The non-negative spectrum of a digraph

Proposition 1

Let D be a digraph with common out-neighbor partition {Bi}i=1k . If we renumber the vertices of D such that we enumerate the vertices of 𝓑₁ first, those of 𝓑₂ next, and so on, then N_out assumes block diagonal form N_out(D) = diag(B₁, …, B_k), where B_i denotes the block associated with the class 𝓑_i.

Example 1

Figure 1 shows an example digraph. The gray vertices form the class 𝓑(0) = {0, 1, 2, 3, 4}, by virtue of the zig-zag trails

Figure 1

Common out-neighbor partition of a digraph.

1. 4 → 7 ← 0 → 1 ← 2 → 0 ← 1,

2. 0 → 2 ← 3.

Note that non-bold numbers are merely “helpers” (common out-neighbors) that establish the zig-zag relation. The black vertices form a class 𝓑(5) = {5, 6}. The white vertices form a singleton class each.

The vertex numbers have already been chosen such that they match the common out-neighbor partition. Hence, with respect to this numbering, the matrix N_out assumes block diagonal form with blocks of sizes 5, 2, 1, 1, 1, 1. This is shown in Figure 2.

Figure 2

Block diagonal form according to common out-neighbor partition.

Theorem 1

Let N_out(D) = diag(B₁, …, B_r), according to the common out-neighbor partition {Bi}i=1r of a given digraph D. For any eigenvector x of B_i (for eigenvalue λ), x ⊗ e_i is an N-eigenvector of D (for N-eigenvalue λ).

Proof

□

Theorem 2

Let N_out(D) = diag(B₁, …, B_r), according to the common out-neighbor partition {Bi}i=1r of a given digraph D. If x is an N-eigenvector of D (for N-eigenvalue λ) that is non-zero on the vertex v ∈ 𝓑_i, then x|_𝓑i is an eigenvector of B_i (for eigenvalue λ). Here, x|_𝓑i ∈ ℝ^|𝓑_i| denotes the restriction of x to the vertices of class 𝓑_i.

Proof

We have

and

Since N_out(D)x = λx by assumption, it follows that B_ix|_𝓑i = λx|_𝓑i for all i = 1, …, r. If x is non-zero on the vertex v ∈ 𝓑_i, then x|_𝓑i ≠ 0, so it is an eigenvector of B_i for eigenvalue λ.□

Corollary 1

If x is a nowhere-zero N-eigenvector of D for N-eigenvalue λ, then λ is a common eigenvalue of all blocks B_i, i = 1, …, r, of N_out.

Corollary 2

Given a digraph D with n vertices, a unit vector e_i ∈ ℝⁿ is an N-eigenvector of D if and only if the unique vertex v on which e_i is non-zero does not have any common out-neighbors with other vertices. The corresponding N-eigenvalue equals the out-degree of v.

Theorem 3

Let D₁, D₂ be two disjoint digraphs and S₁, S₂ two sets of sinks of D₁ and D₂, respectively. Further, let D′ be the digraph obtained by connecting all the vertices of S₁ ∪ S₂ to a new sink η. Then,

where s = | S₁ ∪ S₂|.

Proof

First of all, observe that the vertices of S₁ and S₂ each form singleton blocks in D₁ and D₂, respectively. By connecting these vertices to the new sink η they will be united to a block S₁ ∪ S₂ of D′ (with η being the unique common out-neighbor of any two vertices in this block). Apart from that, all other classes of D₁ and D₂ are also classes of the common out-neighbor partition of D′ (with exactly the same blocks as before). For j ∈ {1, 2}, let Nout(Dj)=diag⁡(B1(j),…,Brj(j)), according to the common out-neighbor partition {Bi(j)}i=1rj of D_j. Let s_j = | S_j|. Then Nout(Dj)=diag⁡(B1(j),…,Brj−sj(j),0sj×sj). Hence we may number the vertices of D′ such that

Observing χ(1_s×s, x) = x^s–1(x – s) and keeping in mind that the loss of s sinks effectively contributes a corrective factor x^–s to the N-characteristic polynomial, the result now follows easily by comparing the three block diagonal forms.□

Theorem 4

Let D₁ and D₂ be two disjoint digraphs. Choose two arbitrary vertices u of D₁ and v of D₂. Join D₁ and D₂ by connecting u, v to a new sink η and let D′ be the resulting digraph. Then,

where B^(u) and B^(v) are the blocks associated with the classes of u, v in D₁, D₂ (respectively) and where i_u, i_v are the respective row/column indices of u and v within these blocks.

