A Complete Characterization of Bidegreed Split Graphs with Four Distinct α-Eigenvalues

Abstract

It is a well-known fact that a graph of diameter d has at least $d+1$ eigenvalues. A graph is d-extremal (resp. $d_{\alpha }$-extremal) if it has diameter d and exactly $d+1$ distinct eigenvalues (resp. $\alpha$-eigenvalues), and a graph is split if its vertex set can be partitioned into a clique and a stable set. Such graphs have a diameter of at most three. If all vertex degrees in a split graph are either $\tilde{d}$ or $\widehat{d}$, then we say it is $(\tilde{d},\widehat{d})$-bidegreed. In this paper, we present a complete classification of the connected bidegreed $3_{\alpha }$-extremal split graphs using the association of split graphs with combinatorial designs. This result is a natural generalization of Theorem 4.6 proved by Goldberg et al. and Proposition 3.8 proved by Song et al., respectively.

1. Introduction

We considered only loopless finite graphs and use [1] for the terminologies and notations not defined in this paper. Let $G=\left(V\right(G),E(G\left)\right)$ be a graph with $n=\left|V\right(G\left)\right|$ vertices and $m=\left|E\right(G\left)\right|$ edges. Usually, we call the number of edges in G, which are incident to a vertex $w\in V\left(G\right)$, its degree and denoted by $d_G\left(w\right)$. We use the symbol $d_G(x,y)$ to denote the distance between two distinct vertices x and y of G, which is equal to the length of the shortest $(x,y)$-path in G. The diameter of G is the maximal distance among all pairs of vertices, formally $diam\left(G\right)=max\left\{d_G(s,t)\right|s,t\in V\left(G\right)\}$. The symbol $G_1\circ G_2$ denotes the corona of graphs $G_1$ and $G_2$, which is obtained by joining each vertex of $G_1$ to all the vertices of a copy of $G_2$. The identity matrix and all-ones matrix are denoted by I and J (each column is an all-ones vector j), respectively. The adjacency matrix of a graph G is defined as $A={\left(a_{ij}\right)}_{n\times n}$, where:

$$a_{ij}=\left\{\begin{array}{c}1,\mathrm{if}\mathrm{vertex}{v}_i\mathrm{is}\mathrm{adjacent}\mathrm{to}\mathrm{vertex}{v}_j\hfill \\ \\ 0,\mathrm{otherwise}.\hfill \end{array}\right.$$

Let $D={\left(d_{ij}\right)}_{n\times n}$ be the diagonal matrix where $d_{ii}$ is the degree of the vertex $v_i$ in G for $i=1,2,\cdots ,n$. The Laplacian matrix and signless Laplacian matrix is defined as $L=D-A$ and $Q=D+A$, respectively. It was reported in [2] that both L and Q have non-negative real eigenvalues. In 2017, Nikiforov introduced a new family of matrices [3]:

$$A_{\alpha }\left(G\right)=\alpha D\left(G\right)+(1-\alpha )A\left(G\right),$$

where $\alpha$ is an arbitrary real number lying in the interval $[0,1].$ It is routine to check that this kind of matrix is the convex combination of $A\left(G\right)$ and $D\left(G\right)$, and we call it the $\alpha$-adjacency matrix in the following discussion. In particular, if $\alpha =0$, then $A_{\alpha }\left(G\right)$ is exactly the adjacency matrix of $G,$ and $A_{\alpha }\left(G\right)=\frac{1}{2}Q\left(G\right)$ if $\alpha =\frac{1}{2}$. We encourage the interested reader to consult [4,5,6,7,8,9,10,11,12,13] and the references therein for more mathematical properties of $A_{\alpha }\left(G\right)$.

In what follows, we use $K_n$ to denote the complete graph with n vertices and denote by ${\mathcal{M}}_{n\times m}$ the set of matrices with n rows and m columns. In particular, if $n=m$, we write it to be ${\mathcal{M}}_n$ for short.

Let M be a matrix of order n; we use $Spec\left(M\right)$ to denote the multiset of eigenvalues of M and use $Spe{c}^{+}\left(M\right)$ to denote the set of distinct eigenvalues of matrix M. The symbols $rank\left(M\right)$ and $tr\left(M\right)$ denote the rank and trace of a matrix M, respectively. For simplicity, $\eta \left(M\right)=|Spe{c}^{+}\left(M\right)|$ is the number of distinct eigenvalues of a matrix M. It is a basic precept of spectral graph theory that low values of $\eta \left(M\right)$ indicate the presence of a special structure in the graph G; see some details from Table 1.

Table 1. Graphs with a small number of distinct eigenvalues.

$\eta \left(G\right)=2$	$\eta \left(G\right)=3$ (G is regular)	$\eta \left(G\right)=4$ (G is regular bipartite)
$G=m{K}_n$	G is strongly regular	G is the incident graph of a design
Doob, 1970 [14]	Shrikhande et al., 1965 [15]	Cvetković et al., 1980 [16]

Lemma 1

([17]). Let M be a real symmetric matrix with λ as its unique eigenvalue, then $M=\lambda I$.

A matrix is called $\omega$-stochastic if all its row sums are equal to the same number $\omega$. Obviously, each all-ones matrix is stochastic whether it is rectangular or square.

Lemma 2

([17]). Let M be a real symmetric ω-stochastic matrix. If x is a γ-eigenvector of M for some $\gamma \ne \omega$, then $j^{T}x=0$.

Lemma 3

([17]). Let $M\in {\mathcal{M}}_n$ be a real symmetric ω-stochastic matrix with $\omega >0$, and let $\sigma \in \mathbb{R}$ such that $n\sigma \ne \omega$, then $rank\left(M\right)=rank(M-\sigma J)$.

We need the following important concept.

Definition 1

([18]). A matrix $M\in {\mathcal{M}}_n$ ($n\ge 2$) is said to be reducible if there is a permutation matrix $P\in {\mathcal{M}}_n$ such that:

$$P^{T}MP=\left[\begin{array}{cc}B\hfill & C\hfill \\ 0\hfill & D\hfill \end{array}\right],$$

where $B\in {\mathcal{M}}_r$, $D\in {\mathcal{M}}_{n-r}$, $C\in {\mathcal{M}}_{r\times (n-r)}$, and $0\in {\mathcal{M}}_{(n-r)\times r}$ is a zero matrix for $1\le r\le n-1$.

