Proof. From Lemma 2.3(p), we obtain $\ell \in \mathbb {N}^+$ such that $P_\ell \not \in \mathcal {G}(\lambda )$ . Clearly, for every nonempty A, the path extensions $(C, A, \ell )$ and $(D, A, \ell )$ , each of which contains $P_\ell $ as a subgraph, are not in $\mathcal {G}(\lambda )$ . With hindsight, using Lemma 2.3(c1, c2, n1, n2), we choose $m \in \mathbb {N}^+$ such that none of the following graphs

(1) $$ \begin{align} (C_3, V_2(C_3), \ell_0, K_m), (C_3, V_2(C_3), K_m), (\overline{K}_2, V(\overline{K}_2), \ell_0, K_m), (\overline{K}_2, V(\overline{K}_2), K_m), \end{align} $$

where $\ell _0 \in \left \{{0, \dots , \ell + 1}\right \}$ , is in $\mathcal {G}(\lambda )$ .

It suffices to prove that every clique extension and every path-clique extension in the extension families $\mathcal {X}(C, \ell , m)$ and $\mathcal {X}(D, \ell , m)$ contains one of the graphs in (1) as a subgraph. Label the vertices of C and D as in Figure 3, and pick an arbitrary $\ell _0 \in \left \{{0, \dots , \ell - 1}\right \}$ . The clique extension $(C, A, K_m)$ and the path-clique extension $(C, A, \ell _0, K_m)$ respectively contain

$$ \begin{align*} (\overline{K}_2, V(\overline{K}_2), 0, K_m) \text{ and } (\overline{K}_2, V(\overline{K}_2), \ell_0 + 1, K_m) & \text{ when } 0 \in A \text{ and } \left\lvert {A \cap \left\{{1,2,3}\right\}} \right\rvert \le 1;\\ (\overline{K}_2, V(\overline{K}_2), 1, K_m) \text{ and } (\overline{K}_2, V(\overline{K}_2), \ell_0 + 2, K_m) & \text{ when } 0 \not\in A \text{ and } \left\lvert {A \cap \left\{{1,2,3}\right\}} \right\rvert = 1;\\ (\overline{K}_2, V(\overline{K}_2), K_m) \text{ and } (\overline{K}_2, V(\overline{K}_2), \ell_0, K_m) & \text{ when } \left\lvert {A \cap \left\{{1,2,3}\right\}} \right\rvert \ge 2, \end{align*} $$

and the clique extension $(D, A, K_m)$ and the path-clique extension $(D, A, \ell _0, K_m)$ respectively contain

$$ \begin{align*} (\overline{K}_2, V(\overline{K}_2), 0, K_m) \text{ and } (\overline{K}_2, V(\overline{K}_2), \ell_0 + 1, K_m) & \text{ when } A \subseteq \left\{{0,2}\right\} \text{ and }A \neq \varnothing; \\ (C_3, V_2(C_3), 0, K_m) \text{ and } (C_3, V_2(C_3), \ell_0 + 1, K_m) & \text{ when } A \in \left\{{\left\{{1}\right\}, \left\{{3}\right\}}\right\}; \\ (\overline{K}_2, V(\overline{K}_2), K_m) \text{ and } (\overline{K}_2, V(\overline{K}_2), \ell_0, K_m) & \text{ when } A \supseteq \left\{{1,3}\right\}; \\ (C_3, V_2(C_3), K_m) \text{ and } (C_3, V_2(C_3), \ell_0, K_m) & \text{ when } \left\lvert {A \cap \left\{{0,2}\right\}} \right\rvert \ge 1 \text{ and } \left\lvert {A \cap \left\{{1,3}\right\}} \right\rvert = 1. \end{align*} $$

Therefore the extension families $\mathcal {X}(C, \ell , m)$ and $\mathcal {X}(D, \ell , m)$ are both disjoint from $\mathcal {G}(\lambda )$ .

Proof. With hindsight, we choose $N = vd^{\ell }$ , where d is the Ramsey number $R(k, 2^{v + 1}m + v)$ , and v is the number of vertices of F. Suppose that G is a connected graph with more than N vertices that contains no member in $\left \{{S_k}\right \} \cup \mathcal {X}(F, \ell , m)$ as a subgraph. Assume for the sake of contradiction that the subgraph of G induced on a vertex subset, say V, is isomorphic to F.

Since G is a connected graph with more than $vd^{\ell }$ vertices, we claim that either there exists a vertex u at distance at least $\ell $ from V, or the maximum degree of G is at least d. Indeed, assume for the sake of contradiction that every vertex of G is within distance $\ell -1$ from V, and the maximum degree of G is less than d. For $i \in \left \{{0, 1, \dots , \ell -1}\right \}$ , let $V_i$ be the set of vertices at distance i from V. Clearly $V_0 = V$ , and so $ \left \lvert {V_0} \right \rvert = v$ . Using the assumption on the maximum degree of G, one can inductively show that $ \left \lvert {V_i} \right \rvert \le vd^i$ . Therefore the number of vertices in G expressed as $\sum _{i = 0}^{\ell - 1} \left \lvert {V_i} \right \rvert $ is at most $\sum _{i=0}^{\ell - 1}vd^i = v(d^\ell - 1)/(d-1) \le vd^\ell $ , which yields a contradiction.

We break the rest of the proof into two cases.

Case 1: There exists a vertex u at distance at least $\ell $ from V. Thus the subgraph $G[V]$ and a shortest path from V to u contains the path extension $(F, A, \ell )$ as a subgraph for some nonempty $A \subseteq V$ , which is a contradiction.

Case 2: The maximum degree of G is at least d. Let u be a vertex of G with the maximum degree, and let $N(u)$ be the set of neighbors of u in G. Since $S_k$ is not a subgraph of G, the subgraph $G[N(u)]$ cannot have an independent set of size k. Because $d = R(k, 2^{v + 1}m + v)$ , the subgraph $G[N(u)]$ contains a clique of order $2^{v + 1}m + v$ , and so there exists a clique $G[U]$ of size $2^{v + 1}m$ that is vertex-disjoint from F.

We further partition the vertices in U as follows. For every subset A of V, let $U(A)$ be the set of vertices in U that are adjacent to all vertices in A and not adjacent to any vertex in $V \setminus A$ . By the pigeonhole principle, there exists a subset A of V such that $ \left \lvert {U(A)} \right \rvert \ge 2m$ . If A is nonempty, then $G[V \cup U(A)]$ contains the clique extension $(F, A, K_m)$ as a subgraph, which is a contradiction.

Hereafter we may assume that $ \left \lvert {U(\varnothing )} \right \rvert \ge 2m$ , hence the distance between V and $U(\varnothing )$ is at least $2$ . Let $v_0 \dots v_{\ell _0}$ be a shortest path between V and $U(\varnothing )$ , where $v_0$ and $v_{\ell _0}$ are respectively at distance $1$ from V and $U(\varnothing )$ . Let $A_1$ be the nonempty set of vertices in V that are adjacent to $v_0$ , and denote by $v_{\ell _0 + 1}$ an arbitrary vertex in $U(\varnothing )$ that is adjacent to $v_{\ell _0}$ . We may assume that $\ell _0 < \ell - 1$ because otherwise $G[V \cup \left \{{v_0, \dots , v_{\ell _0}, v_{\ell _0 + 1}}\right \}]$ would contain the path extension $(F, A_1, \ell )$ as a subgraph, which is a contradiction. Let $A_2$ be the nonempty set of vertices in $U(\varnothing )$ that are adjacent to $v_{\ell _0}$ . If $ \left \lvert {A_2} \right \rvert \ge m$ , then $G[V \cup \left \{{v_0, \dots , v_{\ell _0}}\right \} \cup A_2]$ contains the path-clique extension $(F, A_1, \ell _0, K_m)$ as a subgraph. Otherwise $ \left \lvert {U(\varnothing ) \setminus A_2} \right \rvert> m$ , and so $G[V \cup \left \{{v_0, \dots , v_{\ell _0}, v_{\ell _0 + 1}}\right \} \cup (U(\varnothing ) \setminus A_2)]$ contains the path-clique extension $(F, A_1, \ell _0 + 1, K_m)$ .

Proof. We obtain $\ell , m \in \mathbb {N}^+$ from Lemmas 2.3 and 2.5 such that the following family

$$ \begin{align*} \mathcal{F}_1 := & \left\{{S_4, P_\ell}\right\} \cup \mathcal{X}(C, \ell, m) \cup \mathcal{X}(D, \ell, m) \\ & \cup \left\{{(C_n, V_2(C_n), \ell_0, K_m)}\colon{n \in \left\{{3, \dots, \ell + 1}\right\}, \ell_0 \in \left\{{0, \dots, \ell - 1}\right\}}\right\} \\ & \cup \left\{{(C_n, V_2(C_n), K_m)}\colon{n \in \left\{{3, \dots, \ell + 1}\right\}}\right\} \\ & \cup \left\{{(K_m, V(K_m), \ell_0, K_m)}\colon{\ell_0 \in \left\{{0, \dots, \ell-1}\right\}}\right\} \end{align*} $$

is disjoint from $\mathcal {G}(\lambda )$ .