Proof

The key observation is that connecting u and v to η will unite the classes of u and v. Further, η as a new sink will form a singleton cell (with an associated zero block). But apart from these two effects the common out-neighbor partition of D′ will be exactly the union of the partitions of D₁ and D₂, with the same associated blocks for the cells. What remains is to determine the block B of u, v in N_out(D′). Suppose that the vertices of D₁ and D₂ are ordered such that their N_out matrices assume block diagonal form according to the respective common out-neighbor partition. Without loss, we may assume that B^(u) is the lower-right block in N_out(D₁) and that B^(v) is the top-left block in N_out(D₂). We index the vertices of D′ such that first we enumerate the vertices of D₁, then those of D₂ (both in the same order as before). Then B is essentially diag(B^(u), B^(v)), but we have to increment the main diagonal for u and v to reflect that both of them now have an additional out-neighbor, further we have to symmetrically place two off-diagonal ones to reflect that both u, v now have a common out-neighbor.□

Corollary 3

Let D₁ and D₂ be two disjoint digraphs. Choose an arbitrary vertex v of D₁ and a sink u of D₂. Join D₁ and D₂ by connecting u, v to a new sink η and let D′ be the resulting digraph. Then,

where B is the block associated with the class of v in D₁ and i is the row (resp. column) index of v within this block.

Theorem 5

(Square Theorem). Let D be a bipartite digraph such that each vertex is either a source or a sink. Let k be the number of sources and l be the number of sinks in D. Further, let G be the underlying undirected graph of D.

1. Given an eigenspace basis for eigenvalue λ ≠ 0 of G, the source parts of these vectors form an N-eigenspace basis for N-eigenvalue λ² of D and their sink parts are all N-eigenvectors for N-eigenvalue 0 of D.

2. Every eigenvector for eigenvalue 0 of G is also an N-eigenvector for N-eigenvalue 0 of D.

3. If the source part of any N-eigenvector for N-eigenvalue 0 of D is not null, then this source part is an eigenvector for eigenvalue 0 of G.

4. Given a basis of ℝ^k+l of eigenvectors of G, an N-eigenspace basis for N-eigenvalue 0 of D can be constructed as follows. Collect the sink parts of all the vectors associated with positive eigenvalues of G, together with the vectors associated with eigenvalue 0. Alternatively, collect the source parts of all basis vectors associated with eigenvalue 0 and determine a maximal linearly independent subset of the resulting set, together with l unit vectors, one for each sink (such that it is non-zero exactly on the considered sink).

5. If η is the nullity of G and ν the N-nullity of D, then 1 ≤ ν – η ≤ min(k, l) and ν ≥ max(k, l).

Proof

We assume that the vertices of D are ordered such that the sources are numbered before the sinks (G shall inherit this vertex order). Since G is bipartite we have

for some matrix B ∈ ℝ^k×l. Hence

where we regard G as a (fully bidirected) digraph.

In order to prove (i), suppose that A(G)(x, y)^T = λ(x, y)^T with x ∈ ℝ^k, y ∈ ℝ^l. Since

we get

so an eigenvector (x, y)^T of G for eigenvalue λ is an N-eigenvector of D for N-eigenvalue λ² ≠ 0 if and only if x ≠ 0 and y = 0, and an N-eigenvector of D for N-eigenvalue 0 if and only if either λ = 0 or both x = 0 and y ≠ 0:

Next, suppose that A(G)(x, y)^T = (0, 0)^T. Then B^Tx = 0 in (1), so that

which shows (ii).

For proving (iii) suppose that N_out(D)(x, y)^T = (0, 0)^T. With respect to the block diagonal form of N_out(D) we immediately deduce BB^Tx = 0. Using Theorem 3.9-4 (f) from [5] it follows that Bx = 0. Therefore,

Now we turn to claims (iv) and (v). Assume that we have determined a basis of ℝ^k+l of eigenvectors of G. With respect to linear independence, note that the spectrum of a bipartite graph is symmetric around zero and that for each eigenvector (x, y)^T for eigenvalue λ of G we have a twin eigenvector (x, –y)^T for –λ (cf. [6]). We modify the given basis as follows. For λ > 0 let E_λ and E_–λ be the two eigenspaces for eigenvalues λ and –λ of G, respectively. Select those vectors (x⁽¹⁾, y⁽¹⁾)^T, …, (x^(r), y^(r))^T from the overall basis that form a basis of E_λ. By suitable linear combination we find that their source and sink parts

form a basis of the space E_λ + E_–λ. In the overall basis we replace the eigenvectors for eigenvalues λ and –λ with these vectors. If we do this for all positive eigenvalues of G, then we still have basis of ℝ^k+l. Consequently, the source parts inserted for any eigenvalue λ of G constitute an N-eigenspace basis for N-eigenvalue λ² of D.

