Denote by \(C_{n}\) the cycle on n vertices. Let \(Guv\) denote the graph which arises from G by deleting the edge \(uv\in E(G)\). Similarly, \(G+uv\) is the graph that arises from G by adding an edge \(uv\notin E(G)\), where \(u, v\in V(G)\). For \(v\in V(G)\), \(N(v)\) denotes the neighborhood of v in G and \(d(v)=\vert N(v)\vert \) denotes the degree of vertex v. A pendant vertex of G is a vertex of degree 1. \(\vert x\vert \) denotes the absolute value of a real number x. The terminology not defined here can be found in [15].
Lemma 2.1
[16]
Let
G
be a graph on
n
vertices, e
be an edge of
G. Then
$$q_{1}(G)\ge q_{1}(Ge)\ge q_{2}(G)\ge q_{2}(Ge)\ge\cdots\ge q_{n}(G) \ge q_{n}(Ge)\ge0. $$
Given \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T} \in R^{n}\), we can define a function on \(V(G)\), that is, each vertex \(v_{i}\) is mapped to \(x_{i} =x(v_{i})\). If x is an eigenvector of \(Q(G)\), then it is defined on G naturally, i.e.
\(x(v)\) is the entry of x corresponding to v. Clearly, for \(x\in R^{n}\),
$$x^{T}Q(G)x =\sum_{uv\in E(G)}\bigl(x(u) + x(v) \bigr)^{2}. $$
Let \(x\in R^{n}\) be an arbitrary unit vector. One can find in [10, 15] that
$$ \kappa(G) \le x^{T}Q(G)x, $$
(1)
with equality if and only if x is an eigenvector corresponding to \(\kappa(G)\).
Let \(G_{1}\) and \(G_{2}\) be two vertexdisjoint connected graphs, and let \(v_{i}\in V(G_{i})\) for \(i=1, 2\). Identifying the \(v_{1}\) with \(v_{2}\) and forming a new vertex u (see [10] for details), the resulting graph is called coalescence of \(G_{1}\) and \(G_{2}\), and denoted by \(G_{1}(v_{1})\diamond G_{2}(v_{2})\) or \(G_{1}(u)\diamond G _{2}(u)\). If a connected graph G can be expressed in the form \(G_{1}(u)\diamond G_{2}(u)\), where \(G_{1}\) and \(G_{2}\) are both nontrivial and connected, then \(G_{1}\) is called a branch of G with root u. Clearly \(G_{2}\) is also a branch of G with root u. Let \(x\in R^{n}\) be a vector defined on \(V(G)\). A branch \(G_{i}\) of G is called a zero branch with respect to x if \(x(v) = 0\) for all \(v \in V(G_{i})\); otherwise it is called a nonzero branch with respect to x.
Lemma 2.2
[10]
Let
G
be a connected graph which contains a bipartite branch
B
with root
u, and
x
be an eigenvector corresponding to
\(\kappa(G)\).

(i)
If
\(x(u) = 0\), then
B
is a zero branch of G with respect to
x.

(ii)
If
\(x(u)\neq0\), then
\(x(v)\neq0\)
for every vertex
\(v\in V(B)\).
Lemma 2.3
[10]
Let
G
be a nonbipartite connected graph, and let
x
be an eigenvector corresponding to
\(\kappa(G)\). Let
T
be a tree, which is a nonzero branch of
G
with respect to
x
and with root
u. Then
\(\vert x(q)\vert < \vert x(p)\vert \)
whenever
p, q
are vertices of
T
such that
q
lies on the unique path from
u
to
p.
Lemma 2.4
[12]
Let
\(G = C(v_{0})\diamond B(v_{0})\) (see Figure
1), where
\(C=v_{0}v _{1}v_{2}\cdots v_{k}u_{k}u_{k1} \cdots u_{1}v _{0}\)
is a cycle of length
\(2k+1\)
and
B
is a nontrivial connected bipartite graph. Let
\(x=( x(v_{0}),x(v_{1}),x(v_{2}),\ldots,x(v_{k}),x(u_{1}), x(u_{2}),\ldots,x(u_{k}),\ldots)^{T}\)
be an eigenvector corresponding to
\(\kappa(G)\). Then

(i)
\(\vert x(v_{0})\vert =\max\{\vert x(w)\vert \vert w\in V(C)\}>0\);

(ii)
\(x(v_{i})=x(u_{i})\)
for
\(i=1, 2, \ldots, k\).
