Let \(\mathcal{U}_{ n}^{k}\) be the set of non-bipartite unicyclic graphs of order *n* with *k* pendant vertices. From [1, 2], we know that \(U_{n}^{k}(3)\) is the unique graph whose least *Q*-eigenvalue attains the minimum among all graphs in \(\mathcal{U}_{ n}^{k}\). In this section, we will determine the first three graphs in \(\mathcal{U}_{ n}^{k}\) ordered according to their least *Q*-eigenvalues.

For \(k=1\), from [1], we know that \(\kappa(U_{n}^{1}(3))<\kappa (U_{n}^{1}(5))<\kappa(U_{n}^{1}(7))<\cdots\) .

### Theorem 3.1

*Let*
\(2\le k\le n-4\). *Among all graphs in*
\(\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\), \(C_{3}^{1}(n-k-1)\)
*is the unique graph whose least*
*Q*-*eigenvalue attains the minimum*.

### Proof

Let *G* be a graph in \(\mathcal{U}_{ n}^{k}\backslash\{ U_{n}^{k}(3)\}\) whose least *Q*-eigenvalue attains the minimum, and \(C_{g}=v_{1}v_{2}\cdots v_{g}v_{1}\) be the unique cycle of *G*. Then *g* is odd, and *G* can be obtained by attaching rooted trees \(T_{1}, \ldots, T_{g}\) to the vertices \(v_{1}, \ldots, v_{g}\) of \(C_{g}\), respectively, where \(T_{i}\) contains the root vertex \(v_{i}\). \(\vert V(T_{i})\vert =1\) means that \(V(T_{i})=\{v_{i}\}\) and in this case \(T_{i}\) is a trivial tree. Let \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) be a unit eigenvector corresponding to \(\kappa(G)\).

First, we show that *G* is the cycle \(C_{g}=v_{1}v_{2}\cdots v_{g}v_{1}\) with only one nontrivial tree attached. Otherwise, we assume that there are more than one nontrivial trees attached at two different vertices of the cycle \(C_{g}\). Let \(v_{t}\) be a vertex of the cycle \(C_{g}\) such that \(\vert x_{t}\vert \geq \vert x_{i}\vert \) for \(i=1, 2, \ldots, g\). By Lemma 2.1, \(x_{t}\neq0\). Let \(v_{l}\) be another vertex of the cycle \(C_{g}\) such that \(\vert V(T_{l})\vert >1\), and let

$$G_{1}=G-\sum_{v\in N_{T_{l}}(v_{l})}v_{l}v+\sum _{v\in N_{T_{l}}(v_{l})}v_{t}v. $$

From \(k\le n-4\), we have \(G_{1}\in\mathcal{U}_{n}^{k} \backslash\{ U_{n}^{k}(3)\}\). By Lemma 2.4, we have \(\kappa(G_{1})<\kappa(G)\), a contradiction. Therefore *G* is the cycle \(C=v_{1}v_{2}\cdots v_{g}v_{1}\) with only one nontrivial tree attached. Without loss of generality, we may assume the nontrivial tree is \(T_{g}\).

Second, we show that \(g=3\). Otherwise, we assume that \(g\ge5\). By Lemma 2.3, we have \(x_{(g-3)/2}=x_{(g+3)/2}\). Let

$$G '=G-v_{(g-1)/2}v_{(g-3)/2}+v_{(g-1)/2}v_{(g+3)/2}. $$

Clearly, \(G '\in\mathcal{U}_{ n}^{k+1}\), and from (1) we have

$$\kappa\bigl(G '\bigr)\le x^{T}Q\bigl(G ' \bigr)x=x^{T}Q(G)x=\kappa(G). $$

Let \(v_{t}\) be a pendant vertex of *G*, and \(y=(y_{1}, y_{2}, \ldots, y_{n})^{T}\) be a unit eigenvector corresponding to \(\kappa(G ')\). By Lemma 2.2, we have \(\vert y_{t}\vert > \vert y_{g}\vert >0\). Let \(G ''=G '-v_{1}v_{g}+v_{1}v_{t}\). It is easy to see that \(G ''\in\mathcal{U}_{ n}^{k}\backslash\{ U_{n}^{k}(3)\}\). By Lemma 2.4, we have \(\kappa(G '')<\kappa(G ')\). Then we have \(\kappa(G '')<\kappa(G)\), a contradiction. Therefore \(g=3\).