We obtain $N_0$ from Lemma 2.6 such that for every connected graph G with more than $N_0$ vertices, if no member in $\left \{{S_4}\right \} \cup \mathcal {X}(C, \ell , m) \cup \mathcal {X}(D, \ell , m)$ is a subgraph of G, then neither is C nor D. With hindsight, we choose $N = \max (N_0, (m+1)^{\ell +2})$ . Suppose that G is a connected graph with more than N vertices, and suppose that no member in $\mathcal {F}_1$ is a subgraph of G. By our choice of $N_0$ , we know that neither C nor D is a subgraph of G.

Theorem 2.7 implies that G is the line graph of another connected graph, say H. Since $P_\ell $ is not a subgraph of G, the path $P_{\ell + 1}$ cannot be a general subgraph of H. In particular, the diameterFootnote ⁶ of H is at most $\ell $ . Let d be the maximum degree of H. If $d < m + 2$ , then H has at most $(m+1)^{\ell +1}$ vertices, and at most $(m+1)^{\ell +2}$ edges, and so G has at most $(m+1)^{\ell +2}$ vertices, which is a contradiction.

We may assume that $d \ge m + 2$ . Let u be a vertex of H with degree d. We claim that H is a tree. Suppose on the contrary that H contains a cycle as a general subgraph. Since H does not contain $P_{\ell + 1}$ as a general subgraph, the length of any cycle in H cannot exceed $\ell + 1$ . Suppose in addition that u is on a cycle in H. Let $C_n$ be a shortest cycle containing u, where $n \in \left \{{3, \dots , \ell + 1}\right \}$ . By the choice of $C_n$ , the vertex u has at least $d - 2 \ge m$ neighbors outside $C_n$ . Thus H contains $C_n$ with $S_m$ attached at u as a general subgraph, and so G contains $(C_n, V_2(C_n), K_m)$ as a subgraph, which is a contradiction. Therefore u is not on any cycle in H. Let $C_n$ be any cycle in H, where $n \in \left \{{3, \dots , \ell + 1}\right \}$ . Let $P_{\ell _0}$ be a shortest path between u and $C_n$ , where $\ell _0 \in \left \{{1, \dots , \ell }\right \}$ . Since u is not on any cycle, u has at least $d - 1> m$ neighbors outside $C_n$ and $P_{\ell _0}$ . Thus H contains $C_n \cup P_{\ell _0}$ with $S_m$ attached at u as a general subgraph, and so G contains $(C_n, V_2(C_n), \ell _0 - 1, K_m)$ as a subgraph, which is a contradiction.

We now view H as a tree rooted at u. We claim that u is the only vertex with degree larger than m. Suppose on the contrary that there is another vertex $u'$ with degree at least $m + 1$ in H. Let $P_{\ell _0}$ be a shortest path between u and $u'$ , where $\ell _0 \in \left \{{1, \dots , \ell }\right \}$ . Thus H contains $P_{\ell _0}$ with two vertex-disjoint stars $S_m$ respectively attached to u and $u'$ as a (general) subgraph, and so G contains $(K_m, V(K_m), \ell _0 - 1, K_m)$ as a subgraph, which is a contradiction. Therefore all the vertices but u in H has degree at most m. After the root u is removed from H, each connected component has diameter at most $\ell $ and degree at most m, and so each connected component has at most $m^{\ell +1} \le N$ vertices.

Proof. Suppose that $s_1, s_2, \dots $ is an infinite sequence of distinct tuples in $\mathbb {N}^n$ . For every $i < j$ , color the edge $s_i s_j$ by $0$ if $s_i < s_j$ , otherwise by any $k \in \left \{{1, \dots , n}\right \}$ such that $s_{i,k}> s_{j,k}$ . The infinite Ramsey’s theorem provides an infinite subset $I \subseteq \mathbb {N}^+$ such that the edges $s_i s_j$ , where $i, j \in I$ and $i\neq j$ , all receive the same color c. Because $(\mathbb {N}, \le )$ does not contain infinite descending chains, it must be the case that $c = 0$ . This implies that $\left \{{s_i}\colon {i \in I}\right \}$ is an infinite ascending chain, and in particular $\left \{{s_1, s_2, \dots }\right \}$ is not an antichain.

Proof of Theorem 1.3 for $\lambda < 2$ .

Let $N \in \mathbb {N}$ and $\mathcal {F}_1$ be given by Lemma 2.8, and set

$$ \begin{align*} \mathcal{F}_0 & := \left\{{G \not\in \mathcal{G}(\lambda)}\colon{G \text{ has at most }N \text{ vertices}}\right\}, \\ \widetilde{\mathcal{F}}_2 & := \left\{{G \not\in \mathcal{G}(\lambda)}\colon{\exists H \in \mathcal{T}_N \text{ s.t.\ }G = L(H)}\right\}, \end{align*} $$

where $\mathcal {T}_N$ is the family of rooted trees such that every connected component obtained from removing the root has at most N vertices. Setting $\mathcal {F}_2$ to be the family of graphs that are minimal in $\widetilde {\mathcal {F}}_2$ under taking subgraphs, one can check that $\mathcal {F}_0 \cup \mathcal {F}_1 \cup \mathcal {F}_2$ is a forbidden subgraph characterization of $\mathcal {G}(\lambda )$ .

It suffices to prove that $\mathcal {F}_2$ is finite. Let $T_1, \dots , T_n$ be an enumeration of rooted trees on at most N vertices. We encode $G \in \mathcal {F}_2$ by $t_G \in \mathbb {N}^n$ as follows. Let H be the rooted tree in $\mathcal {T}_N$ such that $G = L(H)$ . After removing the root u from H, we view each connected component as a tree rooted at the vertex that is a child of u in H. Set $t_G := (t_1, \dots , t_n)$ , where $t_i$ is the number of occurrences of $T_i$ as a connected component in the graph obtained by removing u from H. Because no member of $\mathcal {F}_2$ is a subgraph of any other, one can deduce that $\left \{{t_G}\colon {G \in \mathcal {F}_2}\right \}$ is an antichain in $(\mathbb {N}^n, \le )$ , and so $\mathcal {F}_2$ is finite by Lemma 2.9.

Proof of Theorem 1.3 for $\lambda \in [2, \lambda ^*)$ assuming Theorem 2.10.

It is known that $\mathcal {G}(2)$ has a finite forbidden subgraph characterization – in fact, there are $1,812$ minimal forbidden subgraphs for $\mathcal {G}(2)$ ; see [Reference Bussemaker and Neumaier6]. In view of Theorem 2.10, it suffices to show that if G is a minimal forbidden subgraph for $\mathcal {G}(\lambda )$ , then G is a minimal forbidden subgraph for $\mathcal {G}(2)$ , or G, after removing some vertex, is a connected graph in $\mathcal {G}(\lambda ) \setminus \mathcal {G}(2)$ .

Indeed, let G be a minimal forbidden subgraph for $\mathcal {G}(\lambda )$ . Because $\lambda _1(G) < -\lambda \le -2$ , the graph G contains a minimal forbidden subgraph for $\mathcal {G}(2)$ as a subgraph, say F. Clearly both G and F are connected. We may assume that F is a proper subgraph of G because otherwise we are done already. Consider the graph H obtained after removing v from G, where v is a vertex of G that is furthest from F. Using the connectivity of F, one can then show that H is connected. Finally, $H \in \mathcal {G}(\lambda )$ because of the minimality of G, and $H \not \in \mathcal {G}(2)$ because F is a subgraph of H.

Proof of Corollary 1.4 for $\lambda \in (2,\lambda ^*)$ .

It follows immediately from Theorem 2.10 that $-\lambda $ is not a limit point of the set of smallest eigenvalues of graphs for $\lambda \in (2,\lambda ^*)$

Proof. Pick $\ell _0 \in \mathbb {N}$ such that $\lambda _1(F, A, \ell ) < -\lambda $ for every $\ell \ge \ell _0$ . Clearly $\lambda _1(F, A, \ell , K_m) < -\lambda $ for every $\ell \ge \ell _0$ and every $m \in \mathbb {N}^+$ . It suffices to show that for every $\ell \in \left \{{0, \dots , \ell _0 - 1}\right \}$ there exists $m \in \mathbb {N}^+$ such that $\lambda _1(F, A, \ell , K_m) < -\lambda $ .

Figure 5

Minimal forbidden subgraphs for $\mathcal {D}_\infty $ .