Lemma 2.5
[12]
Let
\(G = G_{1}(v_{2}) \diamond T(u)\)
and
\(G^{*} = G_{1}(v_{1})\diamond T(u)\), where
\(G_{1}\)
is a nonbipartite connected graph containing two distinct vertices
\(v_{1}\), \(v_{2}\), and
T
is a nontrivial tree. If there exists an eigenvector
\(x=( x(v_{1}),x(v_{2}),\ldots,x(v_{k}),\ldots)^{T}\)
corresponding to
\(\kappa(G)\)
such that
\(\vert x(v_{1})\vert > \vert x(v_{2})\vert \)
or
\(\vert x(v_{1})\vert = \vert x(v_{2})\vert > 0\), then
\(\kappa(G^{*})<\kappa(G)\).
For \(k \ge1\), let \(G'\) denote the graph obtained from G by deleting the edge uv, inserting k new vertices \(v_{1}, v_{2}, \ldots, v _{k}\) and adding edges \(uv_{1}, v_{1}v_{2}, \ldots, v_{k1}v_{k}, v _{k}v\). Then \(G'\) is called a ksubdivision graph of G by ksubdividing the edge uv.
Lemma 2.6
[17]
Let
\(G'\)
be a
ksubdivision graph of a graph
G. If
k
is even, then
\(\kappa(G')\le\kappa(G)\).
\(U_{n}^{k}(g)\), showed in Figure 2, denotes the unicyclic graph on n vertices with odd girth g and k pendant vertices, where \(g+l+k=n\). \(U_{n}^{*}(3, \Delta)\), showed in Figure 2, denotes the unicyclic graph on n vertices obtained from the cycle \(C_{3}=v_{1}v _{2}v_{3}v_{1}\) by attaching \(\Delta3\) pendant edges and one pendant path at the vertex \(v_{3}\).
Lemma 2.7
[5, 9]
Among all nonbipartite connected graphs on
n
vertices with
k
pendant vertices, \(U_{n}^{k}(3)\)
is the unique graph whose signless Laplacian eigenvalue attains the minimum.
Lemma 2.8
[5]
Let
\(k\ge2\), and
\(g\ge3\)
be an odd integer. Then
\(\kappa(U_{n}^{k1}(g))< \kappa(U_{n}^{k}(g))\).
Lemma 2.9
Let
\(G = G_{1}(v)\diamond B(v)\)
be a connected graph, where
\(G_{1}\)
is a graph of order
n, and
B
is a bipartite graph of order
s. Then
\(\kappa(G ) \le\kappa(G_{1})\). Moreover, if
\(s>1\), \(G_{1}\)
is nonbipartite and there exists an eigenvector
x
corresponding to
\(\kappa(G_{1})\)
such that
\(x(v)\neq0\), then
\(\kappa(G )<\kappa(G_{1})\).
Proof
Let \(V(G_{1})=\{ v_{1}, v_{2}, \ldots, v_{n} \}\), and \(x=(x(v_{1}), x(v_{2}), \ldots, x(v_{n}))^{T}\) be a unit eigenvector corresponding to \(\kappa(G_{1})\). Then
$$\kappa(G_{1})=\sum_{v_{i}v_{j}\in E(G_{1})} \bigl(x(v_{i})+x(v_{j})\bigr)^{2}. $$
Without loss generality, we may assume \(v=v_{n}\). Let \(V(B)=\{ v_{n}, v_{n+1}, \ldots, v_{n+s1} \}\), and let \((U, W)\) be the two parts of the bipartite graph B, where \(v\in U\). Let \(y=(y(v_{1}), y(v_{2}), \ldots, y(v_{n}), y(v_{n+1}), \ldots, y(v_{n+s1}))^{T}\in R^{n+s1}\) defined on \(V(G)\) satisfy that \(y(v_{i})=x(v_{i})\) for \(i=1, 2, \ldots, n\), \(y(u)=x(v)\) if \(u\in U\), and \(y(u)=x(v)\) if \(u\in W\). Then
$$\begin{aligned}& \Vert y\Vert ^{2}=\sum_{i=1}^{n+s1}y(v_{i})^{2}= \sum_{i=1}^{n}x(v_{i})^{2}+(s1)x(v)^{2} \ge\sum_{i=1}^{n}x(v_{i})^{2}= \Vert x\Vert ^{2}=1, \\& \kappa\bigl(G^{*}\bigr)\le\frac{1}{\Vert y\Vert ^{2}} \sum _{v_{i}v_{j}\in E(G^{*})}\bigl(y(v _{i})+y(v_{j}) \bigr)^{2}\le\frac{1}{\Vert x\Vert ^{2}}\sum_{v_{i}v_{j}\in E(G)} \bigl(x(v _{i})+x(v_{j})\bigr)^{2}=\kappa(G). \end{aligned}$$
Clearly, if \(s>1\), \(G_{1}\) is nonbipartite and \(x(v)\neq0\), we have \(\Vert y\Vert ^{2}>\Vert x\Vert ^{2}\). This implies that \(\kappa(G) <\kappa(G_{1})\). □
Lemma 2.10
Let
\(n\ge9\)
and
\(s\ge0\)
be integer. \(G_{1}\)
and
\(G_{2}\), shown in Figure
3, are two unicyclic graphs of order
n. Then
\(\kappa(G_{2})< \kappa(G_{1})\).