Third, we show that *G* has two pendant neighbors exactly. Otherwise, suppose that *G* has \(r\ge3\) pendant neighbors. Let \(v_{a}\) be a pendant neighbor of *G* such that \(d(v_{3}, v_{a})\) is as large as possible, \(v_{s}\) and \(v_{t}\) be two other pendant neighbors of *G*. Applying Lemma 2.4 to \(v_{s}\) and \(v_{t}\), we may obtain a graph \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\) or \(G '\in \mathcal{U}_{ n}^{k+1}\) such that \(\kappa(G ')<\kappa(G)\). If \(G '\in\mathcal{U}_{ n}^{k}\backslash \{U_{n}^{k}(3)\}\), we have a contradiction. If \(G '\in\mathcal{U}_{ n}^{k+1}\), without loss of generality, we may assume that \(v_{s}\) is a pendant vertex of \(G '\). Let *u* and *w* be two pendant vertices adjacent to \(v_{t}\) of \(G '\), and \(G ''=G '-v_{t}w+uw\). Clearly, \(G ''\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\) and \(\kappa(G '')<\kappa(G ')\). Then we have \(\kappa(G '')<\kappa(G)\), a contradiction. Therefore *G* has two pendant neighbors exactly. Let \(v_{a}\) be a pendant neighbor of *G* such that \(d(v_{3}, v_{a})\) is as large as possible, and \(v_{b}\) be another pendant neighbor of *G*.

Fourth, we show that \(v_{b}\) is in path \(v_{3}-v_{a}\). Otherwise, suppose that \(v_{b}\) is not in path \(v_{3}-v_{a}\). Employing Lemma 2.4 to vertices \(v_{a}\) and \(v_{b}\), we may obtain a graph \(G '\in\mathcal {U}_{ n}^{k+1}\) such that \(\kappa(G ')<\kappa(G)\). Without loss of generality, we may assume that \(v_{b}\) is a pendant vertex of \(G '\). Let *u* and *w* be two pendant vertices adjacent to \(v_{a}\) of \(G '\), and \(G ''=G '-v_{a}w+uw\). Clearly, \(G ''\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\) and \(\kappa(G '')<\kappa(G ')\). Then we have \(\kappa(G '')<\kappa(G)\), a contradiction. Therefore \(v_{b}\) is in path \(v_{3}-v_{a}\).

Fifth, we show that \(v_{a}\) and \(v_{b}\) are adjacent. Otherwise, suppose that \(v_{a}\) and \(v_{b}\) are not adjacent. Let \(v_{c}\in N(v_{b})\) be in path \(v_{b}-v_{a}\), then, by Lemma 2.4, we have \(\vert x_{c}\vert >\vert x_{b}\vert \). Let \(v_{t}\) be the pendant vertex adjacent to \(v_{b}\) and \(G '=G-v_{b}v_{t}+v_{c}v_{t}\). Clearly, \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\) and by Lemma 2.4 we have \(\kappa(G ')<\kappa(G)\), a contradiction. Therefore \(v_{a}\) and \(v_{b}\) are adjacent.

Sixth, we show that \(d(v_{b})=3\). Otherwise, suppose that \(d(v_{b})>3\). Let \(v_{t}\) be the pendant vertex adjacent to \(v_{b}\) and \(G '=G-v_{b}v_{t}+v_{a}v_{t}\). Clearly, \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3)\}\). By Lemma 2.4, we have \(\vert x_{a}\vert >\vert x_{b}\vert \), and by Lemma 2.4, we have \(\kappa(G ')<\kappa(G)\), a contradiction. Therefore \(d(v_{b})=3\).

From the above arguments, we have \(G=C_{3}^{1}(n-k-1)\). □

For \(k=n-3\), \(\mathcal{U}_{ n}^{ n-3}=\{\Delta_{r, s, t} \mid r\ge s\ge t\ge0, r+s+t=n-3 \}\), where \(\Delta_{r, s, t}\) is the graph obtained from the cycle \(C_{3}\) by attaching *r*, *s*, *t* pendent edges to the vertices \(v_{1}\), \(v_{2}\), and \(v_{3}\) of the cycle \(C_{3}\), respectively. By a similar reasoning to that of Theorem 3.1, we can prove the following theorem.