If the determinant $det\left(M\right)>0$, clearly M is irreducible, and if M is reducible, it must have at least $(n-1)$ zero entries.

For simplicity, we define the following set:

$$\widehat{{\mathcal{M}}_n[\omega ,\gamma ]}=\left\{M\right|M\mathrm{is}\mathrm{a}\omega -\mathrm{stochastic}\mathrm{matrix}\mathrm{of}\mathrm{order}n\mathrm{and}Spe{c}^{+}\left(M\right)=\{\omega ,\gamma \}\}.$$

Lemma 4

([19]). $M\in \widehat{{\mathcal{M}}_n[\omega ,\gamma ]}$ is an irreducible non-negative symmetric matrix if and only if $M=\frac{\omega -\gamma }{n}J+\gamma I$ for $\omega >\gamma >0$.

For a real symmetric matrix $M\in {\mathcal{M}}_n$, we call its eigenvalue $\mu$ a main eigenvalue if the eigenspace $\epsilon \left(\mu \right)$ is not orthogonal to the all-ones vector; otherwise, we call $\mu$ the non-main eigenvalue [20,21]. For convenience, an eigenvalue is restricted if it has an eigenvector perpendicular to the all-ones vector j; see more details from [2]. The set of all restricted eigenvalues of M is denoted by $Res\left(M\right)$.

Lemma 5

([17]). Let M be a real symmetric non-negative ω-stochastic matrix. Then, $Spec\left(M\right)\backslash \omega \subseteq Res\left(M\right)$. Furthermore, $\omega \in Res\left(M\right)$ if and only if M is reducible.

It is known from the literature that the designation “Schur-complement” has been applied to matrices of the form $H-G{E}^{-1}F$. Usually, we call ${M|}_E=H-G{E}^{-1}F$ the Schur complement of the nonsingular matrix E in:

$$M=\left[\begin{array}{cc}E\hfill & F\hfill \\ G\hfill & H\hfill \end{array}\right].$$

Lemma 6

([22]). Suppose M has an invertible principal submatrix E, then $rank\left(M\right)=rank\left(E\right)+rank\left(M{|}_E\right)$.

Let $M\in {\mathcal{M}}_n>0$; there must exist a unique vector $x={(x_1,x_2,\cdots ,x_n)}^T$ such that $Mx=\rho \left(M\right)x$ for $x>0$ and ${\sum }_{i=1}^n{x}_i=1$. The unique normalized eigenvector is often called the Perron vector, and $\rho \left(M\right)$ is called its Perron root.

Another useful tool is the principal of Perron’s theorem.

Lemma 7

([18], p. 508). If $M\in {\mathcal{M}}_n$ is non-negative and irreducible and $\rho \left(M\right)$ is the Perron root of M, then each of the following hold:

(1) $\rho \left(M\right)>0$;

(2) $\rho \left(M\right)$ is an algebraically (and hence, geometrically) simple eigenvalue of M;

(3) There exists a positive vector x, such that $Mx=\rho \left(M\right)x$;

(4) $\rho \left(M\right)$ is the unique eigenvalue of maximum modulus, that is $\left|\lambda \right|<\rho \left(M\right)$ for every eigenvalue $\lambda \ne \rho \left(M\right)$.

2. Combinatorial Preliminaries

For a given set $F=\{x_1,x_2,\cdots ,x_l\}$, we use $\mathcal{D}$ to denote the family of k-subsets of F, which is called an $(l,k,\lambda )$-design over F if, for any two elements $x_i,x_j$ of F, there are precisely $\lambda$ sets in $\mathcal{D}$ that contain both $x_i,x_j$ for $i,j=1,2,\cdots ,l$. A design is said to be non-trivial if $k<l$. Each element in F is called a point of $\mathcal{D}$, while the elements in $\mathcal{D}$ are called its blocks. We traditionally use b to denote the number of elements in $\mathcal{D}$. For instance, $\{{\alpha }_1,{\alpha }_2\},\{{\alpha }_1,{\alpha }_3\},\{{\alpha }_2,{\alpha }_3\}$ in Table 2 is respectively the block of ${\mathcal{D}}_2$.

Table 2. Examples of the $(l,k,\lambda )$-design.

$F=\{{\alpha }_1,{\alpha }_2,{\alpha }_3\}$	$(3,1,0)$-design	${\mathcal{D}}_1=\{\left\{{\alpha }_1\right\},\left\{{\alpha }_2\right\},\left\{{\alpha }_3\right\}\}$
	$(3,2,1)$-design	${\mathcal{D}}_2=\{\{{\alpha }_1,{\alpha }_2\},\{{\alpha }_1,{\alpha }_3\},\{{\alpha }_2,{\alpha }_3\}\}$
	$(3,3,1)$-design	${\mathcal{D}}_3=\left\{\{{\alpha }_1,{\alpha }_2,{\alpha }_3\}\right\}$

Actually, every element of F exactly appears in $r=\lambda \frac{l-1}{k-1}$ blocks [23], and this number is usually said to be the replication number of $\mathcal{D}$. In what follows, we sometimes expand the notation of the $(l,k,\lambda )$-design to be $(l,b,r,k,\lambda )$.

Lemma 8

([23]). Let $\mathcal{D}$ be a non-trivial $(l,k,\lambda )$-design with b blocks, then b ≥ l.

A design is $symmetric$ if $b=l$, and it is called non-symmetric if $b>l$.

Definition 2.

Let $\mathcal{D}$ be an $(l,b,r,k,\lambda )$-design over F. The associated split graph $G_{\mathcal{D}}$ has $l+b$ vertices corresponding to the points and blocks of $\mathcal{D}$. Two vertices $p,q$ in $G_{\mathcal{D}}$ are adjacent if one of the following condition holds:

$\left(1\right)$$p,q$ both correspond to points;

$\left(2\right)$ p corresponds to a point in F and q to a block in $\mathcal{D}$ such that $p\in q$.

