Partially ordering the class of invertible trees
Abstract
A tree is invertible if and only if has a perfect matching. In [4], Godsil considers an invertible tree and finds that the matrix has entries in and is the signed adjacency matrix of a graph which contains . In this paper, we give a new proof of this theorem, which gives rise to a partial ordering relation on the class of all invertible trees on vertices. In particular, we show that given an invertible tree whose inverse graph has strictly more edges, we can remove an edge from and add another edge to obtain an invertible tree whose median eigenvalue is strictly greater. This extends naturally to a partial ordering. We find the maximal and minimal elements of this poset and explore the implications about the median eigenvalues of invertible trees.
1 Introduction
A tree is a connected graph with no cycles. It is clear from the expansion of the determinant that the adjacency matrix of a tree is invertible if and only if has a perfect matching. Such a tree is said to be invertible. In [4], Godsil considers an invertible tree and finds that the matrix has entries in and is the signed adjacency matrix of a graph which contains . We will refer the underlying graph of as the inverse graph of an invertible tree.
For a graph , let denote the th largest eigenvalue of . For a graph on vertices, the median eigenvalues of are and . Median eigenvalue of bipartite graphs have been studied in [9, 8]. Median eigenvalues have applications in mathematical chemistry as they are related to the HOMOLUMO separation, see for example [3].
Since the eigenvalues of a bipartite graph are symmetric about , the median eigenvalues of a bipartite graph on vertices are such that . We will consider only the median eigenvalue for bipartite graph on an even number of vertices, since the median eigenvalue of any bipartite graph on an odd number of vertices is and it will suffice to consider only the positive median eigenvalue. For an invertible tree, we see that is an eigenvalue of if and only if is an eigenvalue of . In [4], Godsil characterises the invertible trees on vertices which attain the smallest ; he proved for any tree on vertices
and that equality is attained if and only if .
Subsequently classifications of trees attaining the maximum median eigenvalue has been found in [1], which also characterizes the trees which are isomorphic to their inverse graphs. A generalization for bipartite graphs is studied in [11]. A characterization of all bipartite graphs with a unique perfect matching whose adjacency matrices have inverses diagonally similar to nonnegative matrices is given in [13]. Graph inverses are an interesting topic and have also studied in [12, 7, 10, 14].
In this paper, we give a new proof of Godsil theorem on the inverse of trees, using newer techniques. In the process, we define a partial ordering relation on the class of all invertible trees on vertices. In particular, given an invertible tree whose inverse graph has strictly more edges than , we can obtain a nonisomorphic invertible tree such that . This extends naturally to a partial ordering. We show that the maximal elements of this poset are exactly the trees obtained as the rooted product of a tree on vertices with and the minimal elements of this poset are the path graphs.
2 Preliminaries
In this section, we will state results in the literature which will allow us to consider the inverse of a tree. We will denote the adjacency matrix of a graph by , or when the context is clear. If is a graph such that is invertible, we say itself is invertible for brevity. We will refer to standard texts such as [5] and [2] for further background.
The following theorem can be derived from Lemma 2.1 in Chapter 4 of [5].
Theorem 2.1.
If and are distinct vertices in , then
(2.1) 
where denotes the set of all paths in and is the graph obtained from by deleting all vertices of .
For the diagonal entries we will use:
Theorem 2.2.
If is a vertex of , then
(2.2) 
Let denote the number of perfect matching of a tree . If is invertible, then and has a perfect matching and necessarily has an even number of vertices. Let be a tree on vertices. Since every tree has at most perfect matching, we obtain that
In a tree, there is a unique path between any two distinct vertices and ; we will denote this path by . If is a matching in , we will say a path is alternating path if has edges and contains edges of .
Lemma 2.3.
Let be an invertible tree and let be the unique perfect matching. If be distinct vertices of , then has a perfect matching if and only if is a alternating path.
Proof. If is an edge of , then has a perfect matching if and only if .
Suppose has a perfect matching . Consider the graph where denotes symmetric difference. In , the vertices of have degree . Every other vertex is incident to an edge of both and and so their degree in is either or . Thus, is a path in which alternates between edges of and edges not in and (since are covered by ) has odd length, say and has more edge in than not in .
For the converse, let be an alternating path of length in with end vertices and , which contains edges of . If we take the edges of , we obtain a matching of .
3 Inverses of trees
We will use the material in the previous section to study inverses of trees. We use different tools from those used in the original proof of Godsil’s theorem about the inverses of trees and will give an alternate proof for his theorem. This will also lay the groundwork for the partial order on the class of invertible trees in Section 4.
Theorem 3.1.
Let be an invertible tree and let be the unique perfect matching of . If , then
Proof. Since has an even number of vertices, for any vertex has no perfect matching. We obtain by substituting into (2.2) that for all .
Let be distinct vertices of . Evaluating (2.1) at gives
Since has a perfect matching, we have that . By Lemma 2.3, if and only if is a alternating path. If is a alternating path with vertices, then has a perfect matching and , whence the result follows.