### Theorem 3.2

*Let*
\(n\ge8\), *and*
\(G\in\mathcal{U}_{ n}^{n-3}\backslash\{\Delta _{n-3, 0, 0}, \Delta_{n-4, 1, 0}, \Delta_{n-5, 2, 0}\}\). *Then*

$$\kappa(\Delta_{n-3, 0, 0})< \kappa(\Delta_{n-4, 1, 0})< \kappa ( \Delta_{n-5, 2, 0})< \kappa(G). $$

Next, we will determine the graph in \(\mathcal{U}_{ n}^{k}\backslash\{ U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\) whose least *Q*-eigenvalue attains the minimum.

### Theorem 3.3

*Let*
\(2\le k\le n-5\). *Among all graphs in*
\(\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\), \(C_{3}^{ 1}(n-k-2)\)
*or*
\(C_{3}^{ 2}(n-k-1)\)
*is the graph whose least*
*Q*-*eigenvalue attains the minimum*.

### Proof

Let *G* be a graph in \(\mathcal{U}_{ n}^{k}\backslash\{ U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\) whose least *Q*-eigenvalue attains the minimum, and let \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) be a unit eigenvector corresponding to \(\kappa(G)\). By a similar reasoning to that of Theorem 3.1, we can prove that *G* is the cycle \(C=v_{1}v_{2}v_{3}v_{1}\) with only one nontrivial tree \(T_{3}\) attached at \(v_{3}\), and *G* has two pendant neighbors exactly. Let \(v_{a}\) be a pendant neighbor of *G* such that \(d(v_{3}, v_{a})\) is as large as possible, and \(v_{b}\) be another pendant neighbor of *G*. By a similar reasoning to that of Theorem 3.1, we can prove that \(v_{b}\) is in path \(v_{3}-v_{a}\).

Now we show that \(d(v_{b}, v_{a})\le2\). Otherwise, suppose that \(d(v_{b}, v_{a})\ge3\). Let \(v_{t}\) be the pendant vertex adjacent to \(v_{b}\) and \(v_{c}\in N(v_{b})\) be in path \(v_{b}-v_{a}\). Then, by Lemma 2.4, we have \(\vert x_{c}\vert >\vert x_{b}\vert \). Let \(G '=G-v_{b}v_{t}+v_{c}v_{t}\). Clearly, \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\) and \(\kappa(G ')<\kappa(G)\), a contradiction. Therefore \(d(v_{b}, v_{a})\le2\).

If \(d(v_{b}, v_{a})=2\), then we declare \(d(v_{b})=3\). Otherwise, suppose that \(d(v_{b})\ge4\). Let \(v_{t}\) be the pendant vertex adjacent to \(v_{b}\) and let \(G '=G-v_{b}v_{t}+v_{a}v_{t}\). Clearly, \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\) and \(\kappa(G ')<\kappa(G)\), a contradiction. Therefore \(d(v_{b})=3\) and \(G=C_{3}^{1}(n-k-2)\).

If \(d(v_{b}, v_{a})=1\), then we declare \(d(v_{b})=4\). Otherwise, suppose that \(d(v_{b})\ge5\). Let \(v_{t}\) be the pendant vertex adjacent to \(v_{b}\) and let \(G '=G-v_{b}v_{t}+v_{a}v_{t}\). Clearly, \(G '\in\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\) and \(\kappa(G ')<\kappa(G)\), a contradiction. Therefore \(d(v_{b})=4\) and \(G=C_{3}^{2}(n-k-1)\).

From the above arguments, we have \(G=C_{3}^{1}(n-k-1)\) or \(C_{3}^{2}(n-k-1)\). □

For \(k=n-4\), \(\mathcal{U}_{ n}^{ n-4}=\{C_{3}^{ r, s, t, l} \mid l\ge1, r\ge0, s\ge0, t\ge0, r+s+t+l=n-4 \}\), where \(C_{3}^{ r, s, t, l}\), shown in Figure 4, denotes the unicyclic graph of order *n* with \(n-4\) pendant vertices. \(C_{3}^{ 1}(2)\) and \(C_{3}^{ 2}(3)\), shown in Figure 4, are the unicyclic graphs of order *n* with \(n-4\) pendant vertices.