Proof
Let \(\kappa=\kappa(G_{1})\), and \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) be a unit eigenvector corresponding to κ. Then \(\kappa= \sum_{v_{i}v_{j}\in E(G_{1})}(x_{i}+x_{j})^{2}\) and \(0<\kappa<1\). By Lemmas 2.2 and 2.4, we have \(x_{n}\neq0\). From the eigenvalue equation \(Q(G_{1})x =\kappa x\), we have
$$\begin{aligned}& x_{n1} =(\kappa1)x_{n}, \\& x_{n2} =\bigl(\kappa^{2}3\kappa+1\bigr)x_{n}, \\& x_{n3} =\bigl(\kappa^{3}5\kappa^{2}+6\kappa1 \bigr)x_{n}, \\& x_{n4} =\bigl(\kappa^{4}7\kappa^{3}+15 \kappa^{2}10\kappa+1\bigr)x_{n}, \\& x_{n5} =\bigl(\kappa^{3}6\kappa^{2}+9\kappa1 \bigr)x_{n}. \end{aligned}$$
Let \(y=(y_{1}, y_{2}, \ldots, y_{n})^{T}\in R^{n}\) defined on \(V(G_{2})\) satisfy that
$$\begin{aligned}& y_{n5}=(x_{n3}+x_{n4}+x_{n5}),\\& y_{n}=(x_{n}+x_{n1}+x_{n2}), \end{aligned}$$
and \(y_{i}=x_{i}\) for \(i=1, 2, \ldots, n6, n4, n3, n2, n1\). Then
$$\sum_{v_{i}v_{j} \in E(G_{2})}(y_{i}+y_{j})^{2}= \sum_{v_{i}v_{j} \in E(G_{1})}(x_{i}+x_{j})^{2}= \kappa, $$
and
$$\begin{aligned} \Vert y\Vert ^{2}\Vert x\Vert ^{2} =&\sum _{i=1}^{n} y_{i}^{2}\sum _{i=1}^{n} x_{i}^{2} \\ =& \kappa\bigl(\kappa^{7}10\kappa^{6}+32\kappa^{5}18 \kappa^{4}89\kappa ^{3}+156\kappa^{2}70\kappa+4 \bigr)x_{n}^{2}. \end{aligned}$$
Let \(f(t)=t^{7}10t^{6}+32t^{5}18t^{4}89t^{3}+156t^{2}70t+4\). By a computation, \(f(t)=0\) has five real roots which are approximately equal to −1.7787, 0.0667, 0.6606, 2, 2.0890, respectively. By Lemma 2.9, we have
$$\kappa=\kappa(G_{1})\le\kappa(G_{1}v_{4} \cdotsv_{s+3}). $$
Note that \(G_{1}v_{4}\cdotsv_{s+3}\) is a 2tsubdividing graph of \(G_{3}\) or \(G_{4}\) (shown in Figure 3). By Lemma 2.6, we have
$$\kappa=\kappa(G_{1})\le\kappa(G_{1}v_{4} \cdotsv_{s+3})\le \min\bigl\{ \kappa(G_{3}), \kappa(G_{4}) \bigr\} . $$
By a computation, we have \(\kappa(G_{3})\approx0.0588025\) and \(\kappa(G_{5})\approx0.0426304\). It follows that \(\kappa< 0.0667\). Noting that \(f(0)=4\), we have \(f(\kappa)>0\). It follows that \(\Vert y\Vert ^{2}>\Vert x\Vert ^{2}\).