Actually, $G_{\mathcal{D}}$ is a bidegreed split graph with maximal clique C and stable set S such that $c=\left|C\right|=l$ and $s=\left|S\right|=b$. Alternatively, we have $diam\left(G_{\mathcal{D}}\right)\le 3$.

Lemma 9

([17]). Let $\mathcal{D}$ be a non-trivial $(l,k,\lambda )$-design with associated split graph $G_{\mathcal{D}}$. Suppose that $G_{\mathcal{D}}$ has diameter 3, then $\mathcal{D}$ is non-symmetric and $s>c$.

The incidence matrix of a design $\mathcal{D}$ is the $l\times b$ matrix $B=\left(b_{ij}\right)$ with $b_{ij}=1$ if $x_i$ belongs to the j-th block of $\mathcal{D}$ and $b_{ij}=0$ otherwise.

Lemma 10

([17]). Let B be a $l\times b$ matrix with values in $0,1$ such that each column of B contains exactly k ones. Then, $B{B}^T=\lambda J+(r-\lambda )I$ if and only if B is the incidence matrix of an $(l,b,r,k,\lambda )$-design $\mathcal{D}$.

3. Bidegreed Split Graphs with Four $\mathbf{\alpha }$-Eigenvalues

We call a graph $d_{\alpha }$-extremal if it has diameter d and exactly $d+1$ distinct $\alpha$-eigenvalues. Obviously, $d_0$-extremal graphs must be d-extremal ones. Several examples are illustrated in Table 3.

Table 3. Examples of d-extremal graphs.

1-extremal	2-extremal	3-extremal
$K_n$	strongly regular graphs	distance regular graphs [2]

A graph $G=(V,E)$ is called a split graph if $V=C\cup S$, where C a clique C and S a stable set. For simplicity, we denote by $G=(C,S)$ the split graph and assume that $N\left(S\right)=C$. In 2015, Ghorbani and Azimi presented a characterization all split graphs with at most four distinct eigenvalues [24]. In 2020, Goldberg et al. found that there exists a flaw in Ghorbani’s paper, because more split graphs of diameter three with exactly four distinct eigenvalues were constructed in [17]. We encourage the interested reader to consult [25,26,27] for more properties and information on split graphs.

Our main result is the following:

Proposition 1.

Let G be a connected bidegreed split graph of diameter 3, with clique and stable set sizes $c,s$, respectively. Then, G has exactly four distinct α-eigenvalues if and only if one of the following holds:

(1) $G=K_c\circ K_1$ for $c\ge 2$;

(2) $G=G_{\mathcal{D}}$ for a $(c,s,r,k,\lambda )$-design $\mathcal{D}$ such that $\mathcal{D}$ has at least one pair of disjoint blocks and:

$${\alpha }^2{t}_1+\alpha t_2+t_3=0,$$

where $\alpha \in [0,\frac{1}{d-k+1})\cup (\frac{1}{d-k+1},\frac{1}{2}]$ and:

$$\left\{\begin{array}{c}{t}_1=4c^2({\lambda }^2-r)+4c(2k-c)(d-k)+8ck(2\lambda -1)+16k^2\hfill \\ \\ t_2=-8c^2({\lambda }^2-r)+4\lambda c^2(d-k)-8ck(2\lambda -1)\hfill \\ \\ t_3=4c^2({\lambda }^2-r).\hfill \end{array}\right.$$

The following is a natural consequence of Proposition 1.

Corollary 1

([17,19]). Let G be a connected bidegreed split graph of diameter 3, with clique and stable set sizes $c,s,$ respectively. Then, G has exactly four distinct eigenvalues (resp. signless Laplacian eigenvalues) if and only if one of the following graphs:

$\left(1\right)$$G=K_c\circ K_1$ for $c\ge 2$;

$\left(2\right)$$G=G_{\mathcal{D}}$ for a $(c,s,r,k,\lambda )$-design $\mathcal{D}$ such that $r={\lambda }^2$ (resp. $r=\frac{\lambda k-{\lambda }^2}{\lambda +k-2}$) and that $\mathcal{D}$ has at least one pair of disjoint blocks.

In what follows, we shall pay attention to classifying completely the connected bidegreed $3_{\alpha }$-extremal split graphs by using of some tools of combinatorial designs. We assumed that all vertex degrees in G are either $\tilde{d}$ or $\widehat{d}$ and say that G is $(\tilde{d},\widehat{d})$-bidegreed.

In the subsequent discussion, we assumed that G is a connected split bidegreed graph with diameter 3, since the cases with diameters 1 and 2 are trivial. This means that there are exactly two distinct vertex degrees in $G=(C,S)$. It was reported in [28] that all vertices in C (resp. in S) share the same degree, say d (resp. say k). Without loss of generality, we suppose that each vertex in C is adjacent to $k^{\prime }$ vertices inside S, which yields that $k^{\prime }=d-(c-1)$ and $k^{\prime }=\frac{sk}{c}$. In this paper, we set that $\alpha \in [0,\frac{1}{d-k+1})\bigcup (\frac{1}{d-k+1},\frac{1}{2}]$.

It is routine to check that the $\alpha$-adjacency matrix of G, denoted by $A_{\alpha }\left(G\right)$, can be represented as follows:

$$A_{\alpha }\left(G\right)=\left[\begin{array}{cc}\beta J+(\alpha d-\beta )I& \beta B\\ \beta B^T& \alpha kI\end{array}\right],$$

where the vertices of C are listed first and then those of S and $\beta =1-\alpha$.

The initial assumption of bidegreeness implies that the matrix B satisfies $BJ=k^{\prime }J$ and $B^{T}J=kJ$. Therefore, $B{B}^{T}J=k{k}^{\prime }J$, implying that $B{B}^T$ is $k{k}^{\prime }$-stochastic.

