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Generalized spectral characterization of signed trees

Yizhe Jia  Wei Wanga  Hao Zhangb
aSchool of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an 710049, P. R. China
bSchool of Mathematics, Hunan University, Changsha 410082, P. R. China
Corresponding author: [email protected]
Abstract

Let TT be a tree with an irreducible characteristic polynomial ϕ(x)\phi(x) over \mathbb{Q}. Let Δ(T)\Delta(T) be the discriminant of ϕ(x)\phi(x). It is proved that if 2n2Δ(T)2^{-\frac{n}{2}}\sqrt{\Delta(T)} (which is always an integer) is odd and square free, then every signed tree with underlying graph TT is determined by its generalized spectrum.

Keywords: Graph spectra; Cospectral graphs; Determined by spectrum; Rational orthogonal matrix; Signed graph
Mathematics Subject Classification: 05C50

1 Introduction

It is well known that the spectra of graphs encode a lot of combinatorial information about the given graphs. A major unsolved question in spectral graph theory is: “What kinds of graphs are determined (up to isomorphism) by their spectrum (DS for short)?”. The problem originates from chemistry and was raised in 1956 by Günthard and Primas [1], which relates Hückle’s theory in chemistry to graph spectra. The above problem is also closely related to a famous problem of Kac [12]: “Can one hear the shape of a drum?” Fisher [11] modelled the drum by a graph, and the frequency of the sound was characterized by the eigenvalues of the graph. Hence, the two problems are essentially the same.

It was commonly believed that every graph is DS until the first counterexample (a pair of cospectral but non-isomorphic trees) was found by Collatz and Sinogowitz [2] in 1957. Another famous result on cospectral graphs was given by Schwenk [16], which states that almost every tree is not DS. For more constructions of cospectral graphs, see, e.g., [10, 13, 17]. However, it turns out that showing a given graph to be DS is generally very hard and challenging. Up to now, only a few graphs with very special structures are known to be DS. We refer the reader to [7, 8] for more background and known results.

In recent years, Wang and Xu [19] and Wang [20, 21] considered a variant of the above problem. For a simple graph GG, they defined the generalized spectrum of GG as the spectrum of GG together with that of its complement G¯\bar{G}. A graph GG is said to be determined by its generalized spectrum (DGS for short), if any graph having the same generalized spectrum as GG is necessarily isomorphic to GG.

Let GG be a graph on nn vertices with adjacency matrix A=A(G)A=A(G). The walk-matrix of GG is defined as

W(G)=[e,Ae,,An1e],W(G)=[e,Ae,\ldots,A^{n-1}e],

where ee is the all-one vector. Wang [20, 21] proved the following theorem.

Theorem 1.1 ([20, 21]).

If 2n2det(W)2^{-\lfloor\frac{n}{2}\rfloor}\det(W) is odd and square-free, then GG is DGS.

The problem of spectral determination of ordinary graphs naturally extends to signed graphs. This paper is a continuation along this line of research for signed graphs in the flavour of Theorem 1.1.

Let Δ(T)\Delta(T) be the discriminant of a tree TT (see Section 4 for the definition). The main result of the paper is the following theorem.

Theorem 1.2.

Let TT be a tree on nn vertices with an irreducible characteristic polynomial over \mathbb{Q}. If 2n2Δ(T)2^{-\frac{n}{2}}\sqrt{\Delta(T)} (which is always an integer) is odd and square free, then every signed tree with underlying graph TT is DGS.

As an immediately consequence of Theorem 1.2, we have

Corollary 1.3.

Let TT and TT^{\prime} be two cospectral and non-isomorphic trees with a common irreducible characteristic polynomial. Suppose 2n2Δ(T)2^{-\frac{n}{2}}\sqrt{\Delta(T)} is odd and square free. Then no two signed trees with underlying graphs TT and TT^{\prime} respectively are generalized cospectral.

Example 1.

Let TT and TT^{\prime} be two cospectral non-isomorphic trees on 14 vertices (see Fig. 1) with a common irreducible characteristic polynomial

ϕ(T)=ϕ(T)=1+16x279x4+157x6143x8+63x1013x12+x14.\phi(T)=\phi(T^{\prime})=-1+16x^{2}-79x^{4}+157x^{6}-143x^{8}+63x^{10}-13x^{12}+x^{14}.

It can be easily computed that 27Δ(T)=27Δ(T)=5×11×47545992^{-7}\sqrt{\Delta(T)}=2^{-7}\sqrt{\Delta(T^{\prime})}=5\times 11\times 4754599, which is odd and square-free. Thus, according to Theorem 1.2, every signed tree with underlying graph TT (resp. TT^{\prime}) are DGS. In particular, no two signed trees with underlying graphs TT and TT^{\prime} respectively are generalized cospectral.

Refer to caption
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Figure 1: A pair of cospectral non-isomorphic trees on 14 vertices

Theorem 1.2 shows that whenever the underlying tree TT with nn vertices satisfies a simple arithmetic condition, then all the 2n12^{n-1} signed trees (including TT itself) whose underlying is TT is DGS. That is, the DGS property of all these signed trees only depends on the underlying graph TT. This is somewhat unexpected, since given a pair of trees TT and TT^{\prime}, it seems time consuming even to check whether there exist two signed trees with underlying graphs TT and TT^{\prime} respectively that are generalized cospectral; see Example 1.

We mention that Theorem 1.2 is the best possible in the sense that it is no longer true if 2n2Δ(T)2^{-\frac{n}{2}}\sqrt{\Delta(T)} has a multiple odd prime factor. Moreover, the irreducibility assumption of the characteristic polynomial of the tree is essential which cannot be removed; see Remarks 1 and 2 in Section 4.

The rest of the paper is organized as follows. In Section 2, we give some preliminary results that will be needed in the proof of Theorem 1.2. In Section 3, we give a structure theorem, which plays a key role in the paper. In Section 4, we present the proof of Theorem 1.2. Conclusions and future work are given in Section 5.

2 Preliminaries

For the convenience of the reader, we give some preliminary results that will be needed later in the paper. For more results in spectral graphs theory, we refer to [4, 6]

Let G=(V,E)G=(V,E) be a simple graph. A signed graph is a graph obtained from GG by assigning a sign 11 or 1-1 to every edge according to a mapping σ:E{1,1}\sigma:E\rightarrow\{1,-1\}. We use Γ=(G,σ)\Gamma=(G,\sigma) to denote a signed graph with underlying graph GG and sign function (signature) σ\sigma. We call a signed graph a signed bipartite graph, if its underlying graph is bipartite.

Let UU be a subset of VV such that (U,VU)(U,V\setminus U) is a partition of VV. A switching w.r.t. UU (or VUV\setminus U) is an operation that changes all the signs of edges between UU and VUV\setminus U, while keeps the others unchanged. Two signed graphs Γ\Gamma and Γ\Gamma^{\prime} are switching-equivalent if Γ\Gamma^{\prime} can be obtained from Γ\Gamma by a switching operation, or equivalently, there exists a diagonal matrix DD with all diagonal entry ±1\pm 1 such that DA(Γ)D=A(Γ)DA(\Gamma)D=A(\Gamma^{\prime}). A signed graph is balanced if every cycle contains an even number of edges with sign -1. It is well-known that a signed graph is balanced if and only it is switching equivalent to a unsigned graph.

Let Γ\Gamma be a signed graph with adjacency matrix A(Γ)A(\Gamma). The characteristic polynomial of Γ\Gamma is defined as the characteristic polynomial of A(Γ)A(\Gamma), i.e., ϕ(Γ;x)=det(xIA(Γ))\phi(\Gamma;x)=\det(xI-A(\Gamma)), where II is the identity matrix. Two signed graphs Γ\Gamma and Γ\Gamma^{\prime} with adjacency matrices A(Γ)A(\Gamma) and A(Γ)A(\Gamma^{\prime}) respectively are called generalized cospectral if

det(xIA(Γ))=det(xIA(Γ))anddet(xI(JIA(Γ)))=det(xI(JIA(Γ))),\det(xI-A(\Gamma))=\det(xI-A(\Gamma^{\prime}))~{}{\rm and}~{}\det(xI-(J-I-A(\Gamma)))=\det(xI-(J-I-A(\Gamma^{\prime}))),

where JJ is the all-one matrix and JIA(Γ)J-I-A(\Gamma) formally denotes the complement of Γ\Gamma (it is indeed the complement of Γ\Gamma if every edge of Γ\Gamma has been assigned a positive sign +1). A signed graph Γ\Gamma is said to be determined by the generalized spectrum (DGS for short), if any signed graph that is generalized cospectral with Γ\Gamma is isomorphic to Γ\Gamma.

