Proposition 3.1
Let
G
be a connected graph without pendant vertices, and
\(c(G)\geq 2\). Then
G
does not satisfy the minimal nullity condition.
Proof
Since \(c(G)\geq2\), there are at least two different cycles in G. We distinguish the following two cases.
Case 1. There exists a vertex u on a cycle such that \(c(Gu)\leq c(G)3\).
If G satisfies the minimal nullity condition, then we have
$$\begin{aligned} \eta(G) &=n(G) 2m(G) c(G) \\ &\leq \bigl[n(Gu)+1 \bigr] 2m(Gu)  \bigl[c(Gu)+3 \bigr] \quad \bigl( \because m(Gu)\leq m(G) \bigr) \\ &\leq n(Gu)2m(Gu) c(Gu)2 \\ &\leq\eta(Gu)2 \quad (\mbox{by Theorem 1.2}). \end{aligned}$$
By Lemma 2.4 we know that \(\eta(G) \geq\eta(Gu)1\), a contradiction.
Case 2. For any vertex u on a cycle, \(c(Gu)\geq c(G)2\).
We first prove the following claim.
Claim
Let
G
be a connected graph without pendant vertices. Suppose that for any vertex
u
on a cycle of
G, \(c(Gu)\geq c(G)2\) (\(c(G)\geq2\)). Then there are at most
\(c(G)1\)
vertices of
G
not covered by its maximum matching.
In fact, if a vertex u lies on a cycle of G, then \(c(Gu)\leq c(G)1\). We distinguish the following two cases.
Case a. For any vertex u that lies on a cycle of G, \(c(Gu)= c(G)1\).
By Lemma 2.6 all cycles of G are pairwise vertexdisjoint in this case. By induction on \(c(G)\), when \(c(G)=2\), G becomes \(B(l,x,k)\) (\(x\geq2\)). Since \(B(l,x,k)\) (\(x\geq2\)) has a path as its spanning subgraph, there is at most \(1=c(G)1\) vertex not covered by its maximum matching.
We assume that the claim holds for a connected graph G without pendant vertices and \(c(G)< l\) and all cycles of G are pairwise vertexdisjoint. Then we consider the case \(c(G)=l\) (\(l>2\)).
Since all cycles of G are pairwise vertexdisjoint, there must exist a subgraph \(G_{1}\) that is a lollipop graph such that \(GG_{1}\) is a graph without pendant vertices. The cycles of \(GG_{1}\) are also pairwise vertexdisjoint since it is a subgraph of G. Clearly, \(c(GG_{1})=c(G)1=l1< l\). By the induction assumption there are at most \(c(GG_{1})1\) vertices of \(GG_{1}\) not covered by the maximum matching of \(GG_{1}\). It is easy to see that in the lollipop graph \(G_{1}\), there is at most one vertex not covered by the maximum matching of \(G_{1}\). Then, there are at most \(c(GG_{1})1+1=c(G)1\) vertices of G not covered by its maximum matching.
Case b. There is a vertex u that lies on a cycle of G such that \(c(Gu)= c(G)2\). Similarly to Case a, by induction on \(c(G)\), when \(c(G)=2\), G becomes \(B(l,1,k)\) or \(\theta(l,x,k)\). Since \(B(l,1,k)\) and \(\theta(l,x,k)\) have a path as their spanning subgraph, there is at most \(1=c(G)1\) vertex not covered by their maximum matchings.
We assume that the claim holds for a connected graph G without pendant vertices and \(c(G)< l\) and there is a vertex u that lies on a cycle of G such that \(c(Gu)= c(G)2\). It is suffices to prove the claim in the case \(c(G)=l\) (\(l>2\)).