Let $\mu$ be an eigenvalue of $A_{\alpha }\left(G\right)$ with eigenvector ${(x,y)}^T\ne 0$, then we have:

$$\left[\begin{array}{cc}\beta J+(\alpha d-\beta )I& \beta B\\ \beta B^T& \alpha kI\end{array}\right]\left[\begin{array}{c}x\hfill \\ y\hfill \end{array}\right]=\mu \left[\begin{array}{c}x\hfill \\ y\hfill \end{array}\right],$$

which yields that:

$$\left\{\begin{array}{c}[\beta J+(\alpha d-\beta \left)I\right]x+\beta By=\mu x\hfill \\ \\ \beta B^{T}x+\alpha kIy=\mu y.\hfill \end{array}\right.$$

If $\mu -\alpha k\ne 0$, then we have $y={(\mu -\alpha k)}^{-1}\beta B^{T}x$. Hence:

$$\begin{array}{c}\hfill (\mu -\alpha k)\beta Jx+(\mu -\alpha k)(\alpha d-\beta )x+{\beta }^{2}B{B}^{T}x=(\mu -\alpha k)\mu x.\end{array}$$

(1)

Multiplying both sides of Equation (1) by J on the left, we have:

$$c(\mu -\alpha k)\beta Jx+(\mu -\alpha k)(\alpha d-\beta )Jx+{\beta }^{2}k{k}^{\prime }Jx=(\mu -\alpha k)\mu Jx.$$

Equivalently,

$$\begin{array}{c}\hfill [c(\mu -\alpha k)\beta +(\mu -\alpha k)(\alpha d-\beta )+{\beta }^{2}k{k}^{\prime }-(\mu -\alpha k)\mu ]Jx=0,\end{array}$$

(2)

which implies that:

$$\left\{\begin{array}{c}\left({♣}_1\right)Jx=0\mathrm{or}\hfill \\ \\ \left({♣}_2\right)c(\mu -\alpha k)\beta +(\mu -\alpha k)(\alpha d-\beta )+{\beta }^{2}k{k}^{\prime }-(\mu -\alpha k)\mu =0\mathrm{or}\hfill \\ \\ \left({♣}_3\right)\left\{\begin{array}{c}Jx=0\mathrm{and}\hfill \\ \\ c(\mu -\alpha k)\beta +(\mu -\alpha k)(\alpha d-\beta )+{\beta }^{2}k{k}^{\prime }-(\mu -\alpha k)\mu =0.\hfill \end{array}\right.\hfill \end{array}\right.$$

In particular, if $Jx=0$, then Equality (1) can be written as:

$$\begin{array}{c}\hfill \left({♣}_4\right)(\mu -\alpha k)(\alpha d-\beta )x+{\beta }^{2}B{B}^{T}x=(\mu -\alpha k)\mu x.\end{array}$$

We naturally conclude the following result, which is be used in later proofs.

Proposition 2.

Let $G=(C,S)$ be a connected bidegreed split graph with diameter 3. If μ is an eigenvalue of $A_{\alpha }\left(G\right)$, then at least one of the following holds:

(1) $\mu =\alpha k$;

(2) $\mu \ne \alpha k$ is a root of the quadratic equation $\left({♣}_2\right)$;

(3) $\mu \ne \alpha k$ is a root of the quadratic equation $\left({♣}_4\right)$ for any $x\ne 0$ and $j^{T}x=0$.

Let us emphasize that Item (3) of Proposition 2 implies that $\mu$ must be an eigenvalue of $A_{\alpha }\left(G\right)$. Next, we examine Item (2) of Proposition 2.

The theory of equitable partition is very classical. We briefly recall the needed definitions and set our notation. Let G be a graph; a partition $\pi =({\pi }_1,{\pi }_2,\cdots ,{\pi }_p)$ of its vertex set $V\left(G\right)$ is said to be equitable if any vertex $v\in {\pi }_i$ is adjacent to $a_{ij}^{\pi }$ vertices in ${\pi }_j$, irrespective of the choice of v. The partition can be described by an $p\times p$ matrix $A_{\pi }=\left(a_{ij}^{\pi }\right)$. It is well known that if $\pi$ is an equitable partition of a graph, then the spectrum of the corresponding partition matrix is a subset of the spectrum of the graph’s matrix (where the matrix can be the adjacency, Laplacian, or signless Laplacian—under suitable definitions), and their Perron values are equal.

The initial assumption tells us that $\pi =\{C,S\}$ is an equitable partition of the vertex set. Hence, the quotient matrix of A and L can be respectively given by:

$$A_{\pi }\left(G\right)=\left[\begin{array}{cc}c-1& k^{\prime }\\ k& 0\end{array}\right]$$

and:

$$L_{\pi }\left(G\right)=\left[\begin{array}{cc}d-c+1& -k^{\prime }\\ -k& k\end{array}\right].$$

Evidently, $A_{\alpha }\left(G\right)=\alpha D\left(G\right)+(1-\alpha )A\left(G\right)=\alpha L\left(G\right)+A\left(G\right)$. Hence, by the properties of the equitable partition, we have $A_{\alpha }^{\pi }\left(G\right)=\alpha L_{\pi }\left(G\right)+A_{\pi }\left(G\right)$. This implies that the spectrum of $A_{\alpha }^{\pi }\left(G\right)$ is a subset of $A_{\alpha }\left(G\right)$. Hence, $\mu$ in Item (2) of Proposition 2 must be an eigenvalue of $A_{\alpha }\left(G\right)$.

Another useful tool will be the following certainty.

Proposition 3.

The discriminant of the equation $\left({♣}_2\right)$ is positive, i.e.,

$${\beta }^2{(c-1)}^2+{\alpha }^2{(d-k)}^2+4{\beta }^{2}k{k}^{\prime }+2\alpha \beta (d-k)(c-1)-8\alpha \beta k>0.$$

Proof. 

For convenience, we use ${\Delta }_1$ to denote the right side of the inequality in Proposition 3. Note that $d\ge c\ge k+1$ and $k^{\prime }=d-c+1\ge 1$, and it follows that:

$${\Delta }_1\ge {\beta }^2{k}^2+{\alpha }^2{(d-k)}^2+4{\beta }^{2}k{k}^{\prime }-6\alpha \beta k,$$

since $\beta \ge \alpha$ holds for $\alpha \in [0,\frac{1}{2}]$.

