Extreme and statistical properties of eigenvalue indices of simple connected graphs

The adjacency matrix $A$ of $K_{m_{1},\dots,m_{k}}$ has the block form:

$$A={\bf 1}{\bf 1}^{T}-diag(D_{1},\dots,D_{k}),$$

where ${\bf 1}=(1,\dots,1)^{T}\in\mathbb{R}^{m}$, and $D_{i}$ is the $m_{i}\times m_{i}$ matrix consisting of ones. Now, if $\lambda$ is a nonzero eigenvalue of $A$ with an eigenvector $x=(x_{1},\dots,x_{m})^{T}$ then

$$\alpha-\alpha_{p}=\lambda x_{l},\quad\text{for each}\ \ l=\mu_{p-1}+1,\dots,\mu_{p},\ \quad\alpha_{p}=\sum_{l=1+\mu_{p-1}}^{\mu_{p}}x_{l},\quad\mu_{p}=\sum_{r=1}^{p}m_{r},$$

(2)

for $p=1,\dots,k$. Here $\alpha=\sum_{p=1}^{k}\alpha_{p}=\sum_{j=1}^{m}x_{j}$. For example, if $p=1$ then $\sum_{j=1}^{m_{1}}x_{j}=\alpha m_{1}/(\lambda+m_{1})$ provided that $\lambda\not=-m_{1}$. Similarly, we can proceed with the remaining parts $m_{2},\dots,m_{k}$. In the case $\alpha=0$ we have $\lambda\in\{-m_{1},\dots,-m_{k}\}$. In the case $\alpha\not=0$ we conclude $\lambda\not\in\{-m_{1},\dots,-m_{k}\}$, and the eigenvalue $\lambda$ satisfies the rational equation:

$$\psi(\lambda)=1,\quad\text{where}\ \psi(\lambda)=\sum_{i=1}^{k}\frac{m_{i}}{\lambda+m_{i}}.$$

(3)

Conversely, if $\lambda\not\in\{-m_{1},\dots,-m_{k}\}$ satisfies $\psi(\lambda)=1$ then it is easy to verify that the nontrivial vector $x\in\mathbb{R}^{m}$,

$$x=(\underbrace{y_{1},\dots,y_{1}}_{m_{1}\ \text{times}},\underbrace{y_{2},\dots,y_{2}}_{m_{2}\ \text{times}},\dots,\underbrace{y_{k},\dots,y_{k}}_{m_{k}\ \text{times}})^{T},\quad\text{where}\ y_{i}=\frac{m_{i}}{\lambda+m_{i}},$$

is an eigenvector of $A$, i.e. $Ax=\lambda x$.

In what follows, we shall derive necessary bounds on eigenvalues of $A$. Suppose to the contrary that $\lambda<-m_{k}$ is an eigenvalue of $A$. Then $\lambda+m_{i}\leq\lambda+m_{k}<0$ for any $i=1,\dots,k$, and so $\psi(\lambda)<0<1$. Therefore, $\lambda\geq-m_{k}$ for any eigenvalue $\lambda\in\sigma(A)$. To derive an upper bound for the positive eigenvalue of $A$ we introduce an auxiliary function $\phi(\xi_{1},\dots,\xi_{k})=\sum_{i=1}^{k}\frac{\xi_{i}}{\lambda+\xi_{i}}$ where $\lambda>0$ is fixed. The function $\phi:\mathbb{R}^{k}\to\mathbb{R}$ is concave. Using the Lagrange function $\mathscr{L}(\xi,\mu)=\phi(\xi_{1},\dots,\xi_{k})-\mu\sum_{i=1}^{k}\xi_{i}$ it is easy to verify that $\phi$ achieves the unique constrained maximum in the set $\{\xi\in\mathbb{R}^{k},\sum_{i=1}^{k}\xi_{i}=m\}$ at the point $\hat{\xi}=(m/k,\dots,m/k)^{T}$. Therefore, for any $\lambda>0$ we have

$$\psi(\lambda)=\sum_{i=1}^{k}\frac{m_{i}}{\lambda+m_{i}}=\phi(m_{1},\dots,m_{k})\leq\phi(m/k,\dots,m/k)=\frac{m}{\lambda+m/k}.$$

If $\lambda>0$ is a positive eigenvalue of $A$ then $\psi(\lambda)=1$ and so $\lambda+m/k\leq m$, that is, $0<\lambda\leq m-m/k$. Therefore, $\sigma(A)\subset[-m_{k},m-m/k]$.

