 Research
 Open Access
On eigenvalue inequalities of a matrix whose graph is bipartite
 Abdullah Alazemi^{1}Email author,
 Milica Anđelić^{1} and
 Slobodan K. Simić^{2, 3}
https://doi.org/10.1186/s1366001920012
© The Author(s) 2019
 Received: 30 October 2018
 Accepted: 18 February 2019
 Published: 27 February 2019
Abstract
We consider the set of real zero diagonal symmetric matrices whose underlying graph, if not told otherwise, is bipartite. Then we establish relations between the eigenvalues of such matrices and those arising from their bipartite complement. Some accounts on interval matrices are provided. We also provide a partial answer to the still open problem posed in (Zhan in SIAM J. Matrix Anal. Appl. 27:851–860, 2006).
Keywords
 Bipartite complement of matrix
 Eigenvalue bounds
 Interlacing property
 Interval matrices
MSC
 15A18
 15A42
 05C50
1 Introduction
Throughout the paper, we assume that the eigenvalues of a matrix A, that is, the roots of its characteristic polynomial \(\varphi_{A}(\lambda)=\det(\lambda IA)\) (or, occasionally, its underlying graph \(G = G(A)\)) are ordered in nonincreasing way (\(\lambda_{i}(A) \geq \lambda_{j}(A)\) for \(i < j\)).
We will frequently exploit the wellknown Courant–Weyl inequalities (see, e.g., [2, p. 239]).
Theorem 1.1
The rest of the paper is organized as follows. In Sect. 2, we introduce the concept of a bipartite complement of a real symmetric matrix with zero diagonal whose underlying graph is bipartite, and then we show that the oddindexed eigenvalues of the initial matrix interlace the evenindexed eigenvalues of its bipartite complement (and vice versa). In this section, we also show that the bounds on the second largest eigenvalue of matrices in \(\mathcal{M}_{n}\) whose offdiagonal entries are in the (closed) interval \([0,1]\), as given in [3], can be significantly improved in the case where the underlying graph of a matrix in question is bipartite. Also, we extend these results (i.e., bounds) in several directions. First, we prove that these bounds hold for all matrices in \(\mathcal{M}_{n}\), not only for those with \(a_{ij} \in[0,1]\); second, we prove that similar bounds hold for the other eigenvalues (not only for the second largest one). We also deduce, in the nonbipartite case, similar properties for a real symmetric matrix with zero diagonal with respect to its complement. In Sect. 3, we focus on matrices whose underlying graphs are bipartite and all nonzero entries are in a given interval. We provide both upper and lower bounds on the largest eigenvalue of these matrices. If the corresponding matrices are nonnegative, then we also provide upper and lower bounds on the second largest eigenvalue. Here we give a partial answer to the problem posed in [4], that is, we determine extremal values for the second largest eigenvalue of a real symmetric matrix whose entries lie in a nonnegative interval.
2 Matrices with bipartite underlying graph
We first establish relations between the eigenvalues of A and \(A^{b}\). Recall that the spectrum of a symmetric matrix with bipartite underlying graph is symmetric with respect to the origin (see, e.g., [1, p. 56]).
Theorem 2.1
Proof
This result tells that the even and oddindexed eigenvalues of A interlace the odd and evenindexed eigenvalues of \(A^{b}\), respectively. We now deduce when the first inequality in (2.3) becomes equality under the constraint that all entries of A belong to \([0, 1]\).
Theorem 2.2
Proof
We now prove that \(G(A^{b})\) is disconnected. Suppose on the contrary that \(G(A^{b})\) is connected. Then \(A^{b}\) is irreducible. If \(\mathbf{x} =(\mathbf{y}^{t},\mathbf{z}^{t})^{t}\) is the eigenvector corresponding to \(\lambda_{n}(A^{b})\), then \((\mathbf{y}^{t}, \mathbf{z}^{t})^{t}\) is the eigenvector corresponding to \(\lambda_{1}(A^{b})\). Therefore \((\mathbf{y}^{t},\mathbf{z}^{t})^{t}\) is a scalar multiple of the Perron eigenvector of \(A^{b}\). (Note that both A and \(A^{b}\) are nonnegative.) This implies that all entries of y must be either strictly positive or strictly negative. Hence their sum cannot be equal to 0 as expected (see (2.7)). A similar argument holds for z, and we arrive at a contradiction.