A polynomial f(x)[x]f(x)\in{\mathbb{Q}[x]} is irreducible if it cannot be factored into two polynomials with rational coefficients of lower degree. Let f(x)[x]f(x)\in{\mathbb{Q}[x]} be an irreducible polynomial with degree nn and α\alpha be one of its root. Then (α)={c0+c1α++cn1αn1:ci,0in1}\mathbb{Q}(\alpha)=\{c_{0}+c_{1}\alpha+\cdots+c_{n-1}\alpha^{n-1}:c_{i}\in{\mathbb{Q}},~{}0\leq i\leq n-1\} is a number field which is isomorphic to [x]/(f(x))\mathbb{Q}[x]/(f(x)) and is obtained by adding α\alpha to \mathbb{Q}; see e.g. [9].

An orthogonal matrix QQ is a square matrix such that QTQ=InQ^{\textup{T}}Q=I_{n}. It is called rational if every entry of QQ is a rational number, and regular if each row sum of QQ is 11, i.e., Qe=eQe=e, where ee is the all-one column vector. Denote by RO()n{}_{n}(\mathbb{Q}) the set of all nn by nn regular orthogonal matrices with rational entries.

In 2006, Wang and Xu [19] initiated the study of the generalized spectral characterization of graphs. For two generalized cospectral graphs GG and HH, they obtained the following result (see also [5]), which plays a fundamental role in their method.

Theorem 2.1 ([5],[19]).

Let GG be a graph. Then there exists a graph HH such that GG and HH are generalized cospectral if and only if there exists a regular orthogonal matrix QQ such that

QTA(G)Q=A(H).Q^{\textup{T}}A(G)Q=A(H). (1)

Moreover, if detW(G)0\det W(G)\neq 0, then QROn()Q\in\textup{RO}_{n}(\mathbb{Q}) is unique and Q=W(G)W1(H)Q=W(G)W^{-1}(H).

A graph GG with detW(G)0\det W(G)\neq 0 is called controllable (see [14]), denoted by 𝒢n\mathcal{G}_{n} the set of all controllable graphs on nn vertices. For a graph G𝒢nG\in\mathcal{G}_{n}, define

𝒬(G):={QROn():QTA(G)Q=A(H)forsomegraphH}.\mathcal{Q}(G):=\{Q\in\textup{RO}_{n}(\mathbb{Q}):~{}Q^{\textup{T}}A(G)Q=A(H)~{}{\rm for~{}some~{}graph}~{}H\}.

Then according to Theorem 2.1, it is easy to obtain the following

Theorem 2.2 ([19]).

Let GG be a controllable graph. Then GG is DGS if and only if the set 𝒬(G)\mathcal{Q}(G) contains only permutation matrices.

The above theorems extend naturally to signed graphs. By Theorem 2.2, finding out the possible structure of all Q𝒬(G)Q\in\mathcal{Q}(G) is a key to determine whether a (signed) graph GG is DGS.

Notations: We use ene_{n} (or ee if there is no confusion arises) to denote an nn-dimensional column all-one vector, and JJ the all-one matrix. For a vector α=(a1,a2,,an)Tn\alpha=(a_{1},a_{2},\ldots,a_{n})^{\textup{T}}\in{\mathbb{R}^{n}}, we use α2=(a12+a22++an2)1/2||\alpha||_{2}=(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2})^{1/2} to denote the Euclidean norm of α\alpha.

3 A structure theorem for QQ

The key observation of this paper is the following theorem which shows that for two generalized cospectral signed bipartite graphs with a common irreducible characteristic polynomial, the regular rational orthogonal matrix carried out the similarity of their adjacency matrices has a special structure.

Theorem 3.1.

Let Γ\Gamma and Γ~\tilde{\Gamma} be two generalized cospectral signed bipartite graphs with a common irreducible characteristic polynomial ϕ(x)\phi(x) over \mathbb{Q}. Suppose that the adjacency matrices of Γ\Gamma and Γ~\tilde{\Gamma} are given as follows, respectively:

A=A(Γ)=[OMMTO],A~=A(Γ~)=[OM~M~TO].A=A(\Gamma)=\left[\begin{matrix}O&M\\ M^{\rm T}&O\end{matrix}\right],~{}\tilde{A}=A(\tilde{\Gamma})=\left[\begin{matrix}O&\tilde{M}\\ \tilde{M}^{\rm T}&O\end{matrix}\right].

Then there exists a regular orthogonal matrix QQ such that QTAQ=A~Q^{\rm T}AQ=\tilde{A}, where

Q=[Q1OOQ2]orQ=[OQ1Q2O]Q=\left[\begin{matrix}Q_{1}&O\\ O&Q_{2}\end{matrix}\right]~{}{\rm or}~{}Q=\left[\begin{matrix}O&Q_{1}\\ Q_{2}&O\end{matrix}\right]

with Q1Q_{1} and Q2Q_{2} being regular rational orthogonal matrices, respectively.

Corollary 3.2.

The matrix QQ in Theorem 3.1 is the unique rational orthogonal matrix such that QTAQ=A~Q^{\textup{T}}AQ=\tilde{A}.

Proof.

The irreducibility assumption of the characteristic polynomial of AA implies that Γ\Gamma is controllable. Then the corollary follows immediately from Theorem 2.1.

To give the proof of Theorem 3.1, we need several lemmas below.

Lemma 3.3.

Let Γ\Gamma and Γ~\tilde{\Gamma} be two generalized cospectral signed graphs with adjacency matrices AA and A~\tilde{A}, respectively. Then eT(λIA)1e=eT(λIA~)1ee^{\rm T}(\lambda I-A)^{-1}e=e^{\rm T}(\lambda I-\tilde{A})^{-1}e.

Proof.

It can be easily computed that

det(λI(A+tJ))\displaystyle\det(\lambda I-(A+tJ))
=\displaystyle= det(λIA)det(It(λIA)1eeT)\displaystyle\det(\lambda I-A)\det(I-t(\lambda I-A)^{-1}ee^{\rm T})
=\displaystyle= det(λIA)(1teT(λIA)1e).\displaystyle\det(\lambda I-A)(1-te^{\rm T}(\lambda I-A)^{-1}e).

Similarly, det(λI(A~+tJ))=det(λIA~)(1teT(λIA~)1e)\det(\lambda I-(\tilde{A}+tJ))=\det(\lambda I-\tilde{A})(1-te^{\rm T}(\lambda I-\tilde{A})^{-1}e). Thus, the lemma follows.

Lemma 3.4 ([15]).

(λIA)1=i=1nξiξiTλλi(\lambda I-A)^{-1}=\sum_{i=1}^{n}\frac{\xi_{i}\xi_{i}^{\rm T}}{\lambda-\lambda_{i}}, where ξi\xi_{i}’s are normalized eigenvectors of AA associated with λi\lambda_{i}, for in\leq i\leq n.

Lemma 3.5 ([18]).

Let A=(aij)A=(a_{ij}) be a symmetric integral matrix with an irreducible characteristic polynomial ϕ(x)\phi(x). Let λ1,,λn\lambda_{1},\ldots,\lambda_{n} be the distinct eigenvalues of AA. Then there exist polynomials ϕi(x)[x]\phi_{i}(x)\in{\mathbb{Q}[x]} with degϕi<n\deg\phi_{i}<n such that the eigenvectors ξi\xi_{i} of AA associated with λi\lambda_{i} can be expressed as

ξi=(ϕ1(λi),ϕ2(λi),,ϕn(λi))T{\xi}_{i}=(\phi_{1}(\lambda_{i}),\phi_{2}(\lambda_{i}),\ldots,\phi_{n}(\lambda_{i}))^{\rm T}

for 1in1\leq i\leq n.

Proof.