Combining the above arguments, we have
$$\kappa(G_{2}) \le\frac{1}{\Vert y\Vert ^{2}}\sum_{v_{i}v_{j} \in E(G_{2})}(y _{i}+y_{j})^{2}< \frac{1}{\Vert x\Vert ^{2}}\sum _{v_{i}v_{j} \in E(G_{1})}(x _{i}+x_{j})^{2}= \kappa(G_{1}). $$
This completes the proof. □
Lemma 2.11
Let
\(n\ge9\), and
\(U_{n}^{n5}(3)\), \(U_{n}^{n4}(3)\), \(U_{n}^{*}(3, n4)\), \(U_{n}^{*}(3, n3)\)
be shown in Figure
2. Then
$$\kappa\bigl(U_{n}^{n5}(3)\bigr)< \kappa\bigl(U_{n}^{*}(3, n4)\bigr),\qquad \kappa \bigl(U_{n}^{n4}(3)\bigr)< \kappa \bigl(U_{n}^{*}(3, n3)\bigr). $$
Proof
Let \(\kappa=\kappa(U_{n}^{*}(3, n4))\), and \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) be a unit eigenvector corresponding to κ. By Corollary 1.3 of [18], it is easy to see \(\kappa(G)<1/2\). From the eigenvalue equation \(Q(U_{n}^{*}(3, n4))x =\kappa x\), we have \(x_{1}=x_{2}\), \(x_{4}=\cdots=x_{n4}\),
$$\begin{aligned}& (\kappa2)x_{1} =x_{1}+x_{3}, \\& (\kappan+4) x_{3} =2x_{1}+(n7)x_{4}+x_{n3}, \\& (\kappa1) x_{4} =x_{3}, \\& (\kappa2)x_{n3} =x_{3}+x_{n2}, \\& (\kappa2)x_{n2} =x_{n3}+x_{n1}, \\& (\kappa2)x_{n1} =x_{n2}+x_{n}, \\& (\kappa1)x_{n} =x_{n1}. \end{aligned}$$
Since \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) is an eigenvector, \(x\neq0\). It follows that
$$\left \begin{matrix} \kappa3 & 1 & 0& 0& 0& 0& 0 \\ 2 & \kappan+4 &7n & 1& 0& 0& 0 \\ 0 & 1 & \kappa1 & 0& 0 & 0& 0 \\ 0 & 1 & 0 & \kappa2& 1& 0& 0 \\ 0 &0 & 0 & 1&\kappa2& 1& 0 \\ 0 &0 & 0 & 0&1&\kappa2& 1 \\ 0 &0 & 0 & 0&0& 1& \kappa1 \end{matrix} \right =0. $$
This implies that κ is the least root of the following equation:
$$\begin{aligned} f(x) \triangleq& x^{7}(n+7)x^{6}+(10n+6)x^{5}(36n48)x^{4}+(55n99)x ^{3} \\ &{}(31n15)x^{2}+(3n+40)x4=0. \end{aligned}$$
Similarly, we can see that \(\kappa(U_{n}^{n5}(3))\) is the least root of the following equation:
$$g(x)\triangleq x^{5}(n+5)x^{4}+(8n6)x^{3}(18n42)x^{2}+(11n28)x4=0. $$
Noting that \(g(0)=4<0\) and
$$f(x)(x1)^{2}g(x)=x(x1) \bigl(x^{3}  nx^{2} (n19)x + 8n  60\bigr)< 0 $$
for \(0< x<1/2\), we have \(g(\kappa)>0\), and so
$$\kappa\bigl(U_{n}^{n5}(3)\bigr)< \kappa=\kappa \bigl(U_{n}^{*}(3, n4)\bigr). $$
By a similar reasoning to above, we can see that \(\kappa(U_{n}^{*}(3, n3))\) and \(\kappa(U_{n}^{n4}(3))\) are the least root of the following equations respectively:
$$\begin{aligned}& h(x)\triangleq x^{6}(n+6)x^{5}+(8n+5)x^{4}(21n18)x^{3}+(19n10)x ^{2}(3n+24)x+4=0, \\& r(x)\triangleq x^{4}(n+4)x^{3}+(6n5)x^{2}(7n12)x+4=0. \end{aligned}$$
Noting that \(r(0)=4>0\) and
$$h(x)(x1)^{2}r(x)=x\bigl(x^{3}  nx^{2} + (n15)x + 4n  28\bigr)>0 $$
for \(0< x<1/2\), we have \(r(\kappa(U_{n}^{*}(3, n3)))<0\), and so
$$\kappa\bigl(U_{n}^{n4}(3)\bigr)< \kappa\bigl(U_{n}^{*}(3, n3)\bigr). $$
This completes the proof. □