Suppose now that there exists a nonzero vector x such that \(A^{b}\mathbf{x} = \lambda_{n}(A^{b})\mathbf{x}\) and (2.7) holds. Since \(A+A^{b}=A(K_{p,q})\), we have \((A+A^{b})\mathbf{x} = A(K_{p,q}) \mathbf{x}\) and \(A\mathbf{x}+A^{b}\mathbf{x}=\mathbf{0}\), since \(A(K_{p,q})\mathbf{x}=\mathbf{0}\) due to (2.7). Therefore \(A\mathbf{x} =\lambda_{n}(A^{b})\mathbf{x} =\lambda_{1}(A^{b})\mathbf{x}\). This means that \(\lambda_{1}(A^{b})\) is the eigenvalue of A, and we only need to prove that it is not equal to \(\lambda_{1}(A)\). Since (2.7) holds, x cannot be a Perron vector, because all entries of a Perron vector can be taken to be nonnegative.
This completes the proof. □
Remark 2.1
By applying Courant–Weyl inequalities to the largest eigenvalue of \(A+A^{b}\), that is, to \(\lambda_{1}(A(K_{p,q}))\), we obtain \(\lambda_{1}(A)+\lambda_{1}(A^{b})\geq\sqrt{pq}\).
Example 2.1
If G is a nearly complete bipartite graph (with only a few edges missing) and \(A_{G}\) is its adjacency matrix, then the bound (2.3) can be very good. For example, for \(G=K_{5,6}e\), \(\lambda _{2}(A_{G})=0.84076\), and \(\lambda_{1}(A_{G}^{b})=1\). For \(G=K_{5,6}e_{1}e_{2}\), where \(e_{1}\) and \(e_{2}\) are two nonadjacent edges in \(K_{5,6}\), \(\lambda _{2}(A_{G})=\lambda_{1}(A_{G}^{b})=1\).
Theorem 2.3
Remark 2.2
Theorem 2.3 generalizes Proposition 2.1 from [3] since it provides interlacing between all the eigenvalues of A (i.e., \(A^{c}\)), not only between the largest and second largest ones.
We conclude this section with the following result.
Proposition 2.4

\(\sigma(A^{b})=\pm(\sqrt{pq}\sqrt{rs}),\pm{\mu_{2}}^{m_{2}}, \ldots, \pm{\mu_{k}}^{m_{k}}\) if \(\sqrt{pq}\sqrt{rs}\neq\lambda_{2} (A)\),

\(\sigma(A^{b})=\pm(\sqrt{pq}\sqrt{rs})^{m_{2}+1},\pm{\mu _{3}}^{m_{3}}, \ldots, \pm{\mu_{k}}^{m_{k}}\) if \(\sqrt{pq}\sqrt{rs}=\lambda_{2} (A)\).
Proof
Since the matrices A and \(A(K_{p,q})\) are symmetric and commute, they can be simultaneously diagonalized by, for example, matrix P, that is, \(P^{1}AP=D_{1}\) and \(P^{1}A(K_{p,q})P=D\). Consequently, \(P^{1}A^{b}P=DD_{1}\). The matrix \(A(K_{p,q})\) has three distinct eigenvalues, namely \(\sqrt{pq}\), 0, \(\sqrt{pq}\) with multiplicities \(1, p+q2, 1\), respectively. Also, \(\lambda_{1}(A)=\sqrt{\lambda _{1}(BB^{t})}=\sqrt{rs}\), since \(BB^{t}\) has the constant row sum equal to rs. Now the result directly follows. □
3 Interval matrices
By an interval matrix we assume a matrix whose all entries lie in some interval, say (closed) interval \([a,b]\) (\(\infty< a < b < +\infty\)). Its diagonal entries in general need not be equal to 0. Let \(\mathcal{S}_{n}[a,b]\) denote the set of all symmetric interval matrices over the interval \([a,b]\). In [4] the range of extremal eigenvalues of a real symmetric interval matrix was considered.
Here we consider the set of symmetric matrices whose all nonzero entries lie in the interval \([a,b]\) and whose underlying graphs are bipartite with bipartition \(U\cup V\), where \(U=p\), \(V=q\), \(p+q=n\), \(p,q>0\). Without loss of generality we may assume that this set, say \(\mathcal{B}'_{p,q}[a,b]\), consists of all matrices of the form $\left[\begin{array}{cc}O& B\\ {B}^{t}& O\end{array}\right]$, where \(B\in[a,b]^{p\times q}\) for some \(p+q=n\), \(p,q>0\).