If $k^{\prime }\ge 2$, it is not difficult to find that ${\Delta }_1>0$. If $k^{\prime }=1$, then we have $d=c$. To complete the proof, we consider the following two simple situations: if $d=c\ge 3$, the conclusion of the proof is straightforward; if $d=c=2$, then we have $k=1$. Hence, $G=P_4$ and ${\Delta }_1>0$. □

4. Proof of Proposition 1

From the previous discussion, one can find that the equation $\left({♣}_2\right)$ has two different roots, say $\rho$ and $\psi$. In what follows, we assumed that $\rho >\psi$. It is routine to check that $\rho \ne \alpha k$ and $\psi \ne \alpha k$.

Proposition 4.

ρ is the Perron root of $A_{\alpha }\left(G\right)$.

Proof. 

Because $A\left(G\right)$ is irreducible and $D\left(G\right)$ is non-negative, we have that $A_{\alpha }\left(G\right)$ is irreducible. We assumed that ${\rho }^{\ast }$ is the Perron root of $A_{\alpha }\left(G\right)$. According to the Perron–Frobenius theorem, ${\rho }^{\ast }$ is an algebraically simple eigenvalue of $A_{\alpha }\left(G\right)$. Therefore, there is a positive vector y such that $A_{\alpha }\left(G\right)y={\rho }^{\ast }y$. In what follows, we shall verify that ${\rho }^{\ast }$ satisfies Item (2) of Proposition 2.

If ${\rho }^{\ast }$ satisfies Item (3) of Proposition 2, then we have $j^{T}x=0$ for any $x={(x_1,x_2)}^T\ne 0$. This implies that $x_1$ and $x_2$ are either positive or negative, which contradicts the initial assumption that y is a positive vector.

If ${\rho }^{\ast }$ satisfies Item (1) of Proposition 2, then we have ${\rho }^{\ast }=\alpha k$. Note that $\rho \ne \alpha k$ and $\psi \ne \alpha k$; it then follows that $\rho <\alpha k$ and $\psi <\alpha k$ since ${\rho }^{\ast }=\alpha k$. In addition, $\rho +\psi =(c-1)\beta +(d+k)\alpha >2\alpha k$, again a contradiction.

From the previous analysis, we obtained that ${\rho }^{\ast }$ must satisfy Item (2) of Proposition 2. Hence, ${\rho }^{\ast }=\rho$ or ${\rho }^{\ast }=\psi$. Consequently, we have that ${\rho }^{\ast }=\rho$ since ${\rho }^{\ast }$ is the Perron root of $A_{\alpha }\left(G\right)$ and $\rho >\psi$. □

Note that, when $j^{T}x=0$ for some non-negative vector x, by Proposition 2 (3), the equation $\left({♣}_4\right)$ can be rewritten as:

$$\begin{array}{c}\hfill {\beta }^{2}B{B}^{T}x=(\mu -\alpha k)\mu x-(\mu -\alpha k)(\alpha d-\beta )x,\end{array}$$

which is equivalent to:

$$\begin{array}{c}\hfill B{B}^{T}x=\frac{\mu -\alpha k}{{\beta }^2}[\mu -(\alpha d-\beta )]x,\end{array}$$

where ${\beta }^2>0$, since $0\le \alpha \le \frac{1}{2}$. For convenience, let $\gamma =\frac{\mu -\alpha k}{{\beta }^2}[\mu -(\alpha d-\beta )]$. It follows that $\gamma$ is a restricted eigenvalue of $B{B}^T$. Consequently, we have $\gamma \ge 0$ since $B{B}^T$ is positive semi-definite.

It is not difficult to find that the discriminant of the quadratic equation:

$$\begin{array}{c}\hfill \left({♣}_5\right)\frac{\mu -\alpha k}{{\beta }^2}[\mu -(\alpha d-\beta )]-\gamma =0\end{array}$$

is equal to ${\Delta }_2^{\gamma }=4{(1-\alpha )}^2\gamma +{[\alpha (d-k+1)-1]}^2\ge 0$. Knowing that:

$$\begin{array}{c}\hfill {\Delta }_2^{\gamma }=0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\gamma =0\mathrm{and}\alpha (d-k+1)=1,\end{array}$$

it immediately yields that ${\Delta }_2^{\gamma }>0$ since $\alpha \ne \frac{1}{d-k+1}$. Hence, the quadratic equation $\left({♣}_5\right)$ has two distinct roots, say ${\mu }_1^{\gamma },{\mu }_2^{\gamma }$. Let $P_{\gamma }=\{{\mu }_1^{\gamma },{\mu }_2^{\gamma }\}$.

Here are some remarks and consequences. If $\gamma \ne 0$, we know that ${\mu }_1^{\gamma }\ne \alpha k$ and ${\mu }_2^{\gamma }\ne \alpha k$, which implies that $A_{\alpha }\left(G\right)$ has two distinct eigenvalues induced by each non-zero restricted eigenvalue $\gamma$. If $\gamma =0$, then ${\mu }_1^0=d-1$ and ${\mu }_2^0=k$.

The following proposition allows us to generalize the result in [19].

Proposition 5.

Let γ be a zero-restricted eigenvalue of $B{B}^T$, then $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$ or $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$.

Proof. 

Knowing that $\gamma \in Res\left(B{B}^T\right)$ and $\gamma =0$, then we have ${\mu }_1^0=\alpha k$ and ${\mu }_2^0=\alpha d-\beta$. Hence, ${\mu }_2^0=\alpha d-\beta$ is an eigenvalue of $A_{\alpha }\left(G\right)$ since $\mu \ne \alpha k$. Next, we shall prove that ${\mu }_1^0=\alpha k$ is also an eigenvalue of $A_{\alpha }\left(G\right)$ or $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$.

Let ${(x,y)}^T$ be an eigenvector of $A_{\alpha }\left(G\right)$ corresponding to ${\mu }_2^0=\alpha d-\beta$. Hence,

$$A_{\alpha }\left(G\right)\left[\begin{array}{c}x\hfill \\ y\hfill \end{array}\right]={\mu }_2^0\left[\begin{array}{c}x\hfill \\ y\hfill \end{array}\right],$$

which is equivalent to the following:

$$\left\{\begin{array}{c}[\beta J+(\alpha d-\beta \left)I\right]x+\beta By=(\alpha d-\beta )x\hfill \\ \\ \beta B^{T}x+\alpha kIy=(\alpha d-\beta )y.\hfill \end{array}\right.$$

That is:

$$\left\{\begin{array}{c}\beta By=0\hfill \\ \\ \beta B^{T}x=[\alpha (d-k)-\beta ]y.\hfill \end{array}\right.$$

For simplicity, we distinguish the following two cases.