Suppose that \(Gu\) has p connected components, say \(H_{1}, H_{2},\ldots,H_{p}\). Let \(n(H_{i})\) and \(e(H_{i})\) be the number of vertices and size of \(H_{i}\), \(i=1,2,\ldots,p\), respectively. Obviously,
$$\begin{aligned}& n(G)=\sum_{i=1}^{p}n(H_{i})+1 , \qquad e(G)=\sum_{i=1}^{p}e(H_{i})+d_{G}(u), \\& c(H_{i})=e(H_{i})n(H_{i})+1 ,\qquad c(G)=e(G)n(G)+1 , \\& c(Gu)=\sum_{i=1}^{p}c(H_{i}). \end{aligned}$$
For each \(H_{i}\), we have \(c(H_{i})< c(G)\), \(i=1,\ldots,p\). Moreover, we have \(d_{G}(u)=p+2\). In fact,
$$\begin{aligned} d_{G}(u)&=e(G)\sum_{i=1}^{p}e(H_{i}) \\ &= \bigl(c(G)+n(G)1 \bigr)\sum_{i=1}^{p}e(H_{i}) \\ &= \bigl(c(Gu)+2 \bigr)+ \Biggl(\sum_{i=1}^{p}n(H_{i})+1 \Biggr)1\sum_{i=1}^{p}e(H_{i}) \\ &= \Biggl(\sum_{i=1}^{p}c(H_{i})+2 \Biggr)+\sum_{i=1}^{p}n(H_{i}) \sum_{i=1}^{p}e(H_{i}) \\ &=\sum_{i=1}^{p} \bigl(e(H_{i})n(H_{i})+1 \bigr)+2+\sum_{i=1}^{p}n(H_{i}) \sum_{i=1}^{p}e(H_{i}) \\ &=\sum_{i=1}^{p}e(H_{i})\sum _{i=1}^{p}n(H_{i})+p+2+\sum _{i=1}^{p}n(H_{i})\sum _{i=1}^{p}e(H_{i}) \\ &=p+2 . \end{aligned}$$
Thus, by the incident relation between u and \(H_{i}\), G must be isomorphic to one of the two graphs shown in Figure 2.
Now, if G is the graph \(G'\) shown in Figure 2, then we distinguish the following two cases.
Case b.1. Each component \(H_{i}\) has no pendant vertices, \(i=1,2,\ldots,p\).
Now assume that for \(i=1,\ldots,s\), \(c(H_{i})=1\) and for \(i=s+1,\ldots,p\), \(c(H_{i})\geq2\). Then \(c(Gu)=\sum_{i=1}^{p}c(H_{i})=s+c(H_{s+1})+\cdots+c(H_{p})\). When \(i\in\{ 1,\ldots,s\}\), since \(c(H_{i})=1\), \(H_{i}\) is a cycle and there is at most one vertex not covered by a maximum matching of \(H_{i}\); when \(i\in\{s+1,\ldots,p\}\), since \(c(H_{i})\geq2\), by the induction assumption there are at most \(c(H_{i})1\) vertices not covered by a maximum matching of \(H_{i}\). Let \(V_{1}\) be a set of the vertices that are not be covered by a maximum matching of G. Then
$$\begin{aligned} \vert V_{1}\vert &\leq\underbrace{1+\cdots+1}_{s}+ \bigl(c(H_{s+1})1 \bigr)+\cdots+ \bigl(c(H_{p})1 \bigr)+ \bigl\vert \{u\} \bigr\vert \\ &=s+c(H_{s+1})+\cdots+c(H_{p})(ps)+1 \\ &=c(G)(ps)1 \quad \bigl( \because c(G)=c(Gu)+2=s+c(H_{s+1})+ \cdots+c(H_{p})+2 \bigr) \\ &\leq c(G)1 \quad ( \because p\geq s ). \end{aligned}$$
Thus, there are at most \(c(G)1\) vertices not overed by a maximum matching of G.
Case b.2. There exists a component, say \(H_{i_{0}}\) (\(i_{0}\in\{ 1,2,\ldots, p\}\)) that has a pendant vertex.
Let \(v_{1}\) be the pendant vertex of \(H_{i_{0}}\), and \(P_{v_{1}v_{2}\cdots v_{x}v_{x+1}}\) be a path in \(H_{i_{0}}\) such that \(v_{1}\) is its end vertex and \(d_{G}(v_{2})=\cdots=d_{G}(v_{x})=2\), \(d_{G}(v_{x+1})>2\). Clearly, \(v_{1}\) is an adjacent vertex of u.
When \(i_{0}=1\), by removing the path \(P_{v_{1}v_{2}\cdots v_{x}}\) from G the resulting graph \(GP_{v_{1}v_{2}\cdots v_{x}}\) is a graph without pendant vertex with \(c(GP_{v_{1}v_{2}\cdots v_{x}})< c(G)\). By the induction assumption there are at most \(c(GP_{v_{1}v_{2}\cdots v_{x}})1\) vertices not covered by a maximum matching of \(GP_{v_{1}v_{2}\cdots v_{x}}\). Recall that there is at most one vertex not covered by a maximum matching of the path \(P_{v_{1}v_{2}\cdots v_{x}}\). Therefore, there are at most \(c(GP_{v_{1}v_{2}\cdots v_{x}})1+1\leq c(G)1\) vertices not covered by a maximum matching of G.