We first determine \(\max\{\lambda_{1}(A): A\in\mathcal {B}'_{p,q}[a,b]\}\) and \(\min\{\lambda_{1}(A): A\in\mathcal{B}'_{p,q}[a,b]\}\), and we also identify matrices that attain these extremal values.
Next, we apply the following results to \(B^{t}B\).
Theorem 3.1
([4])
 (i)If \(a>b\), thenIf n is even, then the equality holds if and only if A is permutation similar to$$\lambda_{1}(A)\leq \textstyle\begin{cases} n(ba)/2 & \textit{if }n\textit{ is even}, \\ (nb\sqrt{b^{2}+(n^{2}1)a^{2}})/2 & \textit{if }n\textit{ is odd}. \end{cases} $$If n is odd, then the equality holds if and only if A is permutation similar to$\left[\begin{array}{cc}b{J}_{\frac{n}{2}}& a{J}_{\frac{n}{2}}\\ a{J}_{\frac{n}{2}}& b{J}_{\frac{n}{2}}\end{array}\right].$$\left[\begin{array}{cc}b{J}_{\frac{n1}{2}}& a{J}_{\frac{n1}{2},\frac{n+1}{2}}\\ a{J}_{\frac{n+1}{2},\frac{n1}{2}}& b{J}_{\frac{n+1}{2}}\end{array}\right].$
 (ii)
If \(a\leq b\), then \(\lambda_{1}(A)\leq nb\). If \(a< b\), then the equality holds if and only if \(A=bJ_{n}\).
If \(a= b\), then the equality holds if and only if A is permutation similar tofor some k with \(1\leq k\leq n\).$\left[\begin{array}{cc}b{J}_{k}& a{J}_{k,nk}\\ a{J}_{nk,k}& b{J}_{nk}\end{array}\right]$
Theorem 3.2
([4])

If \(0< a< b\), then \(\lambda_{1}(A)\geq na\) with equality if and only if \(A=aJ_{n}\).

If \(a\leq0< b\), then \(\lambda_{1}(A) \geq a\) with equality if and only if \(A=aI_{n}\).
 (1)
If \(0< a< b\), then \(\lambda_{1}(A)\leq\sqrt{pq}b\). The equality holds if and only if \(B=bJ_{pq}\). Also, \(\lambda_{1}(A)\geq\sqrt{pq}a\). The equality holds if and only if \(B=aJ_{pq}\).
 (2)
If \(a< b< 0\), then \(\lambda_{1}(A)\leq\sqrt{pq}a\). The equality holds if and only if \(B=aJ_{pq}\). Here \(\lambda_{1}(A)\geq \sqrt{pq}b\). The equality holds if and only if \(B=bJ_{pq}\).
 (3)If \(a\leq0< b\), \(a\leq b\), then \(\lambda_{1}(A)\leq\sqrt {pq}b\). If \(a=b\), then the equality holds if and only if B is permutation similar tofor some k with \(1\leq k\leq n\). If \(a< b\), then the holds if and only if \(B=bJ_{pq}\). In this case, \(\lambda_{1}(A)\geq0\). The equality holds if and only if \(B=O_{pq}\), since \(B^{t}B\) is positive semidefinite.$\left[\begin{array}{cc}b{J}_{k}& a{J}_{k,pk}\\ a{J}_{pk,k}& b{J}_{pk}\end{array}\right]$
 (4)
If \(a<0\leq b\), \(a> b\), then \(\lambda_{1}(A)\leq\sqrt {pq}a\). The equality holds if and only if \(B=aJ_{pq}\). Also, \(\lambda_{1}(A)\geq0\). The equality holds if and only if \(B=O_{pq}\).
Next, we determine upper and lower bounds on the second largest eigenvalue of bipartite interval matrices with given cardinality of color classes in the case where \(0\leq a< b\) or \(a< b\leq0\).
Recall first the following:
Theorem 3.3
([2, p. 238])
Next, we provide an upper bound for the second largest eigenvalue of a nonnegative matrix with entries in a given interval.