Case 1. The equation $By=0$ only has zero solutions.

In this case, we have $B^{T}x=\frac{1}{\beta }[\alpha (d-k)-\beta ]y=0$ for $x\ne 0$. Accordingly, the column vectors of $B^T$ are linearly dependent, while the column vectors of B are linearly independent since $By=0$ for $y=0$. Hence, $rank\left(B\right)=rank\left(B^T\right)=rank\left(B{B}^T\right)=rank\left(B^{T}B\right)=s<c$.

If $B{B}^T$ is irreducible, then one can find that $\gamma$ is not unique. In other words, there must exist two distinct eigenvalues, say ${\gamma }_1,{\gamma }_2$, such that ${\gamma }_1,{\gamma }_2\in Res\left(B{B}^T\right)$. Hence, $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$.

If $B{B}^T$ is reducible, in conjunction with the fact that $B{B}^T$ is $k{k}^{\prime }$-stochastic, it follows from Lemma 5 that $k{k}^{\prime }\in Res\left(B{B}^T\right)$. Obviously, $\rho \notin P_{k{k}^{\prime }}$ and $\psi \notin P_{k{k}^{\prime }}$. It is not difficult to find that $\rho \ne \alpha d-\beta$. In the sequel, we shall prove that $\psi \ne \alpha d-\beta$. Actually,

$$\psi =\alpha d-\beta \mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}d=c\mathrm{and}\alpha =\frac{d}{d+1}.$$

Bearing in mind that $\alpha \in [0,\frac{1}{d-k+1})\bigcup (\frac{1}{d-k+1},\frac{1}{2}]$, then we have $\alpha =\frac{d}{d+1}=\frac{1}{2}$. This implies that $d=c=1$, a contradiction to the fact that $diam\left(G\right)=3$. Hence, we have $\psi \ne \alpha d-\beta$. This means that:

$$\{{\mu }_1^{k{k}^{\prime }},{\mu }_2^{k{k}^{\prime }},\alpha d-\beta ,\rho ,\psi \}\subset Spe{c}^{+}\left(A_{\alpha }\left(G\right)\right),$$

which indicates $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$.

Case 2. The equation $By=0$ have non-zero solutions.

Let $y\ne 0$ be a vector such that $By=0$. Direct calculations show that:

$$A_{\alpha }\left(G\right)\left[\begin{array}{c}0\hfill \\ y\hfill \end{array}\right]=\alpha k\left[\begin{array}{c}0\hfill \\ y\hfill \end{array}\right],$$

and consequently, we obtain that $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$ with respect to ${(0,y)}^T$.

As desired, we complete the proof of Proposition 5. □

We naturally conclude with some relations of the (restricted) eigenvalues, which are illustrated in the following table.

Proposition 6.

$\eta \left(A_{\alpha }\left(G\right)\right)\ge min\left\{2\right|Res\left(B{B}^T\right)|+1,5\}$.

Proof. 

Suppose ${\gamma }_1,{\gamma }_2,\cdots ,{\gamma }_l$ are the restricted eigenvalues of $B{B}^T$. To complete the proof, we distinguish the following two cases.

Case 1.$\forall i\in \{1,2,\cdots ,l\}$ such that ${\gamma }_i\ne 0$.

Obviously, each ${\gamma }_i$ could produce two different eigenvalues of $A_{\alpha }\left(G\right)$, say ${\mu }_1^{{\gamma }_1},{\mu }_2^{{\gamma }_1}$. Hence, the contributions of ${\gamma }_1,{\gamma }_2,\cdots ,{\gamma }_l$ to $\eta \left(Q\right)$ equal $2l$. By Table 4, $\rho \notin {\cup }_{i=1}^l\{{\mu }_1^{{\gamma }_i},{\mu }_2^{{\gamma }_i}\}$ is also an eigenvalue of $A_{\alpha }\left(G\right)$. Hence, we have $\eta \left(A_{\alpha }\left(G\right)\right)\ge 2\left|Res\left(B{B}^T\right)\right|+1$.

Table 4. Possible relations of (restricted) eigenvalues.

①	$\forall \gamma$	$\gamma \in Res\left(B{B}^T\right)$	$\rho \notin P_{\gamma }$
②	$\forall \gamma$	$\gamma \ne 0$	$\alpha k\notin P_{\gamma }$
③	$\forall \gamma$	$\gamma =k{k}^{\prime }$	$\psi \notin P_{\gamma }$
④	$\forall {\gamma }_1,{\gamma }_2$	${\gamma }_1\ne {\gamma }_2$	$P_{{\gamma }_1}\cap P_{{\gamma }_2}=\varnothing$

Case 2.$\exists i_0\in \{1,2\cdots ,l\}$ such that ${\gamma }_{i_0}=0$.

It follows that the contributions of $\{{\gamma }_1,{\gamma }_2,\cdots ,{\gamma }_l\}\backslash {\gamma }_{i_0}$ to $\eta \left(A_{\alpha }\left(G\right)\right)$ equals $2(l-1)$. For ${\gamma }_{i_0}=0$, by Proposition 5, we know that $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$ or $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$. If $\eta \left(A_{\alpha }\left(G\right)\right)\ge 5$, then we have $\eta \left(A_{\alpha }\left(G\right)\right)\ge min\left\{2\right|Res\left(B{B}^T\right)|+1,5\}$. If $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$, then $A_{\alpha }\left(G\right)$ has two extra distinct eigenvalues $\alpha d-\beta$ and $\rho$, which will not be contained in ${\bigcup }_{i=1,i\ne i_0}^l\{{\mu }_1^{{\gamma }_i},{\mu }_2^{{\gamma }_i}\}$. Hence, $\eta \left(A_{\alpha }\left(G\right)\right)\ge 2(l-1)+3=2\left|Res\left(B{B}^T\right)\right|+1$.