Let λ1\lambda_{1} be an eigenvalue of AA with corresponding eigenvector ξ1\xi_{1}. Consider the linear system of equations (λ1IA)ξ1=0(\lambda_{1}I-A)\xi_{1}=0. By Gaussian elimination, there exist xi(λ1)x_{i}\in{\mathbb{Q}(\lambda_{1})} such that ξ1=(x1,x2,,xn)T\xi_{1}=(x_{1},x_{2},\ldots,x_{n})^{\textup{T}}. Note (λ1)\mathbb{Q}(\lambda_{1}) is a number field. There exist polynomials ϕi(x)[x]\phi_{i}(x)\in{\mathbb{Q}[x]} with degϕi<n\deg\phi_{i}<n such that xi=ϕi(λ1)x_{i}=\phi_{i}(\lambda_{1}).

By the kk-th equation of (λ1IA)ξ1=0(\lambda_{1}I-A)\xi_{1}=0, we have ψ(λ1):=j=1nak,jϕj(λ1)λ1ϕj(λ1)=0\psi(\lambda_{1}):=\sum_{j=1}^{n}a_{k,j}\phi_{j}(\lambda_{1})-\lambda_{1}\phi_{j}(\lambda_{1})=0, for 1kn1\leq k\leq n. Note ψ(x)[x]\psi(x)\in{\mathbb{Q}[x]} and ψ(λ1)=0\psi(\lambda_{1})=0. By the irreducibility of ϕ(x)\phi(x), we have ϕ(x)\phi(x) divides ψ(x)\psi(x). Thus, ψ(λi)=0\psi(\lambda_{i})=0 for for 1in1\leq i\leq n, and ξi=(ϕ1(λi),ϕ2(λi),,ϕn(λi))T{\xi}_{i}=(\phi_{1}(\lambda_{i}),\phi_{2}(\lambda_{i}),\ldots,\phi_{n}(\lambda_{i}))^{\rm T} is an eigenvector associated with λi\lambda_{i}.

Next, we collect some simple facts about the relationships of eigenvalues/eigenvectors between the adjacency matrix AA of a signed bipartite graph Γ\Gamma and its bipartite-adjacency matrix MM.

Lemma 3.6.

Let Γ\Gamma be a signed bipartite graph with an irreducible characteristic polynomial over \mathbb{Q}. Let the adjacency matrix of Γ\Gamma be A=A(Γ)=[OMMTO]A=A(\Gamma)=\left[\begin{matrix}O&M\\ M^{\textup{T}}&O\end{matrix}\right]. Suppose that [uv]\left[\begin{matrix}u\\ v\end{matrix}\right] is an eigenvector of AA associated with an eigenvalue λ\lambda. Then

  1. 1.

    λ2\lambda^{2} is an eigenvalue of MMTMM^{\textup{T}} and MTMM^{\textup{T}}M with corresponding eigenvectors uu and vv, respectively;

  2. 2.

    uu and vv have the same length, i.e., u2=v2||u||_{2}=||v||_{2};

  3. 3.

    [uv]\left[\begin{matrix}u\\ -v\end{matrix}\right] (resp. [uv]\left[\begin{matrix}-u\\ v\end{matrix}\right]) is an eigenvector of AA associated with eigenvalue λ-\lambda;

  4. 4.

    The characteristic polynomials of MMTMM^{\textup{T}} (resp. MTMM^{\textup{T}}M) is irreducible over \mathbb{Q}.

Proof.

Note that the characteristic polynomial ϕ(x)\phi(x) of AA is irreducible, it follows that zero can never be an eigenvalue of AA, and hence MM must be a square matrix of order m:=n/2m:=n/2.

Let λ0\lambda\neq 0 be any eigenvalue of AA with corresponding eigenvector [uv]\left[\begin{matrix}u\\ v\end{matrix}\right]. Then

A[uv]=[MvMTu]=λ[uv]{Mv=λu,MTu=λv.A\left[\begin{matrix}u\\ v\end{matrix}\right]=\left[\begin{matrix}Mv\\ M^{\textup{T}}u\end{matrix}\right]=\lambda\left[\begin{matrix}u\\ v\end{matrix}\right]\Longleftrightarrow\left\{\begin{aligned} Mv=\lambda u,\\ M^{\textup{T}}u=\lambda v.\end{aligned}\right. (2)

Thus, we have u0u\neq 0 and v0v\neq 0, for otherwise we would have u=v=0u=v=0, since λ0\lambda\neq 0. It follows that

MMTu=λ2u,MTMv=λ2v.MM^{\textup{T}}u=\lambda^{2}u,~{}M^{\textup{T}}Mv=\lambda^{2}v.

It follows from Mv=λuMv=\lambda u that uTMv=λuTuu^{\textup{T}}Mv=\lambda u^{\textup{T}}u. By MTu=λvM^{\textup{T}}u=\lambda v we get vTMTu=λvTvv^{\textup{T}}M^{\textup{T}}u=\lambda v^{\textup{T}}v. Note uTMv=(uTMv)T=vTMTuu^{\textup{T}}Mv=(u^{\textup{T}}Mv)^{\textup{T}}=v^{\textup{T}}M^{\textup{T}}u. It follows that λuTu=λvTv\lambda u^{\textup{T}}u=\lambda v^{\textup{T}}v, and hence uTu=vTvu^{\textup{T}}u=v^{\textup{T}}v since λ0\lambda\neq 0.

Note that

A[uv]=[MvMTu]=λ[uv].A\left[\begin{matrix}u\\ -v\end{matrix}\right]=\left[\begin{matrix}-Mv\\ M^{\textup{T}}u\end{matrix}\right]=-\lambda\left[\begin{matrix}u\\ -v\end{matrix}\right].

Hence, [uv]\left[\begin{matrix}u\\ -v\end{matrix}\right] is an eigenvector of AA associated with eigenvalue λ-\lambda. Since the characteristic polynomial of AA is irreducible, the set of all the eigenvalues of AA can be written as {λ1,λ2,,λm,λ1,λ2,,λm}.\{\lambda_{1},\lambda_{2},\ldots,\lambda_{m},-\lambda_{1},-\lambda_{2},\ldots,\lambda_{m}\}.

Hence, the set of all the eigenvalues of MMTMM^{\textup{T}} (or MTMM^{\textup{T}}M) can be write as {λ12,λ22,,λm2}\{\lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{m}^{2}\}. Since ϕ(A;x)=(x2λ12)(x2λm2)\phi(A;x)=(x^{2}-\lambda_{1}^{2})\cdots(x^{2}-\lambda_{m}^{2}) is irreducible over \mathbb{Q}, ϕ(MMT;x)=ϕ(MTM;x)=(xλ12)(xλm2)\phi(MM^{\textup{T}};x)=\phi(M^{\textup{T}}M;x)=(x-\lambda_{1}^{2})\cdots(x-\lambda_{m}^{2}) is also irreducible \mathbb{Q}. ∎

Now, we present the proof of Theorem 3.1.

Proof of Theorem 3.1.

Set m:=n/2m:=n/2. By Lemma 3.6, let λ1,λ2,,λm,λ1,λ2,,λm\lambda_{1},\lambda_{2},\ldots,\lambda_{m},-\lambda_{1},-\lambda_{2},\ldots,-\lambda_{m} be the eigenvalues of AA and A~\tilde{A} with corresponding normalized eigenvectors

12[u1v1],,12[umvm],12[u1v1],,12[umvm],\displaystyle\frac{1}{\sqrt{2}}\left[\begin{matrix}u_{1}\\ v_{1}\end{matrix}\right],\ldots,\frac{1}{\sqrt{2}}\left[\begin{matrix}u_{m}\\ v_{m}\end{matrix}\right],\frac{1}{\sqrt{2}}\left[\begin{matrix}u_{1}\\ -v_{1}\end{matrix}\right],\ldots,\frac{1}{\sqrt{2}}\left[\begin{matrix}u_{m}\\ -v_{m}\end{matrix}\right], (3)
12[u~1v~1],,12[u~mv~m],12[u~1v~1],,12[u~mv~m],\displaystyle\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{u}_{1}\\ \tilde{v}_{1}\end{matrix}\right],\ldots,\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{u}_{m}\\ \tilde{v}_{m}\end{matrix}\right],\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{u}_{1}\\ -\tilde{v}_{1}\end{matrix}\right],\ldots,\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{u}_{m}\\ -\tilde{v}_{m}\end{matrix}\right], (4)

respectively, where ui,u~i,vi,v~inu_{i},\tilde{u}_{i},v_{i},\tilde{v}_{i}\in{\mathbb{R}^{n}} are mm-dimensional unit vectors.