Let $v_0 \dots v_{\ell _0}$ be the path added to F to obtain $(F, A, \ell _0)$ , where the vertex $v_0$ is connected to every vertex in A. Set $\lambda _0 := -\lambda _1(F, A, \ell _0)$ , and let $\boldsymbol {x} \colon V(F) \cup \left \{{v_0, \dots , v_{\ell _0}}\right \} \to \mathbb {R}$ be an eigenvector of $(F, A, \ell _0)$ associated with $-\lambda _0$ .

Now fix $\ell \in \mathbb {N}$ with $\ell < \ell _0$ , and let $m \in \mathbb {N}^+$ be determined later. Set $V_\ell := V(F) \cup \left \{{v_0, \dots , v_\ell }\right \}$ , and identify the vertex set of $(F, A, \ell , K_m)$ with $V_\ell \cup V(K_m)$ . We abuse notation and write $x_i$ in place of $x_{v_i}$ for $i \in \left \{{\ell , \dots , \ell _0}\right \}$ . Define $\tilde {\boldsymbol {x}}\colon V_\ell \cup V(K_m) \to \mathbb {R}$ by

$$\begin{align*}\tilde{x}_v = \begin{cases} x_v & \text{if }v \in V_\ell; \\ x_{\ell+1}/m & \text{if }v \in V(K_m). \end{cases} \end{align*}$$

We claim that $\sum _{v \in V_\ell }x_v^2> 0$ . Indeed, assume for the sake of contradiction that $x_v = 0$ for $v \in V_\ell $ . Using $-\lambda _0x_i = \sum _{u\sim v_i}x_u$ for $i \in \left \{{\ell , \dots , \ell _0-1}\right \}$ , where the sum is taken over all vertices u that are adjacent to the vertex $v_i$ in $(F, A, \ell _0)$ , one can prove inductively that $x_i = 0$ for $i \in \left \{{\ell + 1, \dots , \ell _0}\right \}$ , which contradicts the assumption that $\boldsymbol {x}$ is a nonzero vector.

Because $\sum _{v \in V_\ell } x_v^2> 0$ , clearly $\tilde {\boldsymbol {x}}$ is a nonzero vector. We compute

$$\begin{align*}\tilde{\boldsymbol{x}}^\intercal \tilde{\boldsymbol{x}} = \sum_{v \in V_\ell}x_v^2 + m(x_{\ell + 1}/m)^2. \end{align*}$$

Moreover we can compute $\tilde {\boldsymbol {x}}^\intercal A_{(F, A, \ell , K_m)} \tilde {\boldsymbol {x}}$ as follows

$$\begin{align*}\tilde{\boldsymbol{x}}^\intercal A_{(F, A, \ell, K_m)} \tilde{\boldsymbol{x}} = \sum_{u, v \in V_\ell\colon u \sim v} x_ux_v + 2x_\ell x_{\ell+1} + m(m-1)(x_{\ell+1}/m)^2. \end{align*}$$

Since $\boldsymbol {x}$ is an eigenvector of $(F, A, \ell _0)$ associated with $-\lambda _0$ , we obtain that

$$\begin{align*}\sum_{u, v \in V_\ell\colon u \sim v} x_ux_v + x_\ell x_{\ell + 1} = \sum_{v \in V_\ell}x_v\sum_{u \sim v}x_u = -\lambda_0 \sum_{v \in V_\ell}x_v^2. \end{align*}$$

Thus $\tilde {\boldsymbol {x}}^\intercal A_{(F, A, \ell , K_m)} \tilde {\boldsymbol {x}}$ can be simplified to

$$\begin{align*}\tilde{\boldsymbol{x}}^\intercal A_{(F, A, \ell, K_m)} \tilde{\boldsymbol{x}} = -\lambda_0 \sum_{v \in V_\ell}x_v^2 + x_\ell x_{\ell+1} + m(m-1)(x_{\ell+1}/m)^2. \end{align*}$$

The Rayleigh principle says that $\lambda _1(F, A, \ell , K_m)$ is at most

$$\begin{align*}\frac{-\lambda_0 \sum_{v \in V_\ell} x_v^2 + x_\ell x_{\ell+1} + m(m-1)(x_{\ell+1}/m)^2}{\sum_{v \in V_\ell}x_v^2 + m(x_{\ell + 1}/m)^2}, \end{align*}$$

which, as $m \to \infty $ , approaches

$$\begin{align*}-\lambda_0 + \frac{(x_\ell + x_{\ell+1})x_{\ell+1}}{\sum_{v \in V_\ell}x_v^2}. \end{align*}$$

Here we used the above claim that the denominator in the limit is positive.

Recall that $\lambda _0 = -\lambda _1(F, A, \ell _0)> \lambda \ge 2$ . It suffices to show that $(x_\ell + x_{\ell +1})x_{\ell +1} \le 0$ . In fact, we prove inductively that $(x_i + x_{i+1})x_{i+1} \le 0$ for $i \in \left \{{\ell , \dots , \ell _0 - 1}\right \}$ . The base case where $i = \ell _0 - 1$ follows immediately from $-\lambda _0x_{\ell _0} = x_{\ell _0-1}$ and $\lambda _0> 2$ . For the inductive step, using $-\lambda _0x_{i+1} = x_i + x_{i+2}$ and $\lambda _0> 2$ , we obtain

$$ \begin{align*} (x_i + x_{i+1})x_{i+1} =\ & (-\lambda_0x_{i+1} - x_{i+2} + x_{i+1})x_{i+1} = -(\lambda_0 - 2)x_{i+1}^2 - (x_{i+1}+x_{i+2})x_{i+1} \\ & \le -(x_{i+1}+x_{i+2})x_{i+1} = -(x_{i+1}+x_{i+2})^2 + (x_{i+1}+x_{i+2})x_{i+2} \le (x_{i+1}+x_{i+2})x_{i+2}, \end{align*} $$

which is nonpositive by the inductive hypothesis.

Proof of Theorem 2.10.

Suppose that $\lambda \in [2, \lambda ^*)$ . Let $\mathcal {F}$ denote the set of minimal forbidden subgraphs for the family $\mathcal {D}_\infty $ . Combining Lemmas 2.14 and 2.15, we choose $\ell , m \in \mathbb {N}^+$ such that for every $F \in \mathcal {F}$ , the extension family $\mathcal {X}(F, \ell , m)$ is disjoint from $\mathcal {G}(\lambda )$ . From Lemma 2.3(s), we know that $S_5 \not \in \mathcal {G}(\lambda )$ . In particular, no graph in $\mathcal {G}(\lambda )$ contains any member in the following family

(2) $$ \begin{align} \left\{{S_5}\right\} \cup \bigcup_{F \in \mathcal{F}} \mathcal{X}(F, \ell, m) \end{align} $$

as a subgraph. Using Lemma 2.6, we obtain $N \in \mathbb {N}$ such that for every connected graph G with more than N vertices, if no member in (2) is a subgraph of G, then neither is any $F \in \mathcal {F}$ , and so G is a generalized line graph and is in $\mathcal {G}(2)$ by Theorems 2.12 and 2.13. This implies that every connected graph in $\mathcal {G}(\lambda ) \setminus \mathcal {G}(2)$ has at most N vertices.

Proof of Theorem 1.3 for $\lambda \ge \lambda ^*$ .

Theorem 2.19 implies that $\lim _+\Lambda _1 \supseteq [\lambda ^*, \infty )$ , which implies through Proposition 2.18 that $\mathcal {G}(\lambda )$ has no finite forbidden subgraph characterization for any $\lambda \ge \lambda ^*$ .

Proof of Corollary 1.4 for $\lambda \ge \lambda ^*$ .

It follows immediately from Theorem 2.19 that $-\lambda $ is a limit point of the set of smallest eigenvalues of graphs for $\lambda \ge \lambda ^*$ .

Proof of (a).

Set $\boldsymbol {a} = (a_1, \dots , a_n)$ and $\lambda = \lambda _1(R(\boldsymbol {a}))$ . Let $v_{-2} \dots v_n$ denote the central path of $R(\boldsymbol {a})$ , and let $v_c$ denote the coxswain attached to $v_0$ . Let $\boldsymbol {x}\colon V(R(\boldsymbol {a})) \to \mathbb {R}$ be a unit eigenvector of $R(\boldsymbol {a})$ associated with $\lambda $ . Clearly we have $\lambda x_c = x_0$ , which implies that

(3) $$ \begin{align} x_c^2 = \frac{x_c^2 + x_0^2}{1+\lambda^2} \le \frac{1}{1+\lambda^2}. \end{align} $$

Let L be the rowing graph $R(\boldsymbol {a})$ with $v_c$ removed, and let $\boldsymbol {x}_L$ be $\boldsymbol {x}$ restricted to $V(L)$ . Notice that L is a line graph of a caterpillar tree, and so $\lambda _1(L) \ge -2$ (see Footnote 1 on Page 2). Finally, we bound the smallest eigenvalue of $R(\boldsymbol {a})$ as follows:

$$ \begin{align*} \lambda = \boldsymbol{x}^\intercal A_{R(\boldsymbol{a})} \boldsymbol{x} = \boldsymbol{x}_L^\intercal A_L \boldsymbol{x}_L + 2x_cx_0 \ge & -2\boldsymbol{x}_L^\intercal \boldsymbol{x}_L + 2x_cx_0 \\ & = -2(1-x_c^2) + 2\lambda x_c^2 = -2+(2+2\lambda)x_c^2 \stackrel{(3)}{\ge} -2 + \frac{2+2\lambda}{1+\lambda^2}, \end{align*} $$

which implies that $\lambda \ge -1-\sqrt {2}$ . In the above inequality, we assumed that $\lambda \le -1$ which follows from the fact that $R(\boldsymbol {a})$ has an edge.