In the trivial case of an equipartite graph $K_{m_{1},\dots,m_{k}}$ with $m_{1}=\dots=m_{k}=m/k$ we obtain $\lambda_{-}(A)\geq-m_{k}=-m/k$ and $\lambda_{+}(A)\leq m-m/k$. Thus, $\Lambda^{gap}\leq m$, and $\Lambda^{ind}\leq m-m/k\leq m-1$. This estimate also follows from the results of [17] and [15]. Therefore, for any $1\leq l<k$ we conclude that $\Lambda^{gap}(A)=\lambda_{+}(A)-\lambda_{-}(A)\leq m-m/k-(-m/k)=m$. Similarly, $\Lambda^{ind}(A)\leq m-1$.

Now, consider a non-equipartite graph $K_{m_{1},\dots,m_{k}}$ with $m_{1}=\dots=m_{l}<m_{l+1}\leq\dots\leq m_{k}$ where $1\leq l<k$. Suppose that $l=1$, that is, $1\leq m_{1}<m_{2}\leq\dots\leq m_{k}$. The function $\psi$ is strictly decreasing in the interval $(-m_{2},-m_{1})$ with infinite limits $\pm\infty$ when $\lambda\to-m_{2}$ and $\lambda\to-m_{1}$, respectively. Therefore, there exists a unique eigenvalue $\lambda\in(-m_{2},-m_{1})$ of the matrix $A$. We have $m_{1}+(k-1)m_{2}\leq\sum_{i=1}^{k}m_{i}=m$. Define $\tilde{\lambda}=-m_{1}/k-m_{2}(k-1)/k$. Then $\tilde{\lambda}\geq-m/k$. In what follows we shall prove that $\psi(\tilde{\lambda})\geq 1$. The function $\xi\mapsto\xi/(\tilde{\lambda}+\xi)$ decreases for $\xi>-\tilde{\lambda}$. Therefore

	$\displaystyle\psi(\tilde{\lambda})$	$\displaystyle\geq$	$\displaystyle\frac{m_{1}}{\tilde{\lambda}+m_{1}}+(k-1)\frac{m_{2}}{\tilde{\lambda}+m_{2}}=-\frac{k}{k-1}\frac{m_{1}}{m_{2}-m_{1}}+k(k-1)\frac{m_{1}}{m_{2}-m_{1}}$
		$\displaystyle=$	$\displaystyle\frac{k}{k-1}\frac{(k-1)^{2}m_{2}-m_{1}}{m_{2}-m_{1}}\geq\frac{k}{k-1}>1,$

because $k\geq 2$. Since $\psi$ is strictly decreasing in the interval $(-m_{2},-m_{1})$ we have $-m/k\leq\tilde{\lambda}<\lambda$ because $\psi(\lambda)=1$.

In the case $l\geq 2$ we can apply a simple perturbation argument. Indeed, let us perturb the adjacency matrix $A$ by a small parameter $0<\varepsilon\ll 1$ as follows:

$$A^{\varepsilon}={\bf 1}{\bf 1}^{T}-diag((1-\varepsilon)D_{1},D_{2},\dots,D_{l-1},(1+\varepsilon)D_{l},D_{l+1},\dots,D_{k}).$$

It corresponds to the perturbation $m_{1}^{\varepsilon}=(1-\varepsilon)m_{1},m_{l}^{\varepsilon}=(1+\varepsilon)m_{l}$. All remaining $m_{i}$ remain unchanged for $i\not=1$ and $i\not=l$. Then for the corresponding perturbed function $\psi^{\varepsilon}$ there exists a solution $\lambda^{\varepsilon}\in(m_{1}-\varepsilon,m_{1})$ of the equation $\psi^{\varepsilon}(\lambda^{\varepsilon})=1$. Since the spectrum of $A^{\varepsilon}$ depends continuously on the parameter $\varepsilon\to 0$, we see that $\lambda^{\varepsilon}\to\lambda=-m_{1}=\dots=-m_{l}$ is an eigenvalue of the graph $G_{A}$ provided that $l\geq 2$. In this case $\lambda=-m_{1}\geq-m/k$.