Let be an invertible tree on vertices with and be the unique perfect matching of . Let denote the bipartition of . The above theorem implies that has entries in and can be considered as the adjacency matrix of a signed graph . Let denote the underlying graph of . Observe that if are in the same part of the bipartition, has even number of edges and so . Thus, is a bipartite graph and is also a bipartition of .
Let the edges of be . We may label the vertices of (and thus and as they are defined on the same vertex set) as
where is the end of in and is the end of in . Let be the permutation of which takes to . Observe that is an isomorphism of which fixes . We will retain these definition for the rest of this section.
Lemma 3.2.
For , is an edge in if and only if is a alternating path in .
Proof. This follows immediately from Theorem 3.1.
Lemma 3.3.
Let be the signed graph with as its underlying graph, where the edges of are the positive edges and the edges not in are negative edges. Then contains as a signed subgraph; more precisely, is an edge of , then is an edge of with the same sign.
Proof. Consider an edge of . We wish to show that is an edge of with the same sign as . If , then we get that and which has positive sign in . Now suppose . We see that . We have that
is clearly a alternating path and so .
Observe that is a spanning tree of . Each edge of is one an unique edgecut set of which contains no other edge of ; these cuts are called the fundamental cuts. We say that a fundamental cut of (or of ) is negative if the corresponding edge of is negative in .
Lemma 3.4.
Each edge in is in negative fundamental cuts, where is the distance from to in .
Proof. Let be the unique path from to in . Consider and in . Let be the unique path in from to . Note that the proof of Lemma 3.3 actually implies that the negative edges of are in onetoone correspondence with the alternating paths in of length . Since is an edge of not in , we have that is an alternating paths in of odd length, say . The path has the following sequence of vertices:
Observe that for are each alternating paths in of length and is an edge of for each . We have that for and for are the edges of a path from to in . Since such a path is unique, these are the edges of . Then has edges. The edges of of which are not in and exactly the negative edges of for which is in the fundamental cut.
For the following lemma, we note that we may consider any graph as a signed graph where the sign of every edge is positive.
Lemma 3.5.
As signed graphs, and are switching equivalent.
Proof. Let be obtained from by switching on every negative cut of . We claim that all edges of are positive, in which case would be equal to considered as a signed graph.
It is clear that, in , all edges of now have positive sign. Consider an edge of not in . The sign of is where is the distance from to in . By Lemma 3.4, every negative edge is in an odd number of negative fundmental cuts and every positive edge is in an even number of negative fundmental cuts. After performing the switching described, every edge not in will also have positive sign.
We have now reconstructed the theorem of Godsil. Note that in the original paper, the statement is imprecise about how is contained as a subgraph of its inverse graph.
Theorem 3.6.
[4] If is a tree with a perfect matching , then is a matrix, which can be considered as the adjacency matrix of a signed graph . Further, is switching equivalent to a signed graph with all positive edges, which contains as a spanning tree, where is obtained from by the isomorphism which swaps the ends of every edge of .
Proof. This follows from Lemma 3.5.
4 A relation on invertible trees
Let be a tree with a unique perfect matching . As in Section 3, we will consider the inverse graph of , the underlying graph of the signed graph given by . We will denote the inverse graph of as . Recall the mapping is the involution permutation of the vertices of which swaps the ends of every edge of . An edge of an invertible tree which is in the unique matching are said to be a matching edge, otherwise, it is a nonmatching edge.
We will now define an operation on the class of invertible trees which maps a tree whose inverse graph has strictly more edges than to a tree whose inverse graph has fewer edges that . Suppose is an invertible tree such that contains an edge and let be any nonmatching edge of the fundamental cycle of in . Let be the tree obtained from by adding and removing . It is clear that the edges of remain a perfect matching in . If for some choice of edges and , then we will say that is obtained from by treeexchange.
Figure 2 shows an example of treeexchange; is an invertible tree whose inverse graph contains such that is not an edge of . We have that where . Note that is not an edge of .
Lemma 4.1.
If contains an edge and is obtained from by treeexchange, then is a proper subgraph of .
Proof. First we will show that is a subgraph of . We will then show that is not an edge of . Observe that is also the unique perfect matching of .
Let . Since is an edge of , the path in is an alternating path. Since is not an edge of , has order at least . Observe that the matching edge at is and the matching edge at is . Thus the first edge of is and the last edge of is . We have that , the fundamental cycle of in , is the cycle consisting of with the first and last edge (that is and ) deleted, together with . Observe that is not a alternating cycle, since both the edges preceding and following are also nonmatching edges.
Consider an edge of . We many assume is distinct from , since we know is an edge of both graphs. The path from to in is a alternating path. If is not in , then is also a alternating path in and so is an edge of . Suppose is an edge in . Since is not a matching edge, the edges immediately preceding and following in are matching edges and must thus be and . Let be the path on from to which does not use . If we replace in with , we obtain a walk in from to . Observe that is alternating, since the first and last edges of are nonmatching edges. Since has no cycles, if is not a walk, must contain as a subsequence, for some vertices and . This is not possible since is alternating and thus there is a alternating path from to in and is an edge of .