Proof of (b).

Let $\boldsymbol {x}$ be the vector that assigns $1$ to $v_{-2}$ , $-2$ to $v_{-1}$ , $4$ to $v_0$ , $0$ to $v_1$ , $-2$ to the coxswain, and $-4/a_1$ to every vertex in the clique of size $a_1$ . The Rayleigh principle says that $\lambda _1(R(a_1))$ is at most

$$\begin{align*}\frac{\boldsymbol{x}^\intercal A_{R(a_1)} \boldsymbol{x}}{\boldsymbol{x}^\intercal \boldsymbol{x}} = \frac{-2(1 \cdot 2 + 2 \cdot 4 + 4 \cdot 2 + a_1 \cdot 4 \cdot (4/a_1)) + a_1(a_1-1) \cdot (-4/a_1)^2}{1^2 + (-2)^2 + 4^2 + (-2)^2 + a_1 \cdot (-4/a_1)^2}, \end{align*}$$

which approaches $-52/25 = -2.08$ as $a_1 \to \infty $ .

Proof of (c).

Take $\ell = \lceil 2/\varepsilon \rceil + 5$ . Let $v_{-2} \dots v_{n_1 + \ell + n_2}$ denote the central path of the rowing graph $R(\boldsymbol {a}, 0^{(\ell )}, \boldsymbol {b})$ , where $\boldsymbol {a} = (a_1, \dots , a_{n_1})$ and $\boldsymbol {b} = (b_1, \dots , b_{n_2})$ , and let $\boldsymbol {x}\colon V(R(\boldsymbol {a},0^{(\ell )},\boldsymbol {b})) \to \mathbb {R}$ be a unit eigenvector of $R(\boldsymbol {a},0^{(\ell )},\boldsymbol {b})$ associated with the smallest eigenvalue. Choose $k \in \left \{{0, \dots , \ell - 1}\right \}$ such that $x_{n_1+k}x_{n_1+k+1}$ reaches the minimum in absolute value. In particular, using the inequality $ \left \lvert {x_{n_1+i}x_{n_1+i+1}} \right \rvert \le (x_{n_1+i}^2 + x_{n_1+i+1}^2)/2$ , we obtain

(4) $$ \begin{align} \left\lvert {x_{n_1+k}x_{n_1+k+1}} \right\rvert \le \frac{1}{\ell}\sum_{i = 0}^{\ell - 1} \left\lvert {x_{n_1+i}x_{n_1+i+1}} \right\rvert \le \frac{1}{\ell} \sum_{i=0}^{\ell}x_{n_1+i}^2 \le \frac{1}{\ell} < \frac{\varepsilon}{2}. \end{align} $$

Notice that removing the edge $v_{n_1+k} v_{n_1+k+1}$ disconnects $R(\boldsymbol {a},0^{(\ell )},\boldsymbol {b})$ into two subgraphs, one of which is $R(\boldsymbol {a}, 0^{(k)})$ , while the other is a line graph, denoted L, of a caterpillar tree. Clearly

(5) $$ \begin{align} \lambda_1(R(\boldsymbol{a}, 0^{(k)})) \ge \lambda_1(R(\boldsymbol{a}, 0^{(\ell)})). \end{align} $$

As $\ell \ge 6$ , the graph $\widetilde {E}_8$ in Figure 1 is a proper subgraph of $R(\boldsymbol {a}, 0^{(\ell )})$ , and so $\lambda _1(R(\boldsymbol {a}, 0^{(\ell )})) < -2$ . Together with $\lambda _1(L) \ge -2$ (see Footnote 1 on Page 2), we obtain

(6) $$ \begin{align} \lambda_1(L)> \lambda_1(R(\boldsymbol{a}, 0^{(\ell)})). \end{align} $$

Let $\boldsymbol {x}_R$ and $\boldsymbol {x}_L$ be the unit eigenvector $\boldsymbol {x}$ restricted to $V(R(\boldsymbol {a},0^{(k)}))$ and $V(L)$ . Finally, we bound the smallest eigenvalue of $R(\boldsymbol {a},0^{(\ell )},\boldsymbol {b})$ as follows:

$$ \begin{align*} \lambda_1(R(\boldsymbol{a},0^{(\ell)},\boldsymbol{b})) & \stackrel{\phantom{(4,5,6)}}{=} \boldsymbol{x} A_{R(\boldsymbol{a},0^{(\ell)},\boldsymbol{b})}\boldsymbol{x} \\ & \stackrel{\phantom{(4,5,6)}}{=} \boldsymbol{x}_R^\intercal A_{R(\boldsymbol{a},0^{(k)})} \boldsymbol{x}_R + 2x_{n_1+k}x_{n_1+k+1} + \boldsymbol{x}_L^\intercal A_L\boldsymbol{x}_L \\ & \stackrel{\phantom{(4,5,6)}}{\ge} \lambda_1(R(\boldsymbol{a},0^{(k)})) \boldsymbol{x}_R^\intercal \boldsymbol{x}_R + 2x_{n_1+k}x_{n_1+k+1} + \lambda_1(L)\boldsymbol{x}_L^\intercal \boldsymbol{x}_L \\ & \stackrel{(4,5,6)}{>} \lambda_1(R(\boldsymbol{a},0^{(\ell)}))(\boldsymbol{x}_R^\intercal \boldsymbol{x}_R + \boldsymbol{x}_L^\intercal \boldsymbol{x}_L) - \varepsilon \\ & \stackrel{\phantom{(4,5,6)}}{=} \lambda_1(R(\boldsymbol{a}, 0^{(\ell)})) - \varepsilon.\\[-37pt] \end{align*} $$

Proof of (d).

Set $\boldsymbol {a} = (a_1, \dots , a_n, 0^{(\ell )})$ and $\lambda = \lambda _1(R(\boldsymbol {a}))$ . Let $\boldsymbol {x}\colon V(R(\boldsymbol {a})) \to \mathbb {R}$ be an eigenvector of $R(\boldsymbol {a})$ associated with $\lambda $ . Let $v_{-2}\dots v_{n + \ell }$ denote the central path of the rowing graph $R(\boldsymbol {a})$ . In $R(\boldsymbol {a})$ , let K denote the clique of size $a_n$ attached to both $v_{n-1}$ and $v_n$ , and pick an arbitrary vertex u from K. We clearly have

(7) $$ \begin{align}\begin{aligned} \lambda x_u & = x_{n-1} + \sigma + x_n, \\\lambda x_n & = x_{n-1} + \sigma + x_u + x_{n+1}, \end{aligned}\end{align} $$

where $\sigma = \sum _{u' \in V(K) \setminus \left \{{u}\right \}} x_{u'}$ .The above identities imply that $\lambda (x_u - x_n) = x_n - x_u - x_{n+1}$ , and so

(8) $$ \begin{align} x_u = x_n - \frac{x_{n+1}}{1 + \lambda}. \end{align} $$

Let $R(\tilde {\boldsymbol {a}})$ be the rowing graph obtained from $R(\boldsymbol {a})$ by removing u and attaching a clique, denoted $\tilde {K}$ , to both $v_n$ and $v_{n+1}$ , where $\tilde {\boldsymbol {a}} = (a_1, \dots , a_n-1,a_{n+1},0^{(\ell -1)})$ , and the order $a_{n+1}$ of $\tilde {K}$ is chosen later. In particular, $V(R(\tilde {\boldsymbol {a}})) = V(R(\boldsymbol {a})) \setminus \left \{{u}\right \} \cup V(\tilde {K})$ .