As desired, we complete the proof of Proposition 6. □

Proposition 7.

$rank(A_{\alpha }\left(G\right)-\alpha kI)\le 2c$.

Proof. 

It is routine to check that:

$$A_{\alpha }\left(G\right)-\alpha kI=\left[\begin{array}{cc}\beta J+(\alpha d-\alpha k-\beta )I& \beta B\\ \beta B^T& 0\end{array}\right].$$

Evidently, $\beta J+(\alpha d-\alpha k-\beta )I=\beta J+\left(\alpha \right(d-k+1)I-I$, the eigenvalues of which could be presented by:

$$\left\{\begin{array}{c}\widehat{{\mu }_1}=\beta c+\alpha (d-k+1)-1\hfill \\ \\ \widehat{{\mu }_2}=\alpha (d-k+1)-1\hfill \\ \\ \cdots \cdots \hfill \\ \\ \widehat{{\mu }_c}=\alpha (d-k+1)-1.\hfill \end{array}\right.$$

Clearly, the matrix $\beta J+(\alpha d-\alpha k-\beta )I$ is invertible since the initial hypotheses $\alpha \ne \frac{1}{d-k+1}$. In what follows, we use M to denote the matrix $\beta J+(\alpha d-\alpha k-\beta )I$ of order c.

According to the definition of the Schur complement, we have:

$$rank(A_{\alpha }\left(G\right)-\alpha kI)=rank\left(M\right)+rank\left((A_{\alpha }\left(G\right)-\alpha kI){|}_M\right),$$

where $[A_{\alpha }\left(G\right)-\alpha kI){{|}_M]}_s=-B^T{M}^{-1}B$. Note that ${[J+xI]}^{-1}=\frac{1}{x}I-\frac{1}{x(x+c)}J$, then we have:

$$M^{-1}=\frac{1}{\alpha d-\alpha k-\beta }I-\frac{\beta }{(\alpha d-\alpha k-\beta )(\alpha d-\alpha k-\beta +c\beta )}J$$

and consequently:

$$[A_{\alpha }\left(G\right)-\alpha kI){|}_M]=-\frac{1}{\alpha d-\alpha k-\beta }{\left[B^{T}B-\frac{\beta k^2}{\alpha d-\alpha k-\beta +c\beta }J\right]}_s.$$

Because $B^{T}B$ is $k{k}^{\prime }$-stochastic and $k{k}^{\prime }\ne \frac{s{k}^2\beta }{\alpha d-\alpha k-\beta +c\beta }$, by Lemma 3, we obtain that:

$$rank\left(B^{T}B\right)=rank\left(B^{T}B-\frac{k^2\beta }{\alpha d-\alpha k-\beta +c\beta }J\right),$$

implying that:

$$rank(A_{\alpha }\left(G\right)-\alpha kI){|}_M)=rank\left(B^{T}B\right)=rank\left(B\right)\le c.$$

Hence, $rank(A_{\alpha }\left(G\right)-\alpha kI)\le 2c$. □

Next, we shall prove our main result:

Proof of Proposition 1. 

To complete the proof, we first prove that if G has four distinct eigenvalues, then it must be one of the forms indicated. Note that $\eta \left(G\right)=4$; by Proposition 6, $B{B}^T$ has at most one restricted eigenvalue. Hence, $\left|Res\right(B{B}^T\left)\right|=1$.

In what follows, we are concerned with the following possibilities.

Case 1.$B{B}^T$ is reducible.

According to Lemma 5, one can find that $\rho \left(B{B}^T\right)=k{k}^{\prime }$ is the restricted and unique eigenvalue of $B{B}^T$. It follows from Lemma 1 that $B{B}^T=k{k}^{\prime }I$, which means that $tr\left(B{B}^T\right)=ck{k}^{\prime }$. On the other hand, each of the diagonal elements of $B{B}^T$ equals $k^{\prime }$ since G is a bidegreed split graph. This implies that $tr\left(B{B}^T\right)=c{k}^{\prime }$. Hence, $k=1$.

Let ${Spec}^{+}\left(A_{\alpha }\left(G\right)\right)=\{\rho ,\psi ,{\mu }_1^{k{k}^{\prime }},{\mu }_2^{k{k}^{\prime }}\}$; it follows from Table 4 that $\alpha k\notin {Spec}^{+}\left(A_{\alpha }\left(G\right)\right)$. An easy check shows that $rank(A_{\alpha }\left(G\right)-\alpha kI)=c+s$, otherwise $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$, a contradiction. By Proposition 7, we have $rank(A_{\alpha }\left(G\right)-\alpha kI)=2c$. Accordingly, $s\le c$, and consequently, $s=c$ and $k^{\prime }=1$, since $k^{\prime }=\frac{sk}{c}\le 1$. Thus, $G\cong K_c\circ K_1$.

Case 2.$B{B}^T$ is irreducible.

According to Lemma 5, we know that $\rho \left(B{B}^T\right)=k{k}^{\prime }\notin Res\left(B{B}^T\right)$. Hence, there must exist $\gamma <k{k}^{\prime }$ such that $Res\left(B{B}^T\right)=\left\{\gamma \right\}$. Therefore, $\eta \left(B{B}^T\right)=\left|\{k{k}^{\prime },\gamma \}\right|=2$. By the Perron–Frobenius theorem, $k{k}^{\prime }$ is an algebraically simple eigenvalue of $B{B}^T$, from which we know that the trail of $B{B}^T$ equals $c{k}^{\prime }=k{k}^{\prime }+(c-1)\gamma$. After a simple computation, we readily obtain that $\gamma =k^{\prime }\frac{c-k}{c-1}<k{k}^{\prime }$ and $\gamma >0$. By Lemma 4, we have:

$$B{B}^T=k^{\prime }\frac{k-1}{c-1}J+k^{\prime }\frac{c-k}{c-1}I.$$

For simplicity, let $\lambda =k^{\prime }\frac{k-1}{c-1}$ and $r-\lambda =k^{\prime }\frac{c-k}{c-1}$, then $r=\lambda +(r-\lambda )=k^{\prime }$. Thus, by Lemma 10, we know that B is the incidence matrix of $(c,k,\lambda )$-design $\mathcal{D}$. Hence, $G=G_{\mathcal{D}}$. In addition, $\lambda =k^{\prime }\frac{k-1}{c-1}\ge 0$. If $\lambda =0$, then $k=1$. Hence, $B{B}^T=k{k}^{\prime }I=k^{\prime }I$, which contradicts the fact $\eta \left(B{B}^T\right)=2$. Therefore, we have $\lambda >0$, and consequently, $k\ge 2$.