When \(i_{0}\in\{2, \ldots, p\}\), by removing the path \(P_{v_{1}v_{2}\cdots v_{x}}\) from G the resulting graph \(GP_{v_{1}v_{2}\cdots v_{x}}\) is a graph without pendant vertex, whereas \(c(GP_{v_{1}v_{2}\cdots v_{x}})=c(G)\). It is obvious that \(GP_{v_{1}v_{2}\cdots v_{x}}\) has exactly two components, say \(G_{1}\) and \(G_{2}\), both without pendant vertices and with \(c(G_{i})< c(G)\), \(i=1,2\).
If \(c(G_{i})\geq2\), \(i=1,2\), by the induction assumption there are at most \(c(G_{i})1\) vertices not covered by a maximum matching of \(G_{i}\), respectively. Since there is at most one vertex not covered by a maximum matching of \(P_{v_{1}v_{2}\cdots v_{x}}\), there are at most \(c(G_{1})1+c(G_{2})1+1=c(G)1\) vertices not covered by a maximum matching of G.
If there is one of \(G_{i}\), say \(G_{1}\), with \(c(G_{1})=1\), then \(P_{v_{1}v_{2}\cdots v_{x}}+G_{1}\) (the subgraph induced by \(v_{1},v_{2},\ldots,v_{x}\) and \(V(G_{1})\)) is a lollipop graph. Therefore, there is at most one vertex not covered by a maximum matching of \(P_{v_{1}v_{2}\cdots v_{x}}+G_{1}\). Since \(c(G)\geq3\), we have \(c(G_{2})\geq2\). By the induction assumption there are at most \(c(G_{2})1\) vertices not covered by a maximum matching of \(G_{2}\). Thus, there are at most \(c(G_{2})1+1=c(G)1\) vertices not covered by a maximum matching of G.
In a similar way, we can prove the claim when G is the graph \(G''\) shown in Figure 2.
By the previous claim, if G satisfies the minimal nullity condition, then we have
$$\begin{aligned} \eta(G)&=n(G) 2m(G) c(G) \\ &\leq c(G)1c(G)=1 , \end{aligned}$$
a contradiction. This completes the proof of the lemma. □
By Proposition 3.1 we have that if a connected graph G with \(c(G)\geq2\) satisfies the minimal nullity condition, then G must have a pendant vertex.
We are now ready to prove Theorem 1.4.
Proof
It suffices to prove the theorem for the case where G is connected.
‘⇒’ Let G be a connected graph that satisfies the minimal nullity condition.
When \(c(G)=0\), G is a tree. Then (a), (b), and (c) hold trivially.
When \(c(G) = 1\), G is a unicyclic graph. Let \(C_{l}\) be the unique cycle of G of length l. By Lemma 2.9, \(C_{l}\) is an odd cycle, and \(m(G)=\frac{l1}{2}+m(GC_{l})\). Thus, (a) and (b) hold obviously. Moreover, since \(m(C_{l})=\frac{l1}{2}\) and \(m(GC_{l})=m(T _{G}  v_{C_{l}} )\), we have \(m(G)=m(C_{l})+m(T _{G}  v_{C_{l}} )\). By Lemma 2.12, \(m(T _{G})=m(T _{G}  v_{C_{l}} )\). Thus, (c) holds.
It remains to prove (a), (b), and (c) for G when \(c(G) \geq2\).
Assume that G is a connected graph with \(c(G) \geq2\) that satisfies the minimal nullity condition. By Proposition 3.1, G must have a pendant vertex. Let u be a pendant vertex of G. It follows from Lemma 2.11 that the adjacent vertex v of u does not lie on a cycle of G, and by removing these two vertices u and v the resulting graph \(G_{0}\) also satisfies the minimal nullity condition. Clearly, \(c(G)=c(G_{0})\geq2\).