We define a vector $\tilde {\boldsymbol {x}}\colon V(R(\tilde {\boldsymbol {a}}))\to \mathbb {R}$ as follows:

$$\begin{align*}\tilde{x}_v = \begin{cases} x_n + x_u & \text{if } v = v_n; \\ -(x_n + x_u + x_{n+1})/a_{n+1} & \text{if } v \in V(\tilde{K}); \\ x_v & \text{otherwise}. \end{cases} \end{align*}$$

Because $\boldsymbol {x}$ is a nonzero vector, one can check that so is $\tilde {\boldsymbol {x}}$ . We compute

$$\begin{align*}\tilde{\boldsymbol{x}}^\intercal \tilde{\boldsymbol{x}} = \boldsymbol{x}^\intercal \boldsymbol{x} - x_n^2 - x_u^2 + (x_n + x_u)^2 + \frac{(x_n + x_u + x_{n+1})^2}{a_{n+1}}. \end{align*}$$

Moreover we can compare $\tilde {\boldsymbol {x}}^\intercal A_{R(\tilde {\boldsymbol {a}})} \tilde {\boldsymbol {x}}$ and $\boldsymbol {x}^\intercal A_{R(\boldsymbol {a})} \boldsymbol {x}$ as follows

$$ \begin{align*} \tilde{\boldsymbol{x}}^\intercal A_{R(\tilde{\boldsymbol{a}})} \tilde{\boldsymbol{x}} = \boldsymbol{x}^\intercal A_{R(\boldsymbol{a})} \boldsymbol{x} - & 2x_u\left(x_{n-1} + \sigma + x_n\right) + 2x_u\left(x_{n-1} + \sigma + x_{n+1} \right) \\ & \qquad\qquad - 2(x_n + x_u + x_{n+1})^2 + a_{n+1}(a_{n+1}-1)\left(\frac{x_n + x_u + x_{n+1}}{a_{n+1}}\right)^2, \end{align*} $$

which simplifies via (7) to

$$\begin{align*}\tilde{\boldsymbol{x}}^\intercal A_{R(\tilde{\boldsymbol{a}})} \tilde{\boldsymbol{x}} = \lambda \boldsymbol{x}^\intercal \boldsymbol{x} - 2\lambda x_u^2 + 2x_u (\lambda x_n - x_u) - \left(1 + \frac{1}{a_{n+1}}\right)(x_n + x_u + x_{n+1})^2. \end{align*}$$

The Rayleigh principle says that $\lambda _1(R(\tilde {\boldsymbol {a}}))$ is at most

$$\begin{align*}\frac{\tilde{\boldsymbol{x}}^\intercal A_{R(\tilde{\boldsymbol{a}})} \tilde{\boldsymbol{x}}}{\tilde{\boldsymbol{x}}^\intercal \tilde{\boldsymbol{x}}} = \frac{\lambda \boldsymbol{x}^\intercal \boldsymbol{x} - 2\lambda x_u^2 + 2x_u (\lambda x_n - x_u) - \left(1 + 1/a_{n+1}\right)(x_n + x_u + x_{n+1})^2}{\boldsymbol{x}^\intercal \boldsymbol{x} - x_n^2 - x_u^2 + (x_n + x_u)^2 + (x_n + x_u + x_{n+1})^2/a_{n+1}}, \end{align*}$$

which, as $a_{n+1} \to \infty $ , approaches

$$\begin{align*}\frac{\lambda \boldsymbol{x}^\intercal \boldsymbol{x} - 2\lambda x_u^2 + 2x_u(\lambda x_n - x_u) - (x_n + x_u + x_{n+1})^2}{\boldsymbol{x}^\intercal \boldsymbol{x} - x_n^2 - x_u^2 + (x_n + x_u)^2}. \end{align*}$$

Here we assumed that the denominator in the limit is nonzero. Indeed, suppose on the contrary that $\boldsymbol {x}^\intercal \boldsymbol {x} = x_n^2 + x_u^2 - (x_n + x_u)^2 = -2x_nx_u$ . Because $-2x_nx_u \le x_n^2 + x_u^2 \le x_n^2 + x_u^2 + x_{n+1}^2 \le \boldsymbol {x}^\intercal \boldsymbol {x}$ , it must be the case that $x_n + x_u = 0$ and $x_{n+1} = 0$ . In view of (8), we have $x_n = x_u = 0$ and hence $\boldsymbol {x} = \boldsymbol {0}$ , which is a contradiction.

It suffices to prove that

$$\begin{align*}\lambda \boldsymbol{x}^\intercal \boldsymbol{x} - 2\lambda x_u^2 + 2x_u(\lambda x_n - x_u) - (x_n + x_u + x_{n+1})^2 \le \lambda\left( \boldsymbol{x}^\intercal \boldsymbol{x} - x_n^2 - x_u^2 + (x_n + x_u)^2 \right), \end{align*}$$

which is equivalent to

$$\begin{align*}(x_n+x_u+x_{n+1})^2 + 2(1+\lambda)x_u^2 \ge 0. \end{align*}$$

Using (8), we know that the left hand side of the last inequality is equal to

$$\begin{align*}\left(2x_n + \frac{\lambda}{1+\lambda}x_{n+1}\right)^2 + 2(\lambda+1)\left(x_n-\frac{x_{n+1}}{1+\lambda}\right)^2 = (2\lambda + 6)x_n^2 - \frac{4}{1+\lambda}x_nx_{n+1}+\left(1+\frac{1}{(1+\lambda)^2}\right)x_{n+1}^2. \end{align*}$$

From Lemma 2.22(a), we know that $\lambda \ge -1-\sqrt {2}$ in addition to the assumption that $\lambda \le -2$ . One can check

$$\begin{align*}1 + \frac{1}{(1+\lambda)^2} \ge 2\lambda + 6 \ge \frac{4}{1+\lambda}\cdot \frac{1}{\lambda}> 0, \quad \text{for }\lambda \in [-1-\sqrt2,-2]. \end{align*}$$

We are left to prove $x_n^2 - \lambda x_nx_{n+1} + x_{n+1}^2 \ge 0$ . In fact, we prove inductively that $x_i^2 - \lambda x_ix_{i+1} + x_{i+1}^2 \ge 0$ for $i \in \left \{{n, \dots , n + \ell - 1}\right \}$ . The base case where $i = n + \ell - 1$ follows immediately from $\lambda x_{n+\ell } = x_{n+\ell -1}$ . For the inductive step, using $\lambda x_{i+1} = x_i + x_{i+2}$ , we obtain

$$\begin{align*}x_i^2 - \lambda x_i x_{i+1} + x_{i+1}^2 = (\lambda x_{i+1}-x_{i+2})^2 - \lambda (\lambda x_{i+1}-x_{i+2})x_{i+1} + x_{i+1}^2 = x_{i+1}^2 - \lambda x_{i+1}x_{i+2} + x_{i+2}^2, \end{align*}$$

which is nonnegative by the inductive hypothesis.

Proof of (e).

Take $m = \lceil 5/\varepsilon \rceil + 1$ . Let $v_{-2} \dots v_n$ denote the central path of the rowing graph $R(\boldsymbol {a})$ , where $\boldsymbol {a} = (a_1, \dots , a_n)$ , and let $\boldsymbol {x}\colon V(R(\boldsymbol {a})) \to \mathbb {R}$ be a unit eigenvector associated with the smallest eigenvalue of $R(\boldsymbol {a})$ . Choose $k \in \left \{{1, \dots , m}\right \}$ such that $x_{k-1}$ reaches the minimum in absolute value. In particular,

(9) $$ \begin{align} x_{k-1}^2 \le \frac{1}{m} \sum_{i=0}^{m-1}x_i^2 \le \frac{1}{m} < \frac{\varepsilon}{5}. \end{align} $$

Let $v_{-2} \dots v_{k-1}v_* v_{k} \dots v_n$ denote the central path of $R(\tilde {\boldsymbol {a}})$ , where $\tilde {\boldsymbol {a}} = (a_1, \dots , a_{k-1},0,a_{k}, \dots , a_n)$ . We naturally view the vertex set of $R(\tilde {\boldsymbol {a}})$ as $V(R(\boldsymbol {a})) \cup \{ v_*\}$ , and we extend the unit eigenvector $\boldsymbol {x}\colon V(R(\boldsymbol {a})) \to \mathbb {R}$ to $\tilde {\boldsymbol {x}}\colon V(R(\tilde {\boldsymbol {a}})) \to \mathbb {R}$ by setting $\tilde {x}_* = x_{k-1}$ . The Rayleigh principle says that $\lambda _1(R(\tilde {\boldsymbol {a}}))$ is at most

$$ \begin{align*} \frac{\tilde{\boldsymbol{x}}^\intercal A_{R(\tilde{\boldsymbol{a}})} \tilde{\boldsymbol{x}}}{\tilde{\boldsymbol{x}}^\intercal \tilde{\boldsymbol{x}}} =& \frac{\boldsymbol{x}^\intercal A_{R(\boldsymbol{a})} \boldsymbol{x} + 2x_{k-1}^2}{\boldsymbol{x}^\intercal \boldsymbol{x} + x_{k-1}^2} = \frac{\lambda_1(R(\boldsymbol{a})) + 2x_{k-1}^2}{1 + x_{k-1}^2} \\ &\qquad\qquad = \lambda_1(R(\boldsymbol{a})) + \frac{(2-\lambda_1(R(\boldsymbol{a})))x_{k-1}^2}{1+x_{k-1}^2} \le \lambda_1(R(\boldsymbol{a})) + (2-\lambda_1(R(\boldsymbol{a})))x_{k-1}^2. \end{align*} $$

From Lemma 2.22(a), we obtain

$$ \begin{align*} (2-\lambda_1(R(\boldsymbol{a})))x_{k-1}^2 \le (2 + (1 + \sqrt2))x_{k-1}^2 \stackrel{(9)}{<} \varepsilon.\\[-35pt] \end{align*} $$

Proof of Theorem 2.19.