### Theorem 3.4

*Let*
\(n\ge7\). *Among all graphs in*
\(\mathcal{U}_{ n}^{ n-4}\backslash \{U_{n}^{ n-4}(3), C_{3}^{1}(3)\}\), \(C_{3}^{ 1}(2)\)
*is the unique graph whose least*
*Q*-*eigenvalue attains the minimum*.

### Proof

By a similar reasoning to that of Theorem 3.3, we can prove that \(C_{3}^{ 2}(3)\) or \(C_{3}^{ 1}(2)\) is the graph whose least *Q*-eigenvalue attains the minimum among all graphs in \(\mathcal{U}_{ n}^{ n-4}\backslash\{U_{n}^{ n-4}(3), C_{3}^{1}(3)\}\). Let \(\kappa=\kappa(C_{3}^{ 2}(3))\) and let \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) be an eigenvector corresponding to *κ*. From the eigenvalue equations, we have \(x_{1}=x_{2}\), \(x_{5}=x_{6}\), \(x_{7}=\cdots=x_{n}\),

$$\begin{aligned}& (\kappa-2)x_{1} = x_{1}+x_{3}, \\& (\kappa-5)x_{3} = 2x_{1}+x_{4}+2x_{5}, \\& (\kappa-n+5)x_{4} = x_{3}+(n-6)x_{7}, \\& (\kappa-1)x_{5} = x_{3}, \\& (\kappa-1)x_{7} = x_{4}. \end{aligned}$$

Since *x* is an eigenvector, it follows that \(\kappa=\kappa(C_{3}^{ 2}(3))\) is the least root of the equation

$$f(x)\triangleq\left| \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} x-3 & -1 & 0 & 0 & 0 \\ -2 & x-5 & -1 & -2 & 0 \\ 0 & -1 & x-n+5 & 0 & -n+6 \\ 0 & -1 & 0 & x-1 & 0 \\ 0 & 0 & -1 & 0 & x-1 \end{array}\displaystyle \right| =0. $$

By an easy computation, we can obtain

$$f(x)=x^{5}-(n+5)x^{4}+(9n-17)x^{3}-(19n-65)x^{2}+(7n-16)x-4. $$

Similarly, from the eigenvalue equation, we can prove that \(\kappa(C_{3}^{ 1}(2))\) is the least root of

$$g(x)\triangleq x^{6}-(n+6)x^{5}+(9n-2)x^{4}-(25n-48)x^{3}+(25n-58)x^{2}-(7n-8)x+4=0. $$

By Lemma 2.6, we have \(0<\kappa(C_{3}^{ 2}(3)), \kappa (C_{3}^{ 1}(2))\le4/n\). Note that for \(n\ge12\),

$$(x-1)f(x)-g(x)=x\bigl((n-10)x^{3}-(3n-34)x^{2}+(n-23)x+4 \bigr)>0 $$

for \(0< x\le4/n\). It follows that \(g(\kappa(C_{3}^{ 2}(3)))<0\). This implies that \(\kappa(C_{3}^{ 1}(2))<\kappa(C_{3}^{ 2}(3))\).

For \(7\le n\le11\), by computation, we can verify that \(\kappa(C_{3}^{ 1}(2))<\kappa(C_{3}^{ 2}(3))\).

From the above arguments, we have \(\kappa(C_{3}^{ 1}(2))<\kappa(C_{3}^{ 2}(3))\) for \(n\ge7\). □

Combining Theorem 3.3 and Lemma 2.8, we have the following theorem.

### Theorem 3.5

*Let*
\(3\le k \leq{(n-4)}/{\sqrt{6}}\). *Among all graphs in*
\(\mathcal {U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{ 1}(n-k-1)\}\), \(C_{3}^{ 2}(n-k-1)\)
*is the unique graph whose least*
*Q*-*eigenvalue attains the minimum*.

Combining Theorem 3.3 and Lemma 2.9, we have the following theorem.

### Theorem 3.6

*Let*
\(n\ge120\), \(k > \frac{-3+\sqrt{21}}{2}n\). *Among all graphs in*
\(\mathcal{U}_{ n}^{k}\backslash\{U_{n}^{k}(3), C_{3}^{1}(n-k-1)\}\), \(C_{3}^{1}(n-k-2)\)
*is the unique graph whose least*
*Q*-*eigenvalue attains the minimum*.