∈ S. Once more, because υ ∈ R − S, V₊ is contained in exactly one connected component at υ in G. As υ is on the path from a to b, at least one of a and b belongs to a connected component at υ in G whose pendent vertices are exclusively in V₋. We can assume this to be b without loss of generality, at which point the argument from the previous case applies.

Having reached a contradiction in every case, we conclude that υ ∉ R − S, from which it follows that V₊ ∪ S is connected in G. The proof that V₋ ∪ (R − S) is connected in G is similar.

Theorems 3.1 and 3.3 are implicitly results about splits, a concept we now formally define. Let X ⊂ V and let X₁ and X₂ be nonempty disjoint subsets of X such that X = X₁ ∪ X₂. We call {X₁, X₂} a split of G if there exists a partition {Z₁, Z₂} of V − X such that both X₁ ∪ Z₁ and X₂ ∪ Z₂ induce subgraphs of G that are connected. Theorem 3.1 shows that, under certain conditions, the signs of the entries of Fiedler vectors corresponding to the graph G_{r} are guaranteed to induce splits of the parent graph G. Theorem 3.3 goes further in the case where G is a tree. If G is a tree, {X₁, X₂} is a split in G if and only if there exists at least one edge in E that belongs to every path between arbitrary vertices x₁ ∈ X₁ and x₂ ∈ X₂. When X = P, each split corresponds to exactly one edge in this manner, and vice versa. Theorem 3.3 considers this case and concludes that Fiedler vectors corresponding to the graph G_R can be used to construct bipartitions of P that are splits of G. In short, the valuations of P from Fiedler vectors of L/L_R identify an edge in G by their signs.

4. APPLICATION

Theorem 3.3 is the main result of the paper and has a variety of applications. Our application in this section is to a tree G whose Laplacian has not been directly supplied. As such, let G be a weighted tree with vertex set V and edge set E, and partition V into pendent vertices P and points of articulation R. We suppose that the Laplacian L of G has not been supplied, but that the Laplacian L/L_R of G_R has been supplied instead. The graph G_R is complete and the entries of L/L_R are strictly nonzero; nevertheless, Theorem 3.3 guarantees that Fiedler vectors of L/L_R reveal structure in the tree G.

To illustrate the point and our approach in general, consider a tree G with vertex set V = {1,…, 8} whose Laplacian L is given below:

$$L=\left[\begin{array}{cccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 2& \hfill 0& \hfill 0& \hfill 0& \hfill -2& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 3& \hfill 0& \hfill 0& \hfill 0& \hfill -3& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 2& \hfill 0& \hfill 0& \hfill -2& \hfill 0\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0& \hfill 0& \hfill -1\\ \hfill -1& \hfill -2& \hfill 0& \hfill 0& \hfill 0& \hfill 4& \hfill 0& \hfill -1\\ \hfill 0& \hfill 0& \hfill -3& \hfill -2& \hfill 0& \hfill 0& \hfill 6& \hfill -1\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill -1& \hfill -1& \hfill 3\end{array}\right].$$

The points of articulation in G are R = {6, 7, 8}, while the pendent vertices are P = {1, 2, 3, 4, 5}. We suppose that L has not been supplied and consider the graph G_R whose Laplacian L/L_R is

$$L/L_R=\left[\begin{array}{ccccc}\hfill 0.7258& \hfill -0.5484& \hfill -0.0484& \hfill -0.0323& \hfill -0.0968\\ \hfill -0.5484& \hfill 0.9032& \hfill -0.0968& \hfill -0.0645& \hfill -0.1935\\ \hfill -0.0484& \hfill -0.0968& \hfill 1.4032& \hfill -1.0645& \hfill -0.1935\\ \hfill -0.0323& \hfill -0.0645& \hfill -1.0645& \hfill 1.2903& \hfill -0.1290\\ \hfill -0.0968& \hfill -0.1935& \hfill -0.1935& \hfill -0.1290& \hfill 0.6129\end{array}\right].$$

The algebraic connectivity of G_R, μ = 0.3836, is a simple eigenvalue of L/L_R. Its corresponding Fiedler vector,

$$Y^T=[0.57920.4418-0.4641-0.5008-0.0562]$$ (4.1)

identifies the split {V₊, V₋} = {{1, 2}, {3, 4, 5}} of G, which corresponds to the edge that is incident to 6 and 8 in G. The cut that corresponds to this split partitions the vertices into V₊ ∪ S = {1, 2, 6} and V₋ ∪ (R − S) = {3, 4, 5, 7, 8}, where S = {6} can be constructed as in the proof of Theorem 3.3. The proof of the theorem exploits the relationship between G_R and the three graphs G_{6,7}, G_{6,8}, and G_{7,8} (see Figure 1), and as we have shown, the structure of each of these graphs places bounds on μ and constrains the Fiedler vector Y in some way. For example, in the context of Lemma 2.1, the three connected components C₀ = {1, 2}, C₁ = {3, 4}, C₂ = {5} at 8 in G_{6,7} identify three principal submatrices of L/L_{6,7} whose inverses have Perron values λ₀ = 2.7808, λ₁ = 2.4201 and λ₂ = 1, respectively. The lemma, along with Corollaries 2.1 and 2.2, tells us that μ is a simple eigenvalue of L/L_R bounded strictly between ${\lambda }_0^{-1}=0.3596\text{ and }{\lambda }_1^{-1}=0.4132$. Theorem 3.1, on the other hand, asserts that the Fiedler vector Y must use the same sign to valuate the vertices in C₁ ∪ C₂ = {3, 4, 5}, but it says nothing about the vertices in the Perron component C₀ at 8 in G_{6,7}. For this information we turn to G_{7,8}, whose Perron component at 6 is {3, 4, 5}. Theorem 3.1 now says that Y must use the same sign to valuate the vertices in P − {3, 4, 5} = {1, 2}, which resolves the sign pattern in (4.1) completely. Because G_{6,7}, G_{6,8}, and G_{7,8} collectively carry the structure of the tree G, Theorem 3.3 guarantees this sign pattern will induce a split.

Fig. 1.

The Schur hierarchy linking G and G_{6,7,8}. Arrows are drawn to distinguish pairs of graphs before and after the removal of one vertex by Schur complementation.