Let . Now we will show that is not adjacent to in . Since is an edge of , we have that is and edge of and this would conclude the argument. Since is a nonmatching edge on , we have that and are also edges of . In particular, we may take the path from to on which uses but not . Observe that does not contain or and is hence not a alternating path in . Since it is the unique path from to , we obtain that is not adjacent to in .
We obtain the following corollary.
Corollary 4.2.
If is obtained from by treeexchange, then .
Proof. Since is a proper subgraph of , we obtain that is strictly less than which gives that
where is the number of vertices of and denotes the th eigenvalue of .
Let denote the class of invertible trees on vertices. We define a relation on as follows:

for all ; and

if is isomorphic to a tree obtained from by treeexchange,
and we extend by taking the transitive closure.
Theorem 4.3.
The relation is a partial order relation on .
Proof. Observe that if and is not isomorphic to , then by iterated applications of Corollary 4.2, we see that . This implies that is antisymmetric. Since is transitive by construction, it is a partial order on .
We may consider as a poset under the partial ordering relation . For , the only invertible trees on vertices are the path graph. Figure 3 shows the Hasse diagrams of and .
5 Maximal elements of ordered by
The maximal elements of ordered by are exactly the trees on vertices which are isomorphic to their inverse graphs, which we will classify here. Note that this classification can be obtained using the result of [6] on “symmetric” characteristic polynomials of trees, but we will give direct proof here.
Let be a graph with vertices and let be a disjoint union of rooted graphs . The rooted product of and , denoted , is the graph obtain by identifying with the root vertex of . Figure 3
Lemma 5.1.
If a tree on vertices is isomorphic to its inverse graph, then is the rooted product of a tree on vertices and copies of rooted .
Proof. We will show the statement by proving that every matching edge of has a vertex of degree . Then, if all the vertices of degree are deleted, we obtain a tree on vertices, which shows that is the rooted product as described, where the matched edges form the copies of , rooted at the vertex which does not have degree (which must exist, since is connected).
Let . We know that is the adjacency matrix of the signing of such that its unique matching, , is exactly the set of positive edges. Let be the permutation matrix of ; observe that is also the adjacency matrix of the subgraph of with edge set . We have that
Thus gives
(5.1) 
Let denote the set of vertices adjacent to vertex . Let be the set of vertices matched to neighbours of ; that is to say,
Note that for any vertex . Noting that any vertex in is matched to exactly one neighbour of , we see that
We obtain that
where is the Kronecker delta. Thus
From (5.1) we obtain that
(5.2) 
Suppose there is a matched edge which is not incident to a vertex of degree . Thus has a neighbour which is distinct from and has a neighbour which is distinct from . Since there are no cycles in , we have that and is not adjacent to . We see from (5.2) that
But this is a contradiction, since . Thus, every matched edge is incident to a vertex of degree , which completes the proof.
We see that attaining the maximum must be a maximal element of ordered by . Let be the rooted product of the path with copies of , each rooted at vertex . This graph is sometimes called an elongated caterpillar. We can rederive the following result of [find citation] by showing that the elongated caterpillar attains the maximum amongst all rooted product of trees on with copies of .
Lemma 5.2.
If is a tree on vertices such that
then is isomorphic to .
Proof. By Lemma 4.1 and 5.1, we obtain that is a rooted product of a tree with . Theorem 2.1 of [6], we obtain that
The roots of are such that , where is an eigenvalue of . For each eigenvalue of , we obtain two eigenvalues from the two roots of , which are
The largest eigenvalue is where . Since is isomorphic to its inverse graph, we bet that . Thus, is maximized amongst all graph which are isomorphic to its inverse graph when is minimized, for all trees on vertices. The tree with the smallest is the path on vertices and so must be isomorphic to .
The eigenvalues of the paths are for , and so the eigenvalues of are for , where
Observe that and so .
6 Minimal elements of ordered by
Godsil proved that the path attains the minimum median eigenvalue in . Thus must be a minimal element of under . We will show that the paths are the only minimal elements of under .
Lemma 6.1.
If is a minimal element of under , then has no vertex with degree at least .
Proof. Suppose for a contradiction that is a minimal element of under and is a vertex of with degree at least . Let be the perfect matching of and be the mapping of the vertices of which switching the ends of the edges of . Let be the neighbour of such that . Let be two other neighbours of and let and be such that . Let . See Figure 4 for the subgraphs of and induced by . We will show that is obtained from by treeexchange.
Observe that the path from to in is a alternating path, and so is an edge of by Lemma 3.2. Thus , which is a contradiction.
7 Open Problems
Treeexchange allows us remove and add an edge from an invertible tree, such that the median eigenvalue strictly increases. One can ask if such an operation is possible for an bipartite graph with a unique matching.
Considering the class of invertible trees as a poset, partially ordered by “” as defined in Section 4 gives rise to many questions. In particular, one may ask for a description of the covering relation of this poset and also which values the Moebius function of this poset takes.
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