Note that x⁽ⁱ⁾ ∈ ℝ^k, so the final basis may contain at most k source parts. Likewise, it may contain at most l sink parts. Since the number of introduced source and sink parts is the same, we deduce that the number of positive eigenvalues of G is at most min(k, l). Further, the N-nullity of D exceeds the nullity of G by exactly the number of positive eigenvalues of G. Moreover, G contains not only isolated vertices (actually none at all, because D has only sources and sinks), so there exists at least one positive eigenvalue for G (cf. Corollary 2.7 in [6]). This proves the first part of claim (v).

Next, observe that we may construct a linearly independent set of l sink parts that are N-eigenvectors for N-eigenvalue 0 of D, by simply taking l sink unit vectors (i.e. for each sink choose the unit vector that is non-zero on exactly that sink). Hence the N-nullity of D is at least l. Moreover, we may reverse the orientation of D and apply the same argument again, with the sinks turned into sources and vice versa. Equivalently, we may consider N_out instead of N_in. Since these matrices have the same spectra it follows that the N-nullity of D is at least max(k, l). Now the proof of claim (v) is complete.

Using suitable linear combinations of the sink unit vectors on the vectors of the original eigenspace basis for eigenvalue 0 of G, we may convert them into source parts. This may cause linear dependence among the newly created source parts, so we reduce them to a maximal linearly independent subset. This achieves the basis proposed in the second part of claim (iv).□

Example 2

In order to demonstrate some aspects of Theorem 5 we consider the bipartite digraph depicted in Figure 4. This digraph D has N-spectrum

Figure 4

Example bipartite digraph having only sources and sinks.

Its undirected counterpart G has the traditional spectrum

To illustrate part (i) of the theorem we determine an eigenvector for simple eigenvalue 2.48 of G, see Figure 5. Now we form the source and sinks parts – as shown in Figure 6 – and readily verify that the source part is an N-eigenvector of D for N-eigenvalue 6.14 = (2.48)², whereas the sink part is an N-eigenvector for N-eigenvalue 0. Note that for the simple eigenvalue –2.48 of G we can get an eigenvector by taking the vector from Figure 5 and simply inverting the signs on all the sink vertices. Naturally, the source part remains the same, so we see that the N-eigenvalue 6.14 of D must be simple.

Figure 5

Eigenvector for simple eigenvalue 2.48.

Figure 6

Source and sink parts forming eigenvectors N-eigenvalues 6.14 resp. 0.00.

With respect to part (iv) of the theorem observe that, since G is missing eigenvalue 0, the easiest way of finding an N-eigenspace basis for N-eigenvalue 0 is given by forming a unit vector basis with respect to the seven sinks of D.

Remark 1

From the proof of part (i) of the Square Theorem we also conclude that the nullity of any bipartite graph G with bipartition set sizes k, l is at least k + l – 2 min(k, l) = |k – l|. This is the “Corollary” to Theorem 3 in [7].

Corollary 4

Let P_n be a directed path with n vertices that has zig-zag orientation. Then

Proof

According to [8], the eigenvalues of an undirected path with n vertices are the numbers

Clearly, these numbers are all distinct. For j = 1, …, ⌊n2⌋ we get the positive eigenvalues. So their squares will occur in the N-spectrum of P. Moreover, it contains ⌊n2⌋ additional zero N-eigenvalues. For odd n the underlying undirected path already has a (single) eigenvalue zero, so altogether we have ⌈n2⌉ zero N-eigenvalues.□

Corollary 5

Let C_2n be a cycle with 2n vertices that has zig-zag orientation. Then

Proof

According to [8], the eigenvalues of an undirected cycle with 2n vertices are the numbers

None of these numbers equals zero. For j = 1, …, n we get one item of each pair of eigenvalues λ, –λ. Hence the result follows.□

Corollary 6

Let T be an integral tree. Obtain T′ by zig-zag orienting T. Then T′ is N-integral.

Corollary 7

Given a zig-zag oriented bipartite digraph D, if the underlying undirected graph G has an eigenspace basis for eigenvalue 0 whose vectors assume only values from {0, 1, –1} on the sources, then D has a simply structured N-eigenspace basis for N-eigenvalue 0.

Proof

This follows directly from the second part of claim (iv).□

Corollary 8

Let G be a forest or a unicyclic graph whose cycle length is divisible by 4. Obtain D by zig-zag orienting G. Then D has a simply structured N-eigenspace basis for eigenvalue 0.