Let \(H_{1}, H_{2},\ldots,H_{p}\) be the connected components of \(G_{0}\). Then each \(H_{i}\), \(i=1,\ldots,p\), satisfies the minimal nullity condition. Without loss of generality, we assume that for \(i\in\{1,\ldots,s\}\), \(c(H_{i})\leq1\) and for \(i\in\{ s+1,\ldots,p\}\), \(c(H_{i})\geq2\). By Proposition 3.1, for \(i\in\{s+1,\ldots,p\}\), each \(H_{i}\) must have a pendant vertex, and by Lemma 2.11 the adjacent vertex of each pendant vertex does not lie on a cycle of \(H_{i}\). By removing a pendant vertex of \(H_{i}\) and its adjacent vertex the resulting graph also satisfies the minimal nullity condition. For a connected component with \(c\geq2\) that satisfies the minimal nullity condition, there must be a pendant vertex, and its adjacent vertex does not lie on a cycle. Then we can repeatedly remove a pendant vertex of \(H_{i}\), \(i\in\{s+1,\ldots,p\}\), and its adjacent vertex until all connected components of the resulting graph has \(c\leq1\). Continuing in a similar way, we can remove a pendant vertex and its adjacent vertex repeatedly until all connected components of the resulting graph do not contain pendant vertices. It is obvious that the final resulting graph, say \(G^{*}\), consists only of cycles and isolated vertices. Furthermore, \(G^{*}\) satisfies the minimal nullity condition and keeps \(c(G^{*})=c(G)\). It follows from Lemma 2.10 that \(G^{*}\) is the disjoint union of \(c(G)\) odd cycles, say \(C_{1}, C_{2}, \ldots, C_{c(G)}\), and some isolated vertices. Therefore, the cycles of G are pairwise vertexdisjoint, and each cycle is odd. Thus, (a) and (b) hold.
On the other hand, since in each step we just remove a pendant vertex and its adjacent vertex, we have \(m(G) =\sum_{i=1}^{c(G)}m(C_{i})+m(G  \bigcup_{i=1}^{c(G)}V(C_{i} ))\). Thus, by Lemma 2.12 we have \(m(T _{G} ) = m(T _{G}  W_{G} )\), and thus (c) holds.
‘⇐’ Let G be a connected graph satisfying (a), (b), and (c). We deal with the sufficient part of the proof by induction on \(c(G)\). When \(c(G)=0\), G is a tree. Then, by Lemma 2.8, the result is trivial. When \(c(G) = 1\), G is a unicyclic graph. Let \(C_{l}\) be the unique cycle of G of length l. Then \(C_{l}\) is an odd cycle, and \(m(T _{G})=m(T _{G}  v_{C_{l}} )\). By Lemma 2.12 we have \(m(G)=m(C_{l})+m(T _{G}  v_{C_{l}} )=\frac{l1}{2}+m(GC_{l})\). Furthermore, by Lemma 2.9, \(\eta(G) = n (G)2m(G) c(G)\) in this case.
We assume that the result holds when \(c(G)< l\). Now we consider the case \(c(G)=l\) (\(l\geq2\)).
Since cycles of G are pairwise vertexdisjoint, there exists a cycle C in G such that all cycles of \(GC\) lie in the same component, say \(H _{1} \), of \(G  C\), and the other components of \(G  C\), say \(H_{i}\) (\(i=2,\ldots,p\)), are trees. Moreover, \(n(H_{i})\geq2\) (\(i=2,\ldots,p\)). If not, there is \(i_{0}\in\{2,\ldots,p\}\) such that \(n(H_{i_{0}})=1\), which means that there is a pendant vertex of G such that its adjacent vertex lies on C. But this contradicts Lemma 2.7 because (c) holds for G if and only if the adjacent vertex of any pendant vertex does not lie on a cycle.
If \(p\geq2\), then let u be a pendant vertex of G that belongs to \(\bigcup_{i=2}^{p}V (H_{i} ) \), and let v be its adjacent vertex. Denote by \(G_{0} \) the graph obtained from G by removing u and v. By Lemma 2.7, v does not lie on a cycle of G, and (c) also holds for \(G_{0} \). Repeating this operation, after a finite number of steps, we can obtain a graph \(G^{*}\) that is a disjoint union of a connected component \(G_{1} \) induced by \(V (C) \cup V (H _{1} )\) and some acyclic graphs. Obviously, (c) also holds for \(G^{*}\).
Since G is a connected graph and all cycles are pairwise vertexdisjoint, there is exactly one edge between C and \(H _{1}\) in \(G^{*}\). Let xy is the unique edge between C and \(H _{1} \), where \(x \in V (C)\) and \(y \in V (H _{1})\). Note that (c) holds for the graph \(G^{*}\) if and only if (c) also holds for its each connected component. Therefore, (c) holds for \(G_{1}\), that is, \(m(T_{G_{1}})=m(T_{G_{1}}W_{G_{1}})\). Certainly, (a) and (b) hold for \(G_{1}\) since it is a subgraph of G. Moreover, both \(H _{1}+x\) and \(H _{1}\) also satisfy (a) and (b) since they are subgraphs of \(G_{1}\). Now let us show that (c) also holds for \(H _{1}+x\) and \(H _{1}\), that is,

(1)
\(m(T_{H _{1}+x})=m(T_{H _{1}+x}W_{H _{1}+x})\).