A complete multipartite graph $G_{A}=K_{m_{1},m_{2},\dots,m_{k}}$ has exactly one positive eigenvalue $\lambda_{1}>0$ (cf. Smith [14]). Since $\sum_{i=1}^{m}\lambda_{i}=0$ we have $\Lambda^{pow}(A)=\sum_{i=1}^{m}|\lambda_{i}|=2\lambda_{1}\leq 2(m-m/k)$. The spectrum of the complete graph $K_{m}$ consists of eigenvalues $m-1$, and $-1$ with multiplicity $m-1$. Therefore, $\Lambda^{gap}=m,\Lambda^{ind}=m-1$, as claimed. ∎

According to Smith [14], a simple connected graph has exactly one positive eigenvalue (i.e. $\lambda_{2}(A)\leq 0$) if and only if it is a complete multipartite graph $K_{m_{1},\dots,m_{k}}$ where $1\leq m_{1}\leq\dots\leq m_{k}$ denotes the sizes of parts, $m_{1}+\dots+m_{k}=m$, and $k\geq 2$ is the number of parts (see [14, Theorem 6.7]).

To prove a), let us consider a graph $G_{A}$ different from any complete multipartite graph $K_{m_{1},\dots,m_{k}}$. Therefore, $\lambda_{2}(A)>0$. We combine this information with the result due to D. Powers regarding the second largest eigenvalue $\lambda_{2}(A)$. According to [38] (see also [39], [40]), for a simple connected graph $G_{A}$ on $m$ vertices we have the following estimate for the second largest eigenvalue $\lambda_{2}(A)$:

$$-1\leq\lambda_{2}(A)\leq\lfloor m/2\rfloor-1$$

(see also Cvetković and Simić [13]). Since $\lambda_{2}(A)>0$ we have $0<\lambda_{+}(A)\leq\lambda_{2}(A)\leq\lfloor m/2\rfloor-1$, and $-\sqrt{\lfloor m/2\rfloor\lceil m/2\rceil}\leq\lambda_{min}(A)\leq\lambda_{-}(A)<0$. Hence the spectral gap $\Lambda^{gap}=\lambda_{+}(A)-\lambda_{-}(A)\leq\sqrt{\lfloor m/2\rfloor\lceil m/2\rceil}+\lfloor m/2\rfloor-1$. If $m$ is even, it leads to the estimate $\Lambda^{gap}\leq m-1$. If $m$ is odd, then it is easy to verify $\Lambda^{gap}\leq m-3/2$. Analogously, $\Lambda^{ind}\leq m/2$ if $m$ is even, and $\Lambda^{ind}\leq\sqrt{m^{2}-1}/2$ if $m$ is odd.

The part b) is contained in Section 3 dealing with statistical properties of eigenvalue indices. ∎

Without loss of generality, we may assume that the adjacency matrix $A$ of the graph $K_{m,m}^{-e}$ has the form

$$A=\left(\begin{array}[]{cc}0&\mathbf{1}\mathbf{1}^{T}\\ \mathbf{1}\mathbf{1}^{t}&0\end{array}\right)-\left(\begin{array}[]{c}0\\ e_{1}\end{array}\right)(e_{1},0)-\left(\begin{array}[]{c}e_{1}\\ 0\end{array}\right)(0,e_{1}),$$

where $\mathbf{1}=(1,\dots,1)^{T},e_{1}=(1,0,\dots,0)^{T}\in\mathbb{R}^{m}$. Assume that $\lambda$ is an eigenvalue of $A$, and $(0,0)\not=(x,y)\in\mathbb{R}^{m}\times\mathbb{R}^{m}$ is an eigenvector. Denote $\alpha=\sum_{i=1}^{m}x_{i},\ \beta=\sum_{i=1}^{m}y_{i}$. Then

$$\beta-y_{1}=\lambda x_{1},\quad\alpha-x_{1}=\lambda y_{1},\quad\beta=\lambda x_{i},\quad\alpha=\lambda y_{i},\quad i=2,\dots,m.$$