Proof

Proposition 1 of [18] states that all forests (or, rather, their adjacency matrices) are totally unimodular. Moreover unicyclic graphs are totally unimodular if and only if their cycle length is divisible by 4.□

Corollary 9

Let T be a tree. Obtain D by zig-zag orienting T. Depending on which of the two possible zig-zag orientations was chosen, either every simply structured eigenspace basis for eigenvalue 0 of T is also a simply structured N-eigenspace basis for N-eigenvalue 0 of D or we can take a sink unit vector basis instead.

Proof

It is a consequence of Lemma 19 in [19], that every null space basis of a tree completely vanishes on exactly the same of the two sets of the vertex bipartition of the tree. Depending on the chosen zig-zag orientation, we see that either any simply structured eigenspace basis for eigenvalue 0 of T will also be a simply structured N-eigenspace basis for N-eigenvalue 0 of D or that a sink unit vector basis will serve the purpose.□

Corollary 10

Let G be a unicyclic graph with even cycle length. Obtain D by zig-zag orienting G. If the cycle length of G is divisible by 4 or if there exists not exactly one vertex v on the cycle in G such that v is not covered by all maximum matchings of the unique tree emanating from v, then D has a simply structured N-eigenspace basis for eigenvalue 0.

Proof

Theorem 4.51 in [16] states that the above condition on G exactly characterizes those unicyclic graphs which have a simply structured null space basis.□

Corollary 11

Let G be a unicyclic graph with even cycle length. Then at least one of the zig-zag orientations of G affords a simply structured N-eigenspace basis for N-eigenvalue 0.

Theorem 6

Let D′ be a digraph obtained by performing block separation on a given digraph D of order n. Then

1. The common out-neighbor partition is obtained by extending the partition of D with singleton blocks, one for each newly introduced vertex.

2. Number the vertices of D according to its common out-neighbor partition such that N_out(D) = diag(B₁, …, B_k). Keeping the original vertex order of D for D′ and numbering the newly introduced vertices after the original vertices, we have

   Nout(D)=diag⁡(B1,…,Bk,Bk+1,…,B2n)

   with B_k+1 = … = B_2n = 0_1×1.

3. N(D′, x) = xⁿN(D, x).

Proof

With respect to the matrix N_out(D′) of the resulting graph D′ we find that each of the original vertices v of D has the same number of out-neighbors as before. The newly introduced vertices are all sinks, by construction. Moreover, the number of common out-neighbors of v and some other vertex w is the same as in D if w is one of the original vertices of D and zero otherwise. Hence, using the proposed vertex numbering, the matrix N_out(D) is a principal submatrix of N_out(D′). Clearly, the rest of N_out(D′) contains only zero entries.□

Example 3

Let us revisit Example 1. The result of block separation performed on the digraph presented there is shown in Figure 7. The vertices are labeled so that it is easy to see the pairs of original and new vertices. Note that the results of Theorem 6 remain valid (just changing the counts related to the newly introduced new singleton cells) if we do not duplicate vertices of D that do not have any incoming neighbors. This helps us prevent unnecessary bloat. Even more, we may refrain from duplicating any vertices belonging to singleton cells since this will only lead to zero blocks in N_out.

Figure 7

Result of block separation for the digraph of Figure 1.

Remark 2

By construction, every helper vertex in a component of a block separated digraph is a sink and every vertex of some original cell is a source. Hence the overall block separated digraph is a zig-zag oriented bipartite graph with exactly the same blocks as before, plus some zero blocks. So we may apply the Square Theorem 5 on each component separately to determine its N-eigenvectors and N-eigenspaces. The results can be trivially projected to the original digraph. Hence, the conjunction of block separation and the Square Theorem permits us to fully predict the N-spectral properties of a digraph from the spectral properties of certain associated bipartite graphs.

Example 4

It is easily checked that the example digraph depicted in Figure 4 is isomorphic the largest component of the block separated digraph shown in Figure 7. With respect to Remark 2 we see that Example 2 also demonstrates the combination of block separation and the Square Theorem.

Corollary 12

Among all orientations of a given path (or cycle), the maximum N-spectral radius is achieved by exactly the zig-zag orientations (or the nearly zig-zag orientation if the given graph lacks a zig-zag orientation).