Theorem 3.4
Proof
This completes the proof. □
Theorem 3.5
Proof
Therefore \(\lambda_{2}(A)=\lambda_{1}(\tilde{A})\), and u is the corresponding positive eigenvector (since we assumed that u has no zero entries). From (3.8) it follows that any row of \(B^{t}\) is orthogonal to u, and consequently \(B^{t}=O\) since \(B^{t}\) is a nonnegative matrix (note that \(a=0\) in this case). Assume that \(\mathbf{v}_{1}=[\mathbf{v}^{t}, \mathbf{w}^{t}]\), \(\mathbf{v}\in \mathbb{R}^{k'}\), \(\mathbf{w}\in\mathbb{R}^{nk'}\). Having in mind that \(B=O\), it easily follows that \(C\mathbf{w}=\lambda_{1}(A)\mathbf {w}\), that is, \(\lambda_{1}(A)\in \operatorname{Spec}(C)\), which implies \(\lambda_{1}(A)=\lambda_{1}(C)\).
According to Theorem 3.1, if n is even and \(k'<\frac{n}{2}\), then \(\lambda_{2}(A)=\lambda_{1}(\tilde{A})\leq k'b<\frac{n}{2}b\), a contradiction. Hence \(k'=\frac{n}{2}\). Moreover, again by Theorem 3.1, \(\lambda_{1}(\tilde{A})=\frac{n}{2}b\) if and only if \(\tilde{A}=bJ_{\frac{n}{2}}\). Consequently, \(\frac{n}{2}b=\lambda_{2}(A)\leq\lambda_{1}(A)\leq(nk')b=\frac{n}{2}b\), that is, \(\lambda_{1}(C)=\frac{n}{2}b\), which implies \(C=bJ_{\frac {n}{2}}\). Hence \(A=J(\frac{n}{2}, 0,b)\).
If n is odd and x contains zeros, in a similar way, we can prove that the equality holds if and only if \(A=J(\frac{n1}{2}, 0,b)\).
This completes the proof. □
For the remaining cases, where not both a and b are nonnegative, we conjecture the following:
Conjecture 1
 (i)If \(a< 0\leq b\), then$$\lambda_{2}(A)\leq\max_{\substack{r+s+t=n, \\ k+l=n, r,s,t,k,l\geq1}}\bigl\{ \lambda_{2}\bigl(\hat{A}(r,s,t) \bigr),\lambda_{2}\bigl( \tilde{B}(k,l)\bigr) \bigr\} . $$
 (ii)If \(a< b\leq0\), then$$\lambda_{2}(A)\leq\max_{r+s+t=n, r,s,t\geq1} \lambda_{2} \bigl(\hat{A}(r,s,t)\bigr). $$
We next consider the lower bounds for second largest eigenvalue of the matrix in \(\mathcal{S}_{n}[a,b]\) for \(a< b\).
Therefore we proved the following:
Theorem 3.6
Next, we focus only on bipartite interval matrices belonging to the set \(\mathcal{B}'_{p,q} [0,1]\).
Proposition 3.7
Let \(A\in\mathcal{B}'_{p,q} [0,1]\), \(n\geq3\). Then \(\lambda_{2}(A)\leq q/2\).
Proof
Let \(d_{i}(A)\) be the row sum of ith row of A, \(D=\operatorname {diag}(d_{1}(A), \ldots, d_{n}(A))\), and \(Q=D+A=(DA)+2A\). The matrix \(DA\) is positive semidefinite, and therefore \(\lambda_{2}(A)\leq\lambda_{2}(Q)/2\) (see [2, p. 495]). On the other hand, $\left[\begin{array}{cc}q{I}_{p}& {J}_{pq}\\ {J}_{qp}& p{I}_{q}\end{array}\right]=Q+(\left[\begin{array}{cc}q{I}_{p}& {J}_{pq}\\ {J}_{qp}& p{I}_{q}\end{array}\right]Q)$, and $\left[\begin{array}{cc}q{I}_{p}& {J}_{pq}\\ {J}_{qp}& p{I}_{q}\end{array}\right]Q$ is positive semidefinite. Consequently, $Q\le \left[\begin{array}{cc}q{I}_{p}& {J}_{pq}\\ {J}_{qp}& p{I}_{q}\end{array}\right]$ and \(\lambda_{2}(Q)\leq q\), and thus \(\lambda_{2}(A)\leq q/2\), as required. □
Proposition 3.8
Proof
Declarations
Acknowledgements
The authors acknowledge the support of the Research Sector, Kuwait University. The authors would like to thank the anonymous referees for careful reading and helpful comments and suggestions that lead to the improvement of the original manuscript.
Funding
Not applicable.
Authors’ contributions
All three authors contributed equally in writing this manuscript. They all read and approved the final version.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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