By Lemma 9, we know that $s>c$. Hence, by Proposition 7, $rank(A_{\alpha }\left(G\right)-\alpha kI)\le 2c<s+c$. It then follows that $\alpha k$ must be an eigenvalue of $A_{\alpha }\left(G\right)$. Accordingly, we know that ${Spec}^{+}\left(G\right)=\{\rho ,\alpha k,{\mu }_1^{\gamma },{\mu }_2^{\gamma }\}$. Note that $\psi$ is also an eigenvalue of $A_{\alpha }\left(G\right)$, which is different from $\rho$ and $\alpha k$, then we have $\psi \in \{{\mu }_1^{\gamma },{\mu }_2^{\gamma }\}$. Hence, the smaller root $\psi$ of the quadratic equation $\left({♣}_2\right)$ is:

$$\psi =\frac{\alpha d+\alpha k+c\beta -\beta -\sqrt{{\Delta }_1}}{2},$$

(3)

which is also a root of:

$$\gamma =\frac{\mu -\alpha k}{{\beta }^2}[\mu -(\alpha d-\beta )].$$

(4)

Combining $\left(3\right)$ and $\left(4\right)$, we have:

$$\frac{1}{{\beta }^2}\left(\frac{\alpha d+\alpha k+c\beta -\beta -\sqrt{{△}_1}}{2}-\alpha k\right)\left(\frac{\alpha d+\alpha k+c\beta -\beta -\sqrt{{△}_1}}{2}-\alpha d+\beta \right)-k^{\prime }\frac{c-k}{c-1}=0,$$

(5)

which is equivalent to the following:

$${\alpha }^2{t}_1+\alpha t_2+t_3=0,$$

where:

$$\left({♣}_6\right)\left\{\begin{array}{c}{t}_1=4c^2({\lambda }^2-r)+8[ck(d-k)+2k^2]-2c[c(d-k)-4k](2\lambda -1)-2c^2(d-k)\hfill \\ \\ t_2=-8c^2({\lambda }^2-r)+2c[c(d-k)-4k](2\lambda -1)+2c^2(d-k)\hfill \\ \\ t_3=4c^2({\lambda }^2-r).\hfill \end{array}\right.$$

Next, we shall verify the sufficiency of Proposition 1.

If $G=K_c\circ K_1$ for $c\ge 2$, it follows that:

$$A_{\alpha }\left(G\right)=\left[\begin{array}{cc}\beta J+(\alpha c-\beta )I& \beta I\\ \beta I& \alpha I\end{array}\right].$$

Direct calculations show that it has four distinct eigenvalues:

$$\left\{\begin{array}{c}\widetilde{{\mu }_1}=\frac{c\alpha +c\beta +\alpha -\beta -\sqrt{{△}_3}}{2}\hfill \\ \\ \widetilde{{\mu }_2}=\frac{c\alpha +c\beta +\alpha -\beta +\sqrt{{△}_3}}{2}\hfill \\ \\ \widetilde{{\mu }_3}=\frac{c\alpha +\alpha -\beta -\sqrt{{△}_4}}{2}\hfill \\ \\ \widetilde{{\mu }_4}=\frac{c\alpha +\alpha -\beta +\sqrt{{△}_4}}{2},\hfill \end{array}\right.$$

where:

$${△}_3={(\beta -c\alpha -c\beta -\alpha )}^2-4(c\alpha \beta +c{\alpha }^2-\alpha \beta -{\beta }^2)>0$$

and:

$${△}_4={(\beta -c\alpha -\alpha )}^2-4(c{\alpha }^2-\alpha \beta )>0.$$

If $G=G_{\mathcal{D}}$ for a $(c,s,r,k,\lambda )$-design $\mathcal{D}$ such that $\mathcal{D}$ has at least one pair of disjoint blocks, then we know that the diameter of G is equal to three. By Lemma 9, we have $s>c$. Hence, $rank(A_{\alpha }\left(G\right)-\alpha kI)\le 2c<s+c$ according to Proposition 7, implying that $\alpha k$ is an eigenvalue of $A_{\alpha }\left(G\right)$.

By Lemma 10 and the definition of the bidegreed split graph, we have $B{B}^T=\lambda J+(r-\lambda )I$ and $r=\lambda \frac{c-1}{k-1}=k^{\prime }$. Hence, $\lambda =k^{\prime }\frac{k-1}{c-1}$ and $r-\lambda =k^{\prime }\frac{c-k}{c-1}$. Consequently, we have that each element of $B{B}^T$ is positive since $\lambda >0$ and $r-\lambda >0$. It follows from Definition 1.4 that $B{B}^T$ is irreducible. Direct calculations show that the eigenvalues of $B{B}^T$ are $r+(c-1)\lambda ,r-\lambda ,\dots ,r-\lambda$. Hence, $r+(c-1)\lambda =k{k}^{\prime }$ is the Perron eigenvalue of $B{B}^T$.

By Lemma 5, we have $k{k}^{\prime }\notin Res\left(B{B}^T\right)$, which yields that $r-\lambda$ is the unique restricted eigenvalue of $B{B}^T$. To complete the proof, we let $P_{r-\lambda }=\{{\mu }_1^{r-\lambda },{\mu }_2^{r-\lambda }\}$. Note that ${\alpha }^2{t}_1+\alpha t_2+t_3=0$, then we obtain that $\psi \in P_{r-\lambda }$. Hence, $Spe{c}^{+}\left(A_{\alpha }\left(G\right)\right)=\{\rho ,\alpha k,{\mu }_1^{r-\lambda },{\mu }_2^{r-\lambda }\}$, as desired $\eta \left(A_{\alpha }\left(G\right)\right)=4$.

This completes the proof of Proposition 1. □