Proof

With respect to block separation, note that the N-spectral radius of a given digraph is determined by the maximal spectral radius among the underlying bipartite graphs of the components of the separation digraph. Orienting a graph does not introduce mutual adjacency in the resulting digraph. Therefore, block separation essentially decomposes the given digraph into directed paths (ignoring any isolated vertices). Next we consider the maximum block separation component size for (nearly) zig-zag orientations of paths and cycles. A zig-zag orientation of P_n or a nearly zig-zag orientation C_2n+1 will introduce a directed path of the same order in the block separation digraph, plus some isolated vertices. A zig-zag orientation of C_2n will introduce a directed cycle of the same order, plus some isolated vertices. Obviously, any other orientation of a given path or cycle will result in further decomposition of the maximal components of the block separation digraph. But careful analysis of the eigenvalue formula given in the proof of Corollary 4 (resp. Corollary 5) reveals the well-known fact that σ(P_n–1) < σ(P_n) < σ(C_n) < σ(C_n+1) (for n ≥ 1, setting σ(P₀) := 0). Hence the proof is complete.□

Theorem 7

Let D be a digraph and let M be a loopless multi-graph, both of order n. Define R(D) as the multi-relation of distinct vertices of D having common out-neighbors, i.e. the multiplicity of each pair (v, w) ∈ R(D) equals the number of common out-neighbors of the vertices v ≠ w in D. Then N_out(D) = Q(M) if and only if the following conditions are satisfied:

1. M represents the multi-relation R(D).

2. For every vertex v of D, the out-degree of v equals the number of instances in which there exist vertices w, z ∈ V(D) such that w ≠ v and z is a common out-neighbor of v and w.

Proof

By the definition of the matrix N_out(D), each of its off-diagonal entries specifies the number of common out-neighbors of the vertices associated with the row/column indices of the respective considered entry. So the off-diagonal part of N_out(D) exactly represents R(D). Note here that, by construction, R(D) contains no pairs (v, v).

It follows that the off-diagonal entries of N_out(D) and Q(M) = A(M) + D(M) coincide if and only if A(M) is the adjacency matrix of (the multi-graph associated with) the relation R(D) – which is equivalent to condition (i).

The diagonal entries of N_out(D) are the out-degrees of the respective vertices of D, whereas the diagonal entries of Q(M) are the degrees of the vertices of M. But with respect to R(D) the degree of a vertex v of M in turn equals the number of instances in which there exist vertices w, z ∈ V(D) such that w ≠ v and z is a common out-neighbor of v and w. It follows, under condition (i), that the diagonal entries of N_out(D) and Q(M) coincide if and only if condition (ii) holds.□

Remark 3

Finding pairs (D, M) of digraphs D and loopless multi-graphs M such that N_out(D) = Q(M) is actually very easy. Start with an arbitrary digraph. In view of condition (ii) of Theorem 7, identify all vertices for which the associated diagonal entry of N_out(D) is “too large”, i.e. strictly greater (by a difference of, say, d), than the sum of the other entries in the same row/column. For each such vertex v execute the following step exactly d times: Add an incoming sink w to any of its out-neighbors. This step will create a new instance of common out-neighborship for v, hence extending the aforementioned row/column of N_out by a new entry 1. Moreover, by construction, in the resulting digraph the diagonal entry of N_out associated with w is less than or equal to the sum of the other entries in the same row/column. After carrying out the mentioned process for all identified vertices none of the associated diagonal entries of N_out is too large any more. Further, for every vertex whose associated diagonal entry is too small we may simply add a suitable number of pendant sinks. All in all, we achieve that condition (ii) of Theorem 7 is satisfied. Hence M is now easily derived by means of condition (i).

Proposition 2

Given a digraph D and its block separation digraph D′, create a new undirected multi-graph M as follows:

1. Let M initially have the vertices of D but no edges.

2. Let every duplicated vertex in D′ represent a clique formed by the original neighbors of that vertex. For each duplicated vertex from D′ augment M by introducing new edges for the associated clique, skipping any 1-cliques.

Then M represents R(D).

Proof

The construction yields the desired result since the duplicated vertices in the block separation of D are exactly the sinks of the block separation, which in turn are exactly those vertices of D which act as common out-neighbors. Hence the sinks of the block separation introduce cliques in the relation R(D).□

Example 5

Using the block separation digraph in Figure 7, the graph M constructed in Proposition 2 has vertices 0, …, 10. Vertex 0′ introduces a 2-clique among the vertices {1, 2}. Likewise, vertices 1′, 7′ and 9′ introduce 2-cliques among {0, 2}, {0, 4} and {5, 6}, respectively. The vertex 2′ gives rise to a 3-clique among {0, 3, 4}. Note that, altogether, we get a double edge between vertices 0 and 4.