By Lemma 3.3, we have eT(xIA)1e=eT(xIA~)1ee^{\textup{T}}(xI-A)^{-1}e=e^{\textup{T}}(xI-\tilde{A})^{-1}e. It follows from Lemma 3.4 that

i=1m(12e2mT[uivi])2xλi+i=1m(12e2mT[uivi])2x+λi=i=1m(12e2mT[u~iv~i])2xλi+i=1m(12e2mT[u~iv~i])2x+λi.\sum_{i=1}^{m}\frac{(\frac{1}{\sqrt{2}}e_{2m}^{\textup{T}}\left[\begin{matrix}u_{i}\\ v_{i}\end{matrix}\right])^{2}}{x-\lambda_{i}}+\sum_{i=1}^{m}\frac{(\frac{1}{\sqrt{2}}e_{2m}^{\textup{T}}\left[\begin{matrix}u_{i}\\ -v_{i}\end{matrix}\right])^{2}}{x+\lambda_{i}}=\sum_{i=1}^{m}\frac{(\frac{1}{\sqrt{2}}e_{2m}^{\textup{T}}\left[\begin{matrix}\tilde{u}_{i}\\ \tilde{v}_{i}\end{matrix}\right])^{2}}{x-\lambda_{i}}+\sum_{i=1}^{m}\frac{(\frac{1}{\sqrt{2}}e_{2m}^{\textup{T}}\left[\begin{matrix}\tilde{u}_{i}\\ -\tilde{v}_{i}\end{matrix}\right])^{2}}{x+\lambda_{i}}. (5)

Hence, we have that for each 1im1\leq i\leq m,

{(emTui+emTvi)2=(emTu~i+emTv~i)2,(emTuiemTvi)2=(emTu~iemTv~i)2.\displaystyle\left\{\begin{aligned} (e_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i})^{2}&=&(e_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i})^{2},\\ (e_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i})^{2}&=&(e_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i})^{2}.\end{aligned}\right. (6)

For a fixed ii, we distinguish the following two cases:

Case 1. emTui+emTvie_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i} and emTu~i+emTv~ie_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i} have the same sign (resp. opposite sign), and emTuiemTvie_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i} and emTu~iemTv~ie_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i} have the same sign (resp. opposite sign). It follows from (6) that

{emTui+emTvi=emTu~i+emTv~i,emTuiemTvi=emTu~iemTv~i,or{emTui+emTvi=(emTu~i+emTv~i),emTuiemTvi=(emTu~iemTv~i),\displaystyle\left\{\begin{aligned} e_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i}&=&e_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i},\\ e_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i}&=&e_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i},\end{aligned}\right.~{}{\rm or}~{}\left\{\begin{aligned} e_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i}&=&-(e_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i}),\\ e_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i}&=&-(e_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i}),\end{aligned}\right.

which implies that either i) emTui=emTu~ie_{m}^{\textup{T}}u_{i}=e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=emTv~ie_{m}^{\textup{T}}v_{i}=e_{m}^{\textup{T}}\tilde{v}_{i}; or ii) emTui=emTu~ie_{m}^{\textup{T}}u_{i}=-e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=emTv~ie_{m}^{\textup{T}}v_{i}=-e_{m}^{\textup{T}}\tilde{v}_{i}.

Case 2. emTui+emTvie_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i} and emTu~i+emTv~ie_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i} have the same sign (resp. opposite sign), and emTuiemTvie_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i} and emTu~iemTv~ie_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i} have the opposite sign (resp. same sign). Then

{emTui+emTvi=emTu~i+emTv~i,emTuiemTvi=(emTu~iemTv~i),or{emTui+emTvi=(emTu~i+emTv~i),emTuiemTvi=emTu~iemTv~i,\displaystyle\left\{\begin{aligned} e_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i}&=&e_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i},\\ e_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i}&=&-(e_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i}),\end{aligned}\right.~{}{\rm or}~{}\left\{\begin{aligned} e_{m}^{\textup{T}}u_{i}+e_{m}^{\textup{T}}v_{i}&=&-(e_{m}^{\textup{T}}\tilde{u}_{i}+e_{m}^{\textup{T}}\tilde{v}_{i}),\\ e_{m}^{\textup{T}}u_{i}-e_{m}^{\textup{T}}v_{i}&=&e_{m}^{\textup{T}}\tilde{u}_{i}-e_{m}^{\textup{T}}\tilde{v}_{i},\end{aligned}\right.

which implies that either i) emTui=emTv~ie_{m}^{\textup{T}}u_{i}=e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=emTu~ie_{m}^{\textup{T}}v_{i}=e_{m}^{\textup{T}}\tilde{u}_{i}; or ii) emTui=emTv~ie_{m}^{\textup{T}}u_{i}=-e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=emTu~ie_{m}^{\textup{T}}v_{i}=-e_{m}^{\textup{T}}\tilde{u}_{i}.

Thus, for a fixed ii, we may assume that either emTui=τiemTu~ie_{m}^{\textup{T}}u_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=τiemTv~ie_{m}^{\textup{T}}v_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{v}_{i} or emTui=σiemTv~ie_{m}^{\textup{T}}u_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=σiemTu~ie_{m}^{\textup{T}}v_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{u}_{i}, where τi,σi{1,1}\tau_{i},\sigma_{i}\in\{1,-1\}. Next, we show that uniformly, either emTui=τiemTu~ie_{m}^{\textup{T}}u_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=τiemTv~ie_{m}^{\textup{T}}v_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{v}_{i} for all 1im1\leq i\leq m or emTui=σiemTv~ie_{m}^{\textup{T}}u_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=σiemTu~ie_{m}^{\textup{T}}v_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{u}_{i} for all 1im1\leq i\leq m. This is the key technical part of the proof, which highly depends on the irreducibility assumption of ϕ\phi.

According to Lemma 3.5, the eigenvectors of MMTMM^{\textup{T}} associated with eigenvalues λi2\lambda_{i}^{2} can be expressed as ξi=(ϕ1(λi),ϕ2(λi),,ϕm(λi))T\xi_{i}=(\phi_{1}(\lambda_{i}),\phi_{2}(\lambda_{i}),\ldots,\phi_{m}(\lambda_{i}))^{\textup{T}}, where ϕj(x)[x]\phi_{j}(x)\in{\mathbb{Q}[x]} with degϕj<n\deg\phi_{j}<n.

By Lemma 3.6, uiu_{i} is an eigenvector of MMTMM^{\textup{T}} associated with λi2\lambda_{i}^{2}. Note uiu_{i} is a unit vector. It follows that uiu_{i} and ξi/ξi2\xi_{i}/||\xi_{i}||_{2} differ by at most a sign, i.e., there exists a ϵi{1,1}\epsilon_{i}\in\{1,-1\} such that ui=ϵiξiξi2u_{i}=\epsilon_{i}\frac{\xi_{i}}{||\xi_{i}||_{2}}, and

vi\displaystyle v_{i} =\displaystyle= 1λiMTui\displaystyle\frac{1}{\lambda_{i}}M^{\textup{T}}u_{i}
=\displaystyle= ϵiλiMT(ϕ1(λi),ϕ2(λi),,ϕm(λi))T/ξi2\displaystyle\frac{\epsilon_{i}}{\lambda_{i}}M^{\textup{T}}(\phi_{1}(\lambda_{i}),\phi_{2}(\lambda_{i}),\ldots,\phi_{m}(\lambda_{i}))^{\textup{T}}/||\xi_{i}||_{2}
=\displaystyle= ϵi(φ1(λi),φ2(λi),,φm(λi))T/ξi2,\displaystyle\epsilon_{i}(\varphi_{1}(\lambda_{i}),\varphi_{2}(\lambda_{i}),\ldots,\varphi_{m}(\lambda_{i}))^{\textup{T}}/||\xi_{i}||_{2},

for some φj(x)[x]\varphi_{j}(x)\in{\mathbb{Q}[x]} with degree less than nn, for 1jm1\leq j\leq m. The last equality follows since the entries of the vector 1λiMT(ϕ1(λi),ϕ2(λi),,ϕm(λi))T\frac{1}{\lambda_{i}}M^{\textup{T}}(\phi_{1}(\lambda_{i}),\phi_{2}(\lambda_{i}),\ldots,\phi_{m}(\lambda_{i}))^{\textup{T}} belong to (λi)\mathbb{Q}(\lambda_{i}), which is a number field. Further note that ui2=vi2=1||u_{i}||_{2}=||v_{i}||_{2}=1, we have φ1(λi)2+φ2(λi)2++φm(λi)2=ξi2,\varphi_{1}(\lambda_{i})^{2}+\varphi_{2}(\lambda_{i})^{2}+\cdots+\varphi_{m}(\lambda_{i})^{2}=||\xi_{i}||^{2}, for 1im1\leq i\leq m.