In fact, We have \(m(T_{G_{1}})=m(T_{H _{1}+x})\) since \(T_{G_{1}}\cong T_{H _{1}+x}\), and \(W_{G_{1}}=W_{H _{1}+x}\cup\{v_{C}\}\), where \(v_{C}\) is the vertex in \(W_{G_{1}}\) contracted from the cycle C. Since \(m(T_{G_{1}})=m(T_{G_{1}}W_{G_{1}})\), we have
$$\begin{aligned} m(T_{H _{1}+x})&=m \bigl(T_{H _{1}+x} \bigl(W_{H _{1}+x}\cup \{v_{C}\} \bigr) \bigr) \\ &=m(T_{H _{1}+x}W_{H _{1}+x}v_{C}) \\ &\leq m(T_{H _{1}+x}W_{H _{1}+x}). \end{aligned}$$
On the other hand, it is clear that \(m(T_{H _{1}+x})\geq m(T_{H _{1}+x}W_{H _{1}+x})\). Therefore, \(m(T_{H _{1}+x})=m(T_{H _{1}+x}W_{H _{1}+x})\).

(2)
\(m(T_{H _{1}})=m(T_{H _{1}}W_{H _{1}})\).
In fact, \(T_{G_{1}}\) can be regarded as the graph induced by \(V(T_{H _{1}})\) and the vertex \(v_{C}\). Therefore, \(T_{G_{1}}=T_{H _{1}}+v_{C}\) and \(W_{G_{1}}=W_{H _{1}}\cup\{v_{C}\}\). Since (c) holds for \(G_{1}\), we have \(m(T_{G_{1}})=m(T_{G_{1}}W_{G_{1}})\). Thus,
$$\begin{aligned} m(T_{H _{1}}+v_{C})&=m \bigl(T_{H _{1}}+v_{C} \bigl(W_{H _{1}}\cup\{v_{C}\} \bigr) \bigr) \\ &=m(T_{H _{1}}W_{H _{1}}) \\ &\leq m(T_{H _{1}}). \end{aligned}$$
Clearly, \(m(T_{H _{1}}+v_{C})\geq m(T_{H _{1}})\). Then we have \(m(T_{H _{1}}+v_{C})=m(T_{H _{1}})\). The previous derivation forces the last inequality involved to become an equality. Thus, \(m(T_{H _{1}})=m(T_{H _{1}}W_{H _{1}})\).
Since \(c(H _{1}+x)< c(G_{1})=c(G)\) (respectively, \(c(H _{1})< c(G_{1})=c(G)\)), by the induction assumption, \(H _{1}+x\) (respectively, \(H _{1}\)) satisfies the minimal nullity condition, that is, \(\eta(H _{1}+x) =n(H _{1}+x)2m(H _{1}+x)c(H _{1}+x)\) and \(\eta(H _{1})=n(H _{1})2m(H _{1})c(H _{1})\).