In view of Theorem 2.20, we may assume that $\lambda \in (\lambda ^*, \lambda ')$ . Fix $\varepsilon> 0$ . We assume for the sake of contradiction that no rowing graph has its smallest eigenvalue in $(-\lambda - \varepsilon , -\lambda )$ because otherwise we are done. Define

$$ \begin{align*} & \qquad S := \left\{{(a_1, \dots, a_n) \in \mathbb{N}^n}\colon{n \in \mathbb{N}^+ \text{ and } a_n> 0}\right\}, \\ A := & \left\{{(a_1, \dots, a_n) \in S}\colon{\lambda_1(R(a_1, \dots, a_n, 0^{(\ell)})) < -\lambda \text{ for some }\ell \in \mathbb{N}}\right\}. \end{align*} $$

Claim. For every $m \in \mathbb {N}^+$ there exists $(a_1, \dots , a_n) \in A$ with exactly m nonzero entries such that

(10) $$ \begin{align} (a_1, \dots, a_{k-1}, 0, a_k, \dots, a_n) \not\in A\text{ for every }k \in \left\{{1, \dots, n}\right\}. \end{align} $$

Proof of Claim.

We order the elements in S as follows:

$$ \begin{align*} (a_1, \dots, a_{n_1}) \prec & (b_1, \dots, b_{n_2}) \text{ if and only if}\\ & \qquad(a_1, \dots, a_{n_1}, 0^{(n_2)}) \text{ strictly precedes } (b_1, \dots, b_{n_2}, 0^{(n_1)}) \text{ lexicographically}. \end{align*} $$

Alternatively, $(a_1, \dots , a_{n_1}) \prec (b_1, \dots , b_{n_2})$ if and only if there exists $\delta> 0$ such that $\sum _{i=1}^{n_1} a_ix^i < \sum _{i=1}^{n_2} b_ix^i$ for every $x \in (0, \delta )$ . We shall repeatedly use the fact that for every $\ell \in \mathbb {N}^+$ the set

$$\begin{align*}S_\ell := \left\{{\boldsymbol{a} \in S}\colon{\text{the length of }\boldsymbol{a} \text{ is at most }\ell}\right\} \end{align*}$$

does not contain an infinite descending chain, hence $(S_\ell , \preceq )$ is well-founded, that is, every nonempty subset of $S_\ell $ has a minimal element.Footnote ⁸ This fact can be established by a simple induction on $\ell $ .

In particular, the ordering on S implies that $(a_1, \dots a_{k-1}, 0, a_k, \dots , a_n) \prec (a_1, \dots , a_n)$ when both sequences are in S and $k \in \left \{{1, \dots , n}\right \}$ . Thus it suffices to construct, for every $m \in \mathbb {N}^+$ , a minimal element $\boldsymbol {a}^{(m)}$ of $A_m$ , where

$$\begin{align*}A_m := \left\{{(a_1, \dots, a_n) \in A}\colon{(a_1, \dots, a_n) \text{ has }m\text{ nonzero entries}}\right\}. \end{align*}$$

Our inductive construction additionally requires that $\boldsymbol {a}^{(1)}, \boldsymbol {a}^{(2)}, \boldsymbol {a}^{(3)},\dots $ form a descending chain.

Apply Lemma 2.22(c) to obtain $\ell \in \mathbb {N}^+$ such that

(11) $$ \begin{align} & \lambda_1(R(a_1, \dots, a_{n_1}, 0^{(\ell)}, b_1, \dots, b_{n_2}))> \lambda_1(R(a_1, \dots, a_{n_1}, 0^{(\ell)})) - \varepsilon \nonumber\\ & \qquad\qquad\qquad \text{for every }n_1, n_2 \in \mathbb{N}, (a_1, \dots, a_{n_1}) \in \mathbb{N}^{n_1} \text{ and } (b_1, \dots, b_{n_2}) \in \mathbb{N}^{n_2}. \end{align} $$

In particular, when $n_1 = 0$ , we have

$$\begin{align*}\lambda_1(R(0^{(\ell)}, b_1, \dots, b_{n_2}))> \lambda_1(R(0^{(\ell)})) - \varepsilon \text{ for every }(b_1, \dots, b_{n_2}) \in \mathbb{N}^{n_2}. \end{align*}$$

For the base case where $m = 1$ , note that $R(0^{(\ell )})$ is just $E_{2,\ell }$ defined in Proposition 2.1, which satisfies that

$$\begin{align*}\lambda_1(R(0^{(\ell)})) = \lambda_1(E_{2,\ell})> -\lambda^* > -\lambda. \end{align*}$$

Since no rowing graph has its smallest eigenvalue in $(-\lambda - \varepsilon , -\lambda )$ , the above two inequalities imply

$$\begin{align*}\lambda_1(R(0^{(\ell)}, b_1, \dots, b_{n_2})) \ge -\lambda \text{ for every }(b_1, \dots, b_{n_2}) \in \mathbb{N}^{n_2}, \end{align*}$$

which further implies that the length of every sequence in $A_1$ is at most $\ell $ , that is, $A_1 \subseteq S_{\ell }$ . Moreover, Lemma 2.22(b) implies that $A_1$ is nonempty. Therefore there exists a minimal element of $A_1$ .

For the inductive step, suppose that $m \in \mathbb {N}^+$ , and that $\boldsymbol {a}^{(1)} \succ \dots \succ \boldsymbol {a}^{(m)}$ are minimal elements of $A_1, \dots , A_m$ respectively. Say $\boldsymbol {a}^{(m)} = (b_1, \dots , b_n)$ . Lemma 2.22(d) shows that there exists $b_{n+1} \in \mathbb {N}^+$ such that $\boldsymbol {b}^{(m+1)} := (b_1, \dots , b_{n-1}, b_n-1, b_{n+1}) \in A$ . Since $\boldsymbol {b}^{(m+1)} \prec \boldsymbol {a}^{(m)}$ and $\boldsymbol {a}^{(m)}$ is minimal in $A_m$ , it must be the case that $b_n> 1$ or equivalently $\boldsymbol {b}^{(m+1)} \in A_{m+1}$ . It suffices to show that

$$\begin{align*}\left\{{\boldsymbol{b} \in A_{m+1}}\colon{\boldsymbol{b} \preceq \boldsymbol{b}^{(m+1)}}\right\} \subseteq S_{(m+1)\ell}. \end{align*}$$

Assume for the sake of contradiction that there exists $\boldsymbol {b} \in A_{m+1}$ such that $\boldsymbol {b} \preceq \boldsymbol {b}^{(m+1)}$ and the length of $\boldsymbol {b}$ is more than $(m+1)\ell $ . By the pigeonhole principle, there exists an initial segment, denoted $\boldsymbol {b}'$ , of $\boldsymbol {b}$ such that $(\boldsymbol {b}', 0^{(\ell )})$ is an initial segment of $\boldsymbol {b}$ . We may require in addition that either $\boldsymbol {b}'$ is the empty sequence or the last entry of $\boldsymbol {b}'$ is positive. If $\boldsymbol {b}'$ is the empty sequence, then we can argue similarly to the base case that

$$\begin{align*}\lambda_1(R(\boldsymbol{b}, c_1, \dots, c_{n'})) \ge -\lambda \text{ for every }(c_1, \dots, c_{n'}) \in \mathbb{N}^{n'}, \end{align*}$$

which contradicts $\boldsymbol {b} \in A_{m+1}$ .

We may assume that the last entry of $\boldsymbol {b}'$ is positive. Because $(\boldsymbol {b}', 0^{(\ell )})$ is an initial segment of $\boldsymbol {b}$ , and the last entry of $\boldsymbol {b}$ is positive, the number of nonzero entries in $\boldsymbol {b}'$ , denoted $i \in \mathbb {N}^+$ , is at most m. Since $\boldsymbol {b}' \prec \boldsymbol {b} \preceq \boldsymbol {b}^{(m+1)} \prec \boldsymbol {a}^{(m)} \preceq \boldsymbol {a}^{(i)}$ , and $\boldsymbol {a}^{(i)}$ is minimal in $A_i$ , we know that $\boldsymbol {b}' \not \in A$ , and in particular

$$\begin{align*}\lambda_1(R(\boldsymbol{b}', 0^{(\ell)})) \ge -\lambda. \end{align*}$$

From (11), we know that

$$\begin{align*}\lambda_1(R(\boldsymbol{b}, c_1, \dots, c_{n'}))> \lambda_1(R(\boldsymbol{b}', 0^{(\ell)})) - \varepsilon \text{ for every }(c_1, \dots, c_{n'}) \in \mathbb{N}^{n'}. \end{align*}$$

Since no rowing graph has its smallest eigenvalue in $(-\lambda -\varepsilon , -\lambda )$ , the above two inequalities imply

$$\begin{align*}\lambda_1(R(\boldsymbol{b}, c_1, \dots, c_{n'})) \ge -\lambda \text{ for every }(c_1, \dots, c_{n'}) \in \mathbb{N}^{n'}, \end{align*}$$

which contradicts $\boldsymbol {b} \in A_{m+1}$ .