The above discussions apply similarly to the signed bipartite graph Γ~\tilde{\Gamma} with adjacency matrix A~\tilde{A}. Then we have u~i=ϵi~ξ~iξ~i2\tilde{u}_{i}=\tilde{\epsilon_{i}}\frac{\tilde{\xi}_{i}}{||\tilde{\xi}_{i}||_{2}} for ϵi~{1,1}\tilde{\epsilon_{i}}\in\{1,-1\}, where ξ~i=(ϕ1~(λi),ϕ~2(λi),,ϕ~m(λi))T\tilde{\xi}_{i}=(\tilde{\phi_{1}}(\lambda_{i}),\tilde{\phi}_{2}(\lambda_{i}),\ldots,\tilde{\phi}_{m}(\lambda_{i}))^{\textup{T}}, ϕ~j(x)[x]\tilde{\phi}_{j}(x)\in{\mathbb{Q}[x]} with degϕ~j<n\deg\tilde{\phi}_{j}<n. Moreover, v~i=ϵ~i(φ~1(λi),φ~2(λi),,φ~m(λi))T/ξ~i2\tilde{v}_{i}=\tilde{\epsilon}_{i}(\tilde{\varphi}_{1}(\lambda_{i}),\tilde{\varphi}_{2}(\lambda_{i}),\ldots,\tilde{\varphi}_{m}(\lambda_{i}))^{\textup{T}}/||\tilde{\xi}_{i}||_{2} with φ~j(x)[x]\tilde{\varphi}_{j}(x)\in{\mathbb{Q}[x]} with degree less than nn, and φ~1(λi)2+φ~2(λi)2++φ~m(λi)2=ξ~i2\tilde{\varphi}_{1}(\lambda_{i})^{2}+\tilde{\varphi}_{2}(\lambda_{i})^{2}+\cdots+\tilde{\varphi}_{m}(\lambda_{i})^{2}=||\tilde{\xi}_{i}||^{2}.

Claim 1.

If emTu1=τ1emTu~1e_{m}^{\textup{T}}u_{1}=\tau_{1}e_{m}^{\textup{T}}\tilde{u}_{1} and emTv1=τ1emTv~1e_{m}^{\textup{T}}v_{1}=\tau_{1}e_{m}^{\textup{T}}\tilde{v}_{1} with τ1{1,1}\tau_{1}\in\{1,-1\}, then emTui=τiemTu~ie_{m}^{\textup{T}}u_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=τiemTv~ie_{m}^{\textup{T}}v_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{v}_{i} for some τi{1,1}\tau_{i}\in\{1,-1\}, for all 2im2\leq i\leq m.

Proof.

Actually, it follows from emTu1=τ1emTu~1e_{m}^{\textup{T}}u_{1}=\tau_{1}e_{m}^{\textup{T}}\tilde{u}_{1} that

ϵ1j=1mϕj(λ1)j=1mϕj2(λ1)=τ1ϵ1~j=1mϕ~j(λ1)j=1mϕ~j2(λ1).\epsilon_{1}\frac{\sum_{j=1}^{m}\phi_{j}(\lambda_{1})}{\sqrt{\sum_{j=1}^{m}\phi_{j}^{2}(\lambda_{1})}}=\tau_{1}\tilde{\epsilon_{1}}\frac{\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{1})}{\sqrt{\sum_{j=1}^{m}\tilde{\phi}_{j}^{2}(\lambda_{1})}}. (7)

Taking squares on both sides of (7), it follows that

Φ(λ1):=(j=1mϕj(λ1))2j=1mϕ~j2(λ1)(j=1mϕ~j(λ1))2j=1mϕj2(λ1)=0.\Phi(\lambda_{1}):=(\sum_{j=1}^{m}\phi_{j}(\lambda_{1}))^{2}\sum_{j=1}^{m}\tilde{\phi}_{j}^{2}(\lambda_{1})-(\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{1}))^{2}\sum_{j=1}^{m}\phi_{j}^{2}(\lambda_{1})=0.

Note that ϕ(x)\phi(x) is irreducible and Φ(x)[x]\Phi(x)\in{\mathbb{Q}[x]}. It follows that ϕ(x)Φ(x)\phi(x)\mid\Phi(x). Hence Φ(λi)=0\Phi(\lambda_{i})=0 and emTui=τiemTu~ie_{m}^{\textup{T}}u_{i}=\tau_{i}e_{m}^{\textup{T}}\tilde{u}_{i} for some τi{1,1}\tau_{i}\in\{1,-1\}, for 2im2\leq i\leq m. Similarly, we have emTvi=τ~iemTv~ie_{m}^{\textup{T}}v_{i}=\tilde{\tau}_{i}e_{m}^{\textup{T}}\tilde{v}_{i} for some τ~i{1,1}\tilde{\tau}_{i}\in\{1,-1\}, for 2im2\leq i\leq m. Next, we show that τi\tau_{i} and τ~i\tilde{\tau}_{i} coincide, i.e., τi=τ~i=±1\tau_{i}=\tilde{\tau}_{i}=\pm 1, for all 2im2\leq i\leq m.

In fact, it follows from emTv1=τ1emTv~1e_{m}^{\textup{T}}v_{1}=\tau_{1}e_{m}^{\textup{T}}\tilde{v}_{1} that

ϵ1j=1mφj(λ1)j=1mϕj2(λ1)=τ1ϵ1~j=1mφ~j(λ1)j=1mϕ~j2(λ1).\epsilon_{1}\frac{\sum_{j=1}^{m}\varphi_{j}(\lambda_{1})}{\sqrt{\sum_{j=1}^{m}\phi_{j}^{2}(\lambda_{1})}}=\tau_{1}\tilde{\epsilon_{1}}\frac{\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{1})}{\sqrt{\sum_{j=1}^{m}\tilde{\phi}_{j}^{2}(\lambda_{1})}}. (8)

It is easy to see that all the numerators in Eqs. (7) and (8) are non-zero. For example, if j=1mϕj(λ1)=0\sum_{j=1}^{m}\phi_{j}(\lambda_{1})=0, then j=1mϕj(λi)=0\sum_{j=1}^{m}\phi_{j}(\lambda_{i})=0 for 1im1\leq i\leq m by the irreducibility of ϕ\phi. That is, emTξi=0e_{m}^{\textup{T}}\xi_{i}=0 for 1im1\leq i\leq m, which is ridiculous since ξi\xi_{i} (1im1\leq i\leq m) are eigenvectors of MMTMM^{\textup{T}} constituting a basis of m\mathbb{R}^{m}.

Eq. (8) divides Eq. (7), it follows that

j=1mϕj(λ1)j=1mφj(λ1)=j=1mϕ~j(λ1)j=1mφ~j(λ1),\frac{\sum_{j=1}^{m}\phi_{j}(\lambda_{1})}{\sum_{j=1}^{m}\varphi_{j}(\lambda_{1})}=\frac{\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{1})}{\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{1})}, (9)

or equivalently, Ψ(λ1):=j=1mϕj(λ1)j=1mφ~j(λ1)j=1mφj(λ1)j=1mϕ~j(λ1)=0\Psi(\lambda_{1}):=\sum_{j=1}^{m}\phi_{j}(\lambda_{1})\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{1})-\sum_{j=1}^{m}\varphi_{j}(\lambda_{1})\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{1})=0. By the irreducibility of ϕ(x)\phi(x), we obtain that ϕ(x)Ψ(x)\phi(x)\mid\Psi(x), and hence Ψ(λi)=0\Psi(\lambda_{i})=0 for 2im2\leq i\leq m. So Eq. (9) still holds if we replace λ1\lambda_{1} with any λi\lambda_{i}, i.e.,

j=1mϕj(λi)j=1mφj(λi)=j=1mϕ~j(λi)j=1mφ~j(λi),for2im.\frac{\sum_{j=1}^{m}\phi_{j}(\lambda_{i})}{\sum_{j=1}^{m}\varphi_{j}(\lambda_{i})}=\frac{\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{i})}{\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{i})},~{}{\rm for}~{}2\leq i\leq m. (10)