It follows from (1) and (2) that \(m(T_{H _{1}+x})=m(T_{G})=m(T_{H _{1}}+v_{C})=m(T_{H _{1}})\), that is, \(m(T_{H _{1}+x})=m(T_{H _{1}})\). Since \(m(T_{H _{1}+x})=m(T_{H _{1}+x}W_{H _{1}+x})\) and \(m(T_{H _{1}})=m(T_{H _{1}}W_{H _{1}})\), by Lemma 2.12 we have \(m(H _{1}+x)=\sum_{C\in{H _{1}+x}} m(C)+m(T_{H _{1}+x}W_{H _{1}+x})\) and \(m(H _{1})=\sum_{C\in H _{1}} m(C)+m(T_{H _{1}}W_{H _{1}})\). Then,
$$\begin{aligned} m(H _{1}+x)&=\sum_{C\in{H _{1}+x}}m(C)+m(T_{H _{1}+x}W_{H _{1}+x}) \\ &=\sum_{C\in{H _{1}+x}}m(C)+m(T_{H _{1}+x}) \\ &=\sum_{C\in H _{1}} m(C)+m(T_{H _{1}}) \\ &=\sum_{C\in H _{1}} m(C)+m(T_{H _{1}}W_{H _{1}}) \\ &=m(H _{1}), \end{aligned}$$
that is, \(m(H _{1}+x)=m(H _{1})\). Furthermore, since both \(H _{1}+x\) and \(H _{1}\) satisfy the minimal nullity condition, we have
$$\begin{aligned} \eta(H _{1}+x) &=n(H _{1}+x)2m(H _{1}+x)c(H _{1}+x) \\ &= \bigl(n(H _{1})+1 \bigr)2m(H _{1})c(H _{1}) \\ &=\eta(H _{1})+1. \end{aligned}$$
Then by Lemma 2.5 we have
$$\begin{aligned} \eta(G_{1})&=\eta(H _{1})+\eta(G_{1}H _{1}) \\ &=n(H _{1})2m(H _{1})c(H _{1})+\eta(C) \\ &= \bigl(n(G_{1})n(C) \bigr)2 \bigl(m(G)m(C) \bigr) \bigl(c(G_{1})1 \bigr)+0 \\ &=n(G_{1})2m(G_{1})c(G_{1}), \end{aligned}$$
that is, \(G_{1}\) satisfies the minimal nullity condition. On the other hand, each acyclic graph satisfies the minimal nullity condition. Then \(G^{*}\) also satisfies the minimal nullity condition. It follows from Lemma 2.11 that G satisfies the minimal nullity condition since \(G^{*}\) does. The proof is completed. □
Proposition 3.2
Let
G
be a graph with at least one cycle. If
G
satisfies the minimal nullity condition, then for any vertex
u
that lies on a cycle of
G, we have that
\(G  u\)
also satisfies the minimal nullity condition and
\(\eta(G  u) = \eta(G) \).
Proof
It suffices to prove the result for the case where G is connected. Let G be a connected graph satisfying the minimal nullity condition. By Theorem 1.4 all cycles of G, say \(C_{1},C_{2},\ldots ,C_{c(G)}\), are pairwise vertexdisjoint, and each \(C_{i}\) (\(i=1,\ldots,c(G)\)) is an odd cycle. Let u be any vertex that lies on a cycle of G, say \(C_{i_{0}}\) (\(i_{0}\in\{1,\ldots,c(G)\}\)). Since G satisfies the minimal nullity condition, by Theorem 1.4 we have \(m(T_{G})=m(T_{G}W_{G})\). Then by Lemma 2.12 we have
$$\begin{aligned} m(G)&=\sum_{i=1} ^{c(G)} m(C_{i})+ m \Biggl( G  \bigcup_{i=1} ^{c(G)}C_{i} \Biggr) \\ &=\sum_{i\neq i_{0}}m(C_{i})+m(C_{i_{0}})+m \biggl((G u) \bigcup_{i\neq i_{0}}C_{i}(C_{i_{0}}u) \biggr) \\ &=\sum_{i\neq i_{0}} m(C_{i})+m(C_{i_{0}})+m \biggl((G u) \bigcup_{i\neq i_{0}}C_{i} \biggr)m(C_{i_{0}}u) \\ &=\sum_{i\neq i_{0}} m(C_{i})+m \biggl((G u)  \bigcup_{i\neq i_{0}}C_{i} \biggr) \\ &=\sum_{C_{i} \text{ in }Gu} m(C_{i})+m \biggl((G u)  \bigcup_{C_{i} \text{ in } Gu}C_{i} \biggr) \\ &\leq m(Gu) . \end{aligned}$$
However, \(m(G)\geq m(Gu)\). Thus, we have \(m(G)=m(Gu)\). Furthermore,
$$m(Gu)=\sum_{C_{i} \text{ in } Gu} m(C_{i})+m \biggl((G u) \bigcup_{C_{i} \text{ in } Gu}C_{i} \biggr). $$
By Lemma 2.12 we have \(m(T_{Gu})=m(T_{Gu}W_{Gu})\), that is, \(Gu\) satisfies (c) of Theorem 1.4. Moreover, \(Gu\) also satisfies both (a) and (b) of Theorem 1.4 since \(Gu\) is a subgraph of G. So \(G  u\) satisfies the minimal nullity condition. Then we have
$$\begin{aligned} \eta(G)&=n(G)2m(G)c(G) \\ &= \bigl[n(Gu)+1 \bigr]2m(Gu) \bigl[c(Gu)+1 \bigr] \quad \bigl( \because m(Gu)=m(G) \bigr) \\ &=n(Gu)2m(Gu)c(Gu) \\ &=\eta(Gu) , \end{aligned}$$
and the proof is completed. □