Assuming $\lambda=\pm 1$ leads to an obvious contradiction, as it implies $\alpha=\beta=0$, and $x=0,y=0$. The matrix $A$ has zero eigenvalue $\lambda=0$, with $2(m-1)$ dimensional eigenspace $\{(x,y)\in\mathbb{R}^{m}\times\mathbb{R}^{m},x_{1}=y_{1}=0\}$. Therefore, for $\lambda\not=\pm 1,0$ we have $x_{1}=(\alpha-\beta\lambda)/(1-\lambda^{2}),\ y_{1}=(\beta-\alpha\lambda)/(1-\lambda^{2})$, and $x_{2}=\beta/\lambda,y_{i}=\alpha/\lambda,,\quad i=2,\dots,m$. It results in a system of two linear equations for $\alpha,\beta$:

$$\alpha=\frac{m-1}{\lambda}\beta+\frac{\alpha-\beta\lambda}{1-\lambda^{2}},\quad\beta=\frac{m-1}{\lambda}\alpha+\frac{\beta-\alpha\lambda}{1-\lambda^{2}},$$

which has a non-trivial solution $(\alpha,\beta)\not=(0,0)$ provided that $\lambda\not=\pm 1,0,$ is a solution of the following dispersion equation:

$$\left(\frac{1}{1-\lambda^{2}}-1\right)^{2}-\left(\frac{m-1}{\lambda}-\frac{\lambda}{1-\lambda^{2}}\right)^{2}=0.$$

After rearranging terms, $\lambda$ is a solution of the cubic equation

$$\pm\lambda^{3}+m\lambda^{2}-m+1=0,$$

having roots $\mp 1$ (which are not eigenvalues of $A$), and four other roots $\lambda^{\pm,\pm}$ given as in (4), as claimed. The rest of the proof easily follows. ∎

It is similar to the proof of the previous Proposition 3. Arguing similarly as before, one can show that $\lambda^{(4)}=-1$ is an eigenvalue with multiplicity one. The other nonzero eigenvalues are roots of the cubic equation $\lambda^{2}(1-\lambda)-m(m-2-m\lambda)=0$ which can be transformed into a depressed cubic equation with a positive discriminant $\Delta$. Thus, it has three distinct real eigenvalues $\lambda^{(1),(2),(3)}$. Performing the standard asymptotic analysis, we conclude $\lambda_{+}(A)=\lambda^{(2)}=1-2/m-2/m^{3}+O(m^{-4})$ as $m\to\infty$, as claimed. ∎

Recall the known fact (see, e.g. [16]) that the set of fractional parts $\sqrt{m}-[\sqrt{m}]$ of roots of all positive integers $m$ is dense in the interval $[0,1)$. Hence, there exists an integer $m_{2}$, such that $\sqrt{\delta}\leq\sqrt{m_{2}}-[\sqrt{m_{2}}]\leq\sqrt{\gamma}$. Take $m_{1}:=[\sqrt{m_{2}}]^{2}\leq m_{2}$. Then $\sqrt{\delta}\leq\sqrt{m_{2}}-\sqrt{m_{1}}\leq\sqrt{\gamma}$. By squaring and rearranging terms, we obtain $(m_{1}+m_{2})-c\leq 2\sqrt{m_{1}m_{2}}\leq(m_{1}+m_{2})-d$. Now we take the bipartite graph $K_{m_{1},m_{2}}$, of order $m=m_{1}+m_{2}$. Since $\Lambda^{gap}(K_{m_{1},m_{2}})=2\sqrt{m_{1}m_{2}}$ the claim follows. ∎

Let $G_{A}$ be a bipartite but not complete bipartite graph with adjacency matrix $A$ having null space of dimension $k$. Since $G_{A}$ is not complete bipartite, we have $k\leq m-4$. It follows that $m$ and $k$ have the same parity, so that $m-k=2r