By the previous discussions, we get that

ϵij=1mϕj(λi)j=1mϕj2(λi)=τiϵi~j=1mϕ~j(λi)j=1mϕ~j2(λi),forfor2im.\epsilon_{i}\frac{\sum_{j=1}^{m}\phi_{j}(\lambda_{i})}{\sqrt{\sum_{j=1}^{m}\phi_{j}^{2}(\lambda_{i})}}=\tau_{i}\tilde{\epsilon_{i}}\frac{\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{i})}{\sqrt{\sum_{j=1}^{m}\tilde{\phi}_{j}^{2}(\lambda_{i})}},~{}{\rm for}~{}{\rm for}~{}2\leq i\leq m. (11)
ϵij=1mφj(λi)j=1mϕj2(λi)=τ~iϵi~j=1mφ~j(λi)j=1mϕ~j2(λi),for2im.\epsilon_{i}\frac{\sum_{j=1}^{m}\varphi_{j}(\lambda_{i})}{\sqrt{\sum_{j=1}^{m}\phi_{j}^{2}(\lambda_{i})}}=\tilde{\tau}_{i}\tilde{\epsilon_{i}}\frac{\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{i})}{\sqrt{\sum_{j=1}^{m}\tilde{\phi}_{j}^{2}(\lambda_{i})}},~{}{\rm for}~{}2\leq i\leq m. (12)

Eq. (12) divides Eq. (11), we obtain j=1mϕj(λi)j=1mφj(λi)=τiτ~ij=1mϕ~j(λi)j=1mφ~j(λi),\frac{\sum_{j=1}^{m}\phi_{j}(\lambda_{i})}{\sum_{j=1}^{m}\varphi_{j}(\lambda_{i})}=\frac{\tau_{i}}{\tilde{\tau}_{i}}\frac{\sum_{j=1}^{m}\tilde{\phi}_{j}(\lambda_{i})}{\sum_{j=1}^{m}\tilde{\varphi}_{j}(\lambda_{i})}, together with Eq. (10), we get the conclusion that τi=τ~i=±1\tau_{i}=\tilde{\tau}_{i}=\pm 1 for 2im2\leq i\leq m. ∎

Claim 2.

: If emTu1=σ1emTv~1e_{m}^{\textup{T}}u_{1}=\sigma_{1}e_{m}^{\textup{T}}\tilde{v}_{1} and emTv1=σ1emTu~1e_{m}^{\textup{T}}v_{1}=\sigma_{1}e_{m}^{\textup{T}}\tilde{u}_{1} with σ1{1,1}\sigma_{1}\in\{1,-1\}, then emTui=σiemTv~ie_{m}^{\textup{T}}u_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=σiemTu~ie_{m}^{\textup{T}}v_{i}=\sigma_{i}e_{m}^{\textup{T}}\tilde{u}_{i} for some σi{1,1}\sigma_{i}\in\{1,-1\}, for all 2im2\leq i\leq m.

Proof.

This follows by using the same argument as Claim 1; we omit the details here. ∎

Write

U=[u1,u2,,um],V=[v1,v2,,vm],U=[u_{1},u_{2},\ldots,u_{m}],~{}V=[v_{1},v_{2},\ldots,v_{m}],
U~=[u~1,u~2,,u~m],V~=[v~1,v~2,,v~m].\tilde{U}=[\tilde{u}_{1},\tilde{u}_{2},\ldots,\tilde{u}_{m}],~{}\tilde{V}=[\tilde{v}_{1},\tilde{v}_{2},\ldots,\tilde{v}_{m}].

If the condition of Claim 1 holds, we may replace uiu_{i} and viv_{i} with ui-u_{i} and vi-v_{i} respectively, whenever τi=1\tau_{i}=-1 for 1im1\leq i\leq m. Then we have emTui=emTu~ie_{m}^{\textup{T}}u_{i}=e_{m}^{\textup{T}}\tilde{u}_{i} and emTvi=emTv~ie_{m}^{\textup{T}}v_{i}=e_{m}^{\textup{T}}\tilde{v}_{i} for 1im1\leq i\leq m. Let R=12[UUVV]R=\frac{1}{\sqrt{2}}\left[\begin{matrix}U&U\\ V&-V\end{matrix}\right] and R~=12[U~U~V~V~].\tilde{R}=\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{U}&\tilde{U}\\ \tilde{V}&-\tilde{V}\end{matrix}\right]. Define

Q:=RR~T=[UU~TOOVV~T].Q:=R\tilde{R}^{\textup{T}}=\left[\begin{matrix}U\tilde{U}^{\textup{T}}&O\\ O&V\tilde{V}^{\textup{T}}\end{matrix}\right]. (13)

Then QQ is an orthogonal matrix and

RTAR=R~TA~R~=diag(λ1,,λm,λ1,,λm).R^{\textup{T}}AR=\tilde{R}^{\textup{T}}\tilde{A}\tilde{R}={\rm diag}(\lambda_{1},\ldots,\lambda_{m},-\lambda_{1},\ldots,-\lambda_{m}).

Thus, QTAQ=A~Q^{\textup{T}}AQ=\tilde{A}. Next, it remains to show that QQ is regular, i.e., Qe2m=e2mQe_{2m}=e_{2m}, which is equivalent to U~Tem=UTem\tilde{U}^{\textup{T}}e_{m}=U^{\textup{T}}e_{m} and V~Tem=VTem\tilde{V}^{\textup{T}}e_{m}=V^{\textup{T}}e_{m}. That is, emTui=emTu~i,emTvi=emTv~i,for1im,e_{m}^{\textup{T}}u_{i}=e_{m}^{\textup{T}}\tilde{u}_{i},~{}e_{m}^{\textup{T}}v_{i}=e_{m}^{\textup{T}}\tilde{v}_{i},~{}{\rm for}~{}1\leq i\leq m, which are precisely that we have obtained before, as desired.

If the condition of Claim 2 holds, similarly, we may replace uiu_{i} and viv_{i} with ui-u_{i} and vi-v_{i} respectively, whenever σi=1\sigma_{i}=-1. Then emTui=emTv~ie_{m}^{\textup{T}}u_{i}=e_{m}^{\textup{T}}\tilde{v}_{i} and emTvi=emTu~ie_{m}^{\textup{T}}v_{i}=e_{m}^{\textup{T}}\tilde{u}_{i}, for 1im1\leq i\leq m. Now let R=12[UUVV]R=\frac{1}{\sqrt{2}}\left[\begin{matrix}U&U\\ V&-V\end{matrix}\right] and R~=12[U~U~V~V~].\tilde{R}=\frac{1}{\sqrt{2}}\left[\begin{matrix}\tilde{U}&-\tilde{U}\\ \tilde{V}&\tilde{V}\end{matrix}\right]. Define

Q:=RR~T=[OUV~TVU~TO].Q:=R\tilde{R}^{\textup{T}}=\left[\begin{matrix}O&U\tilde{V}^{\textup{T}}\\ V\tilde{U}^{\textup{T}}&O\end{matrix}\right]. (14)

Then QQ is an orthogonal matrix and still QTAQ=A~Q^{\textup{T}}AQ=\tilde{A} holds. Moreover, it is easy to verify that Qe2m=e2mQe_{2m}=e_{2m}. So QQ is regular.

The proof is complete.

4 Proof of Theorem 1.2

In this section, we present the proof of Theorem 1.2.

Recall that for a monic polynomial f(x)[x]f(x)\in{\mathbb{Z}[x]} with degree nn, the discriminant of f(x)f(x) is defined as:

Δ(f)=1i<jn(αiαj)2,\Delta(f)=\prod_{1\leq i<j\leq n}(\alpha_{i}-\alpha_{j})^{2},

where α1,α2,,αn\alpha_{1},\alpha_{2},\ldots,\alpha_{n} are all the roots of f(x)f(x).