Finally, let m be given by Lemma 2.22(e). The claim provides $(a_1, \dots , a_n) \in A$ with m nonzero entries such that (10) holds. Let $\ell \in \mathbb {N}$ be such that

$$\begin{align*}\lambda_1(R(a_1, \dots, a_n, 0^{(\ell)})) < -\lambda. \end{align*}$$

Since the length of $(a_1, \dots , a_n, 0^{(\ell )})$ is at least m, Lemma 2.22(e) says that there exists $k \in \left \{{1, \dots , m}\right \}$ such that

$$\begin{align*}\lambda_1(R(a_1, \dots, a_{k-1}, 0, a_k, \dots, a_n, 0^{(\ell)})) < \lambda_1(R(a_1, \dots, a_n, 0^{(\ell)})) + \varepsilon, \end{align*}$$

However (10) asserts that $(a_1, \dots , a_{k-1}, 0, a_k, \dots , a_n) \not \in A$ , which implies that

$$\begin{align*}\lambda_1(R(a_1, \dots, a_{k-1}, 0, a_k, \dots, a_n, 0^{(\ell)})) \ge -\lambda. \end{align*}$$

Combining the last three inequalities, we obtain that

$$ \begin{align*} -\lambda - \varepsilon < \lambda_1(R(a_1,\dots, a_n, 0^{(\ell)})) < -\lambda.\\[-35pt] \end{align*} $$

Proof. Let $v_0,v_1,\dots , v_{n-1}$ be the vertices of $C_n^\pm $ , and let $\sigma \colon E(C^\pm _n) \to \left \{{\pm 1}\right \}$ be the signing of $C^\pm _n$ . Suppose that $A^\pm $ is a signed set of $\left \{{v_0,v_{n-1}}\right \}$ . By switching a suitable subset of $\left \{{v_0, v_{n-1}}\right \}$ , we may assume that $A^\pm = \{v_0^+, v_{n-1}^+\}$ . If $C_n^\pm $ has an even number of positive edges, then $C^\pm _n$ is already switching equivalent to the all-negative cycle of length n, whose smallest eigenvalue is $-2$ . If the edge $v_0v_{n-1}$ is negative, then both $(C_n^\pm , A^\pm , \ell , K_m)$ and $(C_n^\pm , A^\pm , K_m)$ contain a subgraph that is switching equivalent to the all-negative triangle $-K_3$ , and so their smallest eigenvalues are at most $\lambda _1(-K_3) = -2$ .

Hereafter we assume that $C_n^\pm $ has an odd number of positive edges and the edge $v_0v_{n-1}$ is positive. Furthermore, by switching a suitable subset of $\left \{{v_1, \dots , v_{n-2}}\right \}$ , we may assume that $v_0v_{n-1}$ is the only positive edge of the signed cycle $C_n^\pm $ .

For (c1), we define a vector $\boldsymbol {x}\colon V(C_n^\pm , A^\pm , \ell , K_m) \to \mathbb {R}$ as follows. Let $v_{n}v_{n+1}\dots v_{n+\ell }$ and $K_m$ be the path and the clique added to $C_n^\pm $ to obtain the path-clique extension $(C_n^\pm , A^\pm , \ell , K_m)$ , where $v_{n}$ is adjacent to $v_0$ and $v_{n-1}$ , and $v_{n+\ell }$ is adjacent to every vertex in $K_m$ . Assign $x_{v_i} = -1$ for $i \in \left \{{0, \dots , n-1}\right \}$ , $x_{v_{n+i}} = 2(-1)^i$ for $i \in \left \{{0, \dots , \ell }\right \}$ , and $x_u = -x_{n+\ell }/m$ for every $u \in V(K_m)$ . We abuse notation and write $x_i$ in place of $x_{v_i}$ for $i \in \left \{{0, \dots , n+\ell }\right \}$ . The Rayleigh principle says that the smallest eigenvalue of $(C_n^\pm , A^\pm , \ell , K_m)$ is at most

$$ \begin{align*} \bigg(-2\sum_{i=0}^{n-2}x_ix_{i+1} + 2x_0x_{n-1} + & 2(x_0 + x_{n-1})x_n + 2\sum_{i=0}^{\ell-1}x_{n+i}x_{n+i+1} - 2x_{n+\ell}^2 + \\ & \qquad\qquad + m(m-1)(x_{n+\ell}/m)^2\bigg) \bigg/ \bigg(\sum_{i=0}^{n-1} x_i^2 + \sum_{i=0}^\ell x_{n+i}^2 + m(x_{n+\ell}/m)^2\bigg), \end{align*} $$

which is equal to $(-2n - 8(\ell +1) - 4/m)/(n + 4(\ell +1) + 4/m)$ , and approaches $-2$ as $m \to \infty $ .

For (c2), we define a vector $\boldsymbol {x} \colon V(C_n^\pm , A^\pm , K_m) \to \mathbb {R}$ as follows. Let $K_m$ be the clique added to $C_n^\pm $ to obtain the clique extension $(C_n^\pm , A^\pm , K_m)$ . Assign $x_{v_i} = -1$ for $i \in \left \{{0, \dots , n-1}\right \}$ and $x_u = -(x_0 + x_{n-1})/m$ for every $u \in V(K_m)$ . We abuse notation and write $x_i$ in place of $x_{v_i}$ for $i \in \left \{{0,\dots , n-1}\right \}$ . The Rayleigh principle says that the smallest eigenvalue of $(C_n^\pm , A^\pm , K_m)$ is at most

$$\begin{align*}\frac{-2\sum_{i=0}^{n-2}x_ix_{i+1} + 2x_0x_{n-1} - 2(x_0 + x_{n-1})^2 + m(m-1)((x_0 + x_{n-1})/m)^2}{\sum_{i=0}^{n-1} x_i^2 + m((x_0 + x_{n-1})/m)^2}, \end{align*}$$

which is equal to $(-2n - 4/m)/(n + 4/m)$ , and approaches $-2$ as $m \to \infty $ .

Proof. From Lemma 2.3(n1, n2, p) and Lemma 2.5, we obtain $\ell , m \in \mathbb {N}^+$ such that $P_\ell \not \in \mathcal {G}^\pm (\lambda )$ , none of the following graphs

(12) $$ \begin{align} (\overline{K}_2, V(\overline{K}_2), 0, K_m), \dots, (\overline{K}_2, V(\overline{K}_2), \ell-1, K_m), \text{ and }(\overline{K}_2, V(\overline{K}_2), K_m) \end{align} $$

is in $\mathcal {G}^\pm (\lambda )$ , and $\mathcal {X}(C, \ell , m) \cup \mathcal {X}(D, \ell , m)$ is disjoint from $\mathcal {G}^\pm (\lambda )$ . Since $P_\ell \not \in \mathcal {G}^\pm (\lambda )$ , no path extension in $\mathcal {X}^\pm (C, \ell , m) \cup \mathcal {X}^\pm (D, \ell , m)$ is in $\mathcal {G}^\pm (\lambda )$ .

We are left to deal with the path-clique extensions and the clique extension

(13) $$ \begin{align} (F, A^\pm, 0, K_m), \dots, (F, A^\pm, \ell-1, K_m), \text{ and }(F, A^\pm, K_m), \end{align} $$

where F is C or D, and $A^\pm $ is the signed set of a nonempty vertex subset A of F. We break the rest of the proof into three cases.

Case 1: $A^\pm $ is all-positive or all-negative. This case follows since the signed graphs in (13), after switching the cut-set between $V(F)$ and its complement in case $A^\pm $ is all-negative, become respectively $(F, A, 0, K_m), \dots , (F, A, \ell -1, K_m)$ , and $(F, A, K_m)$ from the extension family $\mathcal {X}(F, \ell , m)$ , which is disjoint from $\mathcal {G}^\pm (\lambda )$ .

Case 2: There exists an edge $uv$ of F such that $\left \{{u^-, v^+}\right \} \subseteq A^\pm $ . Since the all-negative triangle $-K_3$ , whose smallest eigenvalue is $-2$ , is switching equivalent to a subgraph of $(F, A^\pm , 0)$ , no signed graph in (13) is in $\mathcal {G}^\pm (\lambda )$ .