Then it is clear that Δ(f)\Delta(f) is always an integer for f(x)[x]f(x)\in{\mathbb{Z}[x]}, and Δ(f)=0\Delta(f)=0 if and only if ff has a multiple root. Define the discriminant of a matrix AA, denoted by Δ(A)\Delta(A), as the discriminant of its characteristic polynomial, i.e., Δ(A):=Δ(det(xIA))\Delta(A):=\Delta(\det(xI-A)). The discriminant of a graph GG, denoted by Δ(G)\Delta(G), is defined to be the discriminant of its adjacency matrix.

In [22], Wang and Yu give the following theorem, which is our main tool in proving Theorem 1.2.

Theorem 4.1 ([22]).

Let AA be a symmetric integral matrix. Suppose there exists a rational orthogonal matrix QQ such that QTAQQ^{\rm T}AQ is an integral matrix. If Δ(A)\Delta(A) is odd and square-free, then QQ must be a signed permutation matrix.

However, Theorem 4.1 cannot be used directly, since the Δ(Γ)\Delta(\Gamma) is always a perfect square for a signed bipartite graph Γ\Gamma with an equal size of bipartition, as shown by the following lemma.

Lemma 4.2.

Let Γ\Gamma be a signed bipartite graph with bipartite-adjacency matrix MM, where MM is a square matrix of order m:=n/2m:=n/2. Then Δ(Γ)=2ndet2(M)Δ2(MTM)\Delta(\Gamma)=2^{n}\det^{2}(M)\Delta^{2}(M^{\rm T}M).

Proof.

Let the eigenvalues of Γ\Gamma be ±λ1,±λ2,,±λm.\pm\lambda_{1},\pm\lambda_{2},\ldots,\pm\lambda_{m}. Then the eigenvalues of MTMM^{\textup{T}}M are λ12,λ22,,λm2.\lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{m}^{2}. So we have

Δ(Γ)\displaystyle\Delta(\Gamma) =\displaystyle= 1i<jm(λiλj)21i,jm(λi+λj)21i<jm(λi+λj)2\displaystyle\prod_{1\leq i<j\leq m}(\lambda_{i}-\lambda_{j})^{2}\prod_{1\leq i,j\leq m}(\lambda_{i}+\lambda_{j})^{2}\prod_{1\leq i<j\leq m}(-\lambda_{i}+\lambda_{j})^{2}
=\displaystyle= 2nλ12λ22λm21i<jm(λi2λj2)4\displaystyle 2^{n}\lambda_{1}^{2}\lambda_{2}^{2}\cdots\lambda_{m}^{2}\prod_{1\leq i<j\leq m}(\lambda_{i}^{2}-\lambda_{j}^{2})^{4}
=\displaystyle= 2ndet(MTM)Δ2(MTM)\displaystyle 2^{n}\det(M^{\textup{T}}M)\Delta^{2}(M^{\textup{T}}M)
=\displaystyle= 2n(det(M))2Δ2(MTM).\displaystyle 2^{n}(\det(M))^{2}\Delta^{2}(M^{\rm T}M).

This completes the proof. ∎

Let a0a_{0} be the constant term of the characteristic polynomial of GG defined as above. Then

a0=(1)mdet(MTM)=(1)mdet(M)2.a_{0}=(-1)^{m}\det(M^{\textup{T}}M)=(-1)^{m}\det{{}^{2}}(M).

Note that for a tree with an irreducible characteristic polynomial ϕ(x)\phi(x), the constant term of ϕ(x)\phi(x) is always ±1\pm 1. Thus we have

Corollary 4.3.

Let TT be a tree with an irreducible characteristic polynomial. Then Δ(T)=2nΔ2(MTM)\Delta(T)=2^{n}\Delta^{2}(M^{\rm T}M).

Finally, we are ready to present the proof of Theorem 1.2.

Proof of Theorem 1.2..

Let Γ~\tilde{\Gamma} be any signed graph that is generalized cospectral with Γ=(T,σ)\Gamma=(T,\sigma). We shall show that Γ~\tilde{\Gamma} is isomorphic to Γ\Gamma. Note that Γ~\tilde{\Gamma} has the same number of edges as Γ\Gamma and moreover, the assumption that ϕ(Γ~)=ϕ(Γ)\phi(\tilde{\Gamma})=\phi(\Gamma) is irreducible forces Γ~\tilde{\Gamma} to be connected. Thus, Γ~\tilde{\Gamma} is signed graph whose underlying graph is a tree (say T~\tilde{T}), and Γ~=(T~,σ~)\tilde{\Gamma}=(\tilde{T},\tilde{\sigma}).

Note that both TT and T~\tilde{T} are balanced as signed graphs, we have ϕ(T)=ϕ(Γ)\phi(T)=\phi(\Gamma) and ϕ(T~)=ϕ(Γ~)\phi(\tilde{T})=\phi(\tilde{\Gamma}). Let A(Γ)=D1A(T)D1A(\Gamma)=D_{1}A(T)D_{1} and A(Γ~)=D2A(T~)D2A(\tilde{\Gamma})=D_{2}A(\tilde{T})D_{2}, where D1D_{1} and D2D_{2} are diagonal matrices whose diagonal entries are ±1\pm 1.

By Theorem 2.1, the fact that Γ\Gamma and Γ~\tilde{\Gamma} are generalized cospectral implies that there exists a regular rational orthogonal matrix QQ such that

QTA(Γ)Q=A(Γ~),Q^{\rm T}A(\Gamma)Q=A(\tilde{\Gamma}), (15)

i.e., QT(D1A(T)D1)Q=D2A(T~)D2Q^{\rm T}(D_{1}A(T)D_{1})Q=D_{2}A(\tilde{T})D_{2}, which is equivalent to Q^TA(T)Q^=A(T~),\hat{Q}^{\rm T}A(T)\hat{Q}=A(\tilde{T}), where Q^=D1QD2\hat{Q}=D_{1}QD_{2} is a rational orthogonal matrix.

Let

A(T)=[OMMTO],A(T~)=[OM~M~TO].A(T)=\left[\begin{matrix}O&M\\ M^{\rm T}&O\end{matrix}\right],A(\tilde{T})=\left[\begin{matrix}O&\tilde{M}\\ \tilde{M}^{\rm T}&O\end{matrix}\right].

By Theorem 3.1, assume without loss of generality that Q=[Q1OOQ2]Q=\left[\begin{matrix}Q_{1}&O\\ O&Q_{2}\end{matrix}\right] and Q^=[Q^1OOQ^2].\hat{Q}=\left[\begin{matrix}\hat{Q}_{1}&O\\ O&\hat{Q}_{2}\end{matrix}\right]. Then we have Q^1TMQ^2=M~\hat{Q}_{1}^{\rm T}M\hat{Q}_{2}=\tilde{M}. It follows that

Q^1TMMTQ^1=M~M~TandQ^2TMTMQ^2=M~TM~.\hat{Q}_{1}^{\rm T}MM^{\rm T}\hat{Q}_{1}=\tilde{M}\tilde{M}^{\rm T}~{}{\rm and}~{}\hat{Q}_{2}^{\rm T}M^{\rm T}M\hat{Q}_{2}=\tilde{M}^{\rm T}\tilde{M}.

Note that Δ(MTM)=Δ(MMT)=2n/2Δ(T)\Delta(M^{\rm T}M)=\Delta(MM^{\rm T})=2^{-n/2}\sqrt{\Delta(T)}, which is odd and square-free. Thus, according to Theorem 4.1, both Q^1\hat{Q}_{1} and Q^2\hat{Q}_{2} are signed permutation matrices. It follows that Q=D1Q^Q2Q=D_{1}\hat{Q}Q_{2} is a signed permutation matrix. Moreover, note that QQ is regular. Therefore, QQ is a permutation matrix, and by Eq. (15), we conclude that Γ~\tilde{\Gamma} is isomorphic to Γ\Gamma. The proof is complete.

Remark 1.

The condition of Theorem 1.2 is tight in the sense that Theorem 1.2 is no longer true if 2n2Δ(T)2^{-\frac{n}{2}}\sqrt{\Delta(T)} has a multiple odd prime factor. Let the signed bipartite-adjacency matrices of two signed trees TT and T~\tilde{T} be given as follows, respectively:

M=(100000000110000000101000000001100000010011000000010000000010110000000100000001001),M~=(000000011000110100000100000110000000011000000100000100000000100000000010100001001).{\tiny M=\left(\begin{array}[]{ccccccccc}-1&0&0&0&0&0&0&0&0\\ -1&-1&0&0&0&0&0&0&0\\ 1&0&-1&0&0&0&0&0&0\\ 0&0&1&-1&0&0&0&0&0\\ 0&-1&0&0&-1&1&0&0&0\\ 0&0&0&0&-1&0&0&0&0\\ 0&0&0&0&-1&0&-1&1&0\\ 0&0&0&0&0&0&-1&0&0\\ 0&0&0&0&0&-1&0&0&-1\\ \end{array}\right),\tilde{M}=\left(\begin{array}[]{ccccccccc}0&0&0&0&0&0&0&1&-1\\ 0&0&0&-1&-1&0&1&0&0\\ 0&0&0&-1&0&0&0&0&0\\ -1&-1&0&0&0&0&0&0&0\\ 0&-1&-1&0&0&0&0&0&0\\ -1&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&-1&0&0\\ 0&0&0&0&0&0&0&-1&0\\ -1&0&0&0&0&-1&0&0&1\\ \end{array}\right).}

Then

ϕ(T)=ϕ(T~)=1+22x2162x4+538x6897x8+809x10410x12+116x1417x16+x18,\phi(T)=\phi(\tilde{T})=-1+22x^{2}-162x^{4}+538x^{6}-897x^{8}+809x^{10}-410x^{12}+116x^{14}-17x^{16}+x^{18},

which is irreducible over \mathbb{Q}. However, 29Δ(T)=72×347×3571750512^{-9}\sqrt{\Delta(T)}=7^{2}\times 347\times 357175051, i.e., 29Δ(T)2^{-9}\sqrt{\Delta(T)} has a multiple factor 7 and the condition of Theorem 1.2 is not satisfied. Actually, there indeed exists a regular rational orthogonal matrix Q𝒬(G)Q\in\mathcal{Q}(G) such that A~=QTAQ\tilde{A}=Q^{\textup{T}}AQ, where Q=diag(Q1,Q2)Q={\rm diag}(Q_{1},Q_{2}) and Q1Q_{1} and Q2Q_{2} are given as follows respectively.

Q1=17(112243321223311142224311132431122213341122213331122214112234321223411132112233421),Q2=17(224311132223411132223311142431122213112234321341122213331122214112243321112233421).{\tiny Q_{1}=\frac{1}{7}\left(\begin{array}[]{ccccccccc}-1&-1&-2&-2&4&3&3&2&1\\ -2&-2&3&3&1&-1&-1&4&2\\ 2&2&4&-3&-1&1&1&3&-2\\ 4&-3&1&1&-2&2&2&-1&3\\ -3&4&1&1&-2&2&2&-1&3\\ 3&3&-1&-1&2&-2&-2&1&4\\ 1&1&2&2&3&4&-3&-2&-1\\ 2&2&-3&4&-1&1&1&3&-2\\ 1&1&2&2&3&-3&4&-2&-1\\ \end{array}\right),Q_{2}=\frac{1}{7}\left(\begin{array}[]{ccccccccc}2&2&4&-3&-1&1&1&3&-2\\ 2&2&-3&4&-1&1&1&3&-2\\ -2&-2&3&3&1&-1&-1&4&2\\ 4&-3&1&1&-2&2&2&-1&3\\ 1&1&2&2&3&4&-3&-2&-1\\ -3&4&1&1&-2&2&2&-1&3\\ 3&3&-1&-1&2&-2&-2&1&4\\ -1&-1&-2&-2&4&3&3&2&1\\ 1&1&2&2&3&-3&4&-2&-1\\ \end{array}\right).}

Remark 2.

Theorem 3.1 does not hold without the assumption that the characteristic polynomial of Γ\Gamma is irreducible over \mathbb{Q}, even if Γ\Gamma is controllable. Let Γ\Gamma and Γ~\tilde{\Gamma} be two signed trees with bipartite-adjacency matrices MM and M~\tilde{M} given as follows respectively:

M=(100000000110000100011000000001100000010011000000010000000000100000000110000000011),M~=(100000000111000000001000000001100000000110000010001000010000101000001010000000001).{\tiny M=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&-1&0&0\\ 0&-1&-1&0&0&0&0&0&0\\ 0&0&-1&1&0&0&0&0&0\\ 0&1&0&0&1&-1&0&0&0\\ 0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&-1&0&0\\ 0&0&0&0&0&0&1&1&0\\ 0&0&0&0&0&0&0&-1&-1\\ \end{array}\right),~{}\tilde{M}=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ -1&1&-1&0&0&0&0&0&0\\ 0&0&-1&0&0&0&0&0&0\\ 0&0&1&1&0&0&0&0&0\\ 0&0&0&-1&-1&0&0&0&0\\ 0&-1&0&0&0&1&0&0&0\\ 0&1&0&0&0&0&1&0&1\\ 0&0&0&0&0&-1&0&-1&0\\ 0&0&0&0&0&0&0&0&-1\\ \end{array}\right).}

Q=15(500000000000000000050000000000000000000003211000001122000001122000003211000002311000001122000001122000002311005000000000000000000500000000000000000050000000000000000000000500000000000000000050000000000001132000002211000002211000001132000002211000001123000001123000002211000000000005000000000000000000500000000000000000050000).{\tiny Q=\frac{1}{5}\left(\begin{array}[]{cccccccccccccccccc}5&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&5&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&3&-2&1&-1&0&0&0&0&0&1&-1&2&2\\ 0&0&0&0&0&-1&-1&-2&2&0&0&0&0&0&3&2&1&1\\ 0&0&0&0&0&-2&3&1&-1&0&0&0&0&0&1&-1&2&2\\ 0&0&0&0&0&1&1&2&-2&0&0&0&0&0&2&3&-1&-1\\ 0&0&5&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&5&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&5&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&5&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&5&0&0&0&0&0&0&0\\ 0&0&0&0&0&-1&-1&3&2&0&0&0&0&0&-2&2&1&1\\ 0&0&0&0&0&2&2&-1&1&0&0&0&0&0&-1&1&3&-2\\ 0&0&0&0&0&2&2&-1&1&0&0&0&0&0&-1&1&-2&3\\ 0&0&0&0&0&1&1&2&3&0&0&0&0&0&2&-2&-1&-1\\ 0&0&0&0&0&0&0&0&0&0&0&5&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&5&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&5&0&0&0&0\\ \end{array}\right).}
It is easy to verify that

ϕ(Γ;x)=(1+x)(1+x)(1x+x2)(1+x+x2)(121x2+95x4119x6+60x813x10+x12),\phi(\Gamma;x)=(-1+x)(1+x)(-1-x+x^{2})(-1+x+x^{2})(1-21x^{2}+95x^{4}-119x^{6}+60x^{8}-13x^{10}+x^{12}),

which is reducible over \mathbb{Q} and Γ\Gamma is controllable. Nevertheless, the unique regular rational orthogonal matrix QQ (shown as above) such that QTA(Γ)Q=A(Γ~)Q^{\textup{T}}A(\Gamma)Q=A(\tilde{\Gamma}) is not the form as in Theorem 3.1.

5 Conclusions and Future Work

In this paper, we have given a simple arithmetic condition on a tree TT with an irreducible characteristic polynomial, under which every signed tree with underlying graph TT is DGS. This is a little bit surprising in contrast with Schwenk’s remarkable result stating almost every tree has a cospectral mate.

However, there are several questions remained to be answered. We end the paper by proposing the following questions:

Question 1.

How can Theorem 1.2 be generalized to signed bipartite graphs?

Question 2.

Is it true that every tree with an irreducible characteristic polynomial is DGS?

Question 3.

Is Theorem 3.1 true for controllable bipartite graphs?

For Question 1, the difficulty lies in the fact that for a signed bipartite graph Γ\Gamma, a signed graph Γ~\tilde{\Gamma} generalized cospectral with Γ\Gamma is not necessarily bipartite. For Question 2, we know that it is not true for signed trees. For Question 3, we know that it is not true for controllable signed bipartite graphs. But generally we do not know any single counterexample to Questions 2 and 3. The above questions need further investigations in the future.

Acknowledgments

The research of the second author is supported by National Natural Science Foundation of China (Grant Nos. 11971376 and 12371357) and the third author is supported by Fundamental Research Funds for the Central Universities (Grant No. 531118010622).

The authors would like to thank Professor Huiqiu Lin from East China University of Science and Technology for useful discussions.

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