Case 3: $A^\pm $ is neither all-positive nor all-negative, and no edge $uv$ of F satisfies that $\left \{{u^-, v^+}\right \} \subseteq A^\pm $ . Label the vertices of C and D as in Figure 3. We may assume without loss of generality that $\left \{{1^-, 3^+}\right \} \subseteq A^\pm $ . One can check that, for $F \in \left \{{C, D}\right \}$ , the signed graphs in (12), after switching at the vertex labeled by $1$ , are respectively subgraphs of those in (13), and hence no signed graph in (13) is in $\mathcal {G}^\pm (\lambda )$ .

Proof. Let G be the underlying graph of $G^\pm $ . One can check that G contains neither the claw graph nor the diamond graph as a subgraph. Thus Theorem 2.7 implies that G is a line graph of another graph, denoted H, without isolated vertices. We may identify the vertex set of $G^\pm $ with the edge set of H.

For every $v \in V(H)$ , let $E_v(H)$ be the set of edges that are incident to v in H. Define a signing $\sigma \colon \left \{{(v, e)}\colon {v \in V(H), e \in E_v(H)}\right \} \to \left \{{\pm 1}\right \}$ such that for every $v \in V(H)$ ,

(14) $$ \begin{align} \sigma(v, e_1)\sigma(v, e_2) = A_{G^\pm}(e_1,e_2) \text{ for distinct } e_1, e_2 \in E_v(H). \end{align} $$

Such a signing $\sigma $ can be chosen as follows: for every $v\in V(H)$ , define $\sigma (v, e_0) = 1$ for some arbitrary $e_0 \in E_v(H)$ , and define $\sigma (v, e) = A_{G^\pm }(e_0, e)$ for any other $e \in E_v(H)$ . Since $G^\pm $ does not contain any subgraph that is switching equivalent to the all-negative triangle, it can be seen that (14) holds. Assigning $\sigma (v, e)$ to every incidence $(v, e)$ of H, we obtain a bidirected graph $\vec {H}$ whose signed line graph is $G^\pm $ .

Proof. Denote $\mathcal {C}^\pm _n$ the set of signed cycles of length n. We obtain $\ell , m \in \mathbb {N}^+$ from Lemmas 2.3, 3.2 and 3.4 such that the following family

$$ \begin{align*} \widetilde{\mathcal{F}}_1^\pm :=\ & \left\{{S_4, P_\ell, -K_3}\right\} \cup \mathcal{X}^\pm(C, \ell, m) \cup \mathcal{X}^\pm(D, \ell, m) \\ & \cup \left\{{(C_n^\pm, V_2(C_n), \ell_0, K_m)}\colon{n \in \left\{{3, \dots, \ell + 1}\right\}, C_n^\pm \in \mathcal{C}_n^\pm, \ell_0 \in \left\{{0, \dots, \ell - 1}\right\}}\right\} \\ & \cup \left\{{(C_n^\pm, V_2(C_n), K_m)}\colon{n \in \left\{{3, \dots, \ell + 1}\right\}, C_n^\pm \in \mathcal{C}_n^\pm}\right\} \\ & \cup \left\{{(K_m, V(K_m), \ell_0, K_m)}\colon{\ell_0 \in \left\{{0, \dots, \ell-1}\right\}}\right\} \end{align*} $$

is disjoint from $\mathcal {G}^\pm (\lambda )$ . Let $\mathcal {F}_1^\pm $ be the smallest family, that is closed under switching, containing $\widetilde {\mathcal {F}}_1^\pm $ . Clearly $\mathcal {F}_1^\pm $ is also disjoint from $\mathcal {G}^\pm (\lambda )$ .

We omit the rest of the proof as it, mutatis mutandis, is very much like the one in the proof of Theorem 1.3 for $\lambda < 2$ on Page 7. We point out that one should make use of Lemmas 3.5 and 3.7 in place of Lemma 2.6 and Theorem 2.7 respectively.

Proof of Theorem 1.5 for $\lambda < 2$ .

Let $N \in \mathbb {N}$ and $\mathcal {F}_1^\pm $ be given by Lemma 3.8, and set

$$ \begin{align*} \mathcal{F}_0^\pm & := \left\{{G^\pm \not\in \mathcal{G}^\pm(\lambda)}\colon{G^\pm \text{ has at most }N\text{ vertices}}\right\}, \\ \widetilde{\mathcal{F}}_2^\pm & := \left\{{G^\pm \not\in \mathcal{G}^\pm(\lambda)}\colon{\exists \vec{H} \in \vec{\mathcal{T}}_N \text{ s.t. }G^\pm \text{ is the signed line graph of }\vec{H}}\right\}, \end{align*} $$

where $\vec {\mathcal {T}}_N$ is the family of bidirected rooted tree such that every connected component obtained from removing the root has at most N vertices. Setting $\mathcal {F}_2^\pm $ to be the family of signed graphs that are minimal in $\widetilde {\mathcal {F}}_2^\pm $ under taking subgraphs, one can check that $\mathcal {F}_0^\pm \cup \mathcal {F}_1^\pm \cup \mathcal {F}_2^\pm $ is a forbidden subgraph characterization of $\mathcal {G}^\pm (\lambda )$ .

It suffices to prove that $\mathcal {F}_2^\pm $ is finite. Let $\vec {T}_1, \dots , \vec {T}_n$ be an enumeration of bidirected rooted trees $\vec {T}$ on at most $N+1$ vertices. We encode $G^\pm \in \mathcal {F}_2^\pm $ by $t_{G^\pm } \in \mathbb {N}^n$ as follows. Let $\vec {H}$ be the bidirected rooted tree in $\vec {\mathcal {T}}_N$ such that $G^\pm $ is the signed line graph of $\vec {H}$ . After removing the root u from $\vec {H}$ , let $U_1, U_2, \dots , U_m$ be the vertex sets of the connected components. For $i \in \left \{{1, \dots , m}\right \}$ , we view the subgraph $\vec {H}_i := \vec {H}[\left \{{u}\right \} \cup U_i]$ as a bidirected tree rooted at u. Set $t_{G^\pm } = (t_1, \dots , t_n)$ , where $t_i$ is the number of occurrences of $\vec {T}_i$ in $\vec {H}_1, \dots , \vec {H}_m$ . Because no member of $\mathcal {F}_2^\pm $ is a subgraph of any other, one can deduce that $\left \{{t_{G^\pm }}\colon {G^\pm \in \mathcal {F}_2^\pm }\right \}$ is an antichain in $(\mathbb {N}^n, \le )$ , and so $\mathcal {F}_2^\pm $ is finite by Lemma 2.9.

Proof of Theorem 1.5 for $\lambda \in [2, \lambda ^*)$ .

Using the fact that every minimal forbidden subgraph for $\mathcal {G}^\pm (2)$ has at most $10$ vertices (see [Reference Vijayakumar31]), and Theorem 3.9, one can prove this case by following the argument in the proof of Theorem 1.3 for $\lambda \in [2, \lambda ^*)$ on Page 9.

Proof of Theorem 3.9.

Suppose that $\lambda \in [2, \lambda ^*)$ . Let $\mathcal {F}^\pm $ denote the set of minimal forbidden subgraphs for the family $\mathcal {D}^\pm _\infty $ . According to Theorem 3.11, the family $\mathcal {F}^\pm $ is finite. Combining Lemmas 3.12 and 3.13, we choose $\ell , m \in \mathbb {N}^+$ such that for every $F^\pm \in \mathcal {F}^\pm $ , the signed extension family $\mathcal {X}^\pm (F^\pm , \ell , m)$ is disjoint from $\mathcal {G}^\pm (\lambda )$ . We know that $S_5 \not \in \mathcal {G}^\pm (\lambda )$ and $-K_4 \not \in \mathcal {G}^\pm (\lambda )$ . In particular, no graph in $\mathcal {G}^\pm (\lambda )$ contains a subgraph that is switching equivalent to any member in the following family

(15) $$ \begin{align} \left\{{S_5, -K_4}\right\} \cup \bigcup_{F^\pm \in \mathcal{F}^\pm} \mathcal{X}^\pm(F^\pm, \ell, m). \end{align} $$

Using Lemma 3.5, we obtain $N \in \mathbb {N}$ such that for every connected signed graph $G^\pm $ with more than N vertices, if no member in (15) is switching equivalent to a subgraph of $G^\pm $ , then neither is any $F^\pm \in \mathcal {F}^\pm $ , and so $G^\pm $ is represented by a subset of $D_n$ and is clearly in $\mathcal {G}^\pm (2)$ . This implies that every connected graph in $\mathcal {G}^\pm (\lambda ) \setminus \mathcal {G}^\pm (2)$ has at most N vertices.

Proof of Theorem 1.5 for $\lambda \ge \lambda ^*$ .

If $\mathcal {F}^\pm $ is a finite forbidden subgraph characterization of $\mathcal {G}^\pm (\lambda )$ , then the all-positive signed graphs in $\mathcal {F}^\pm $ would form a finite forbidden subgraph characterization of $\mathcal {G}(\lambda )$ , which contradicts Theorem 1.3 for $\lambda \ge \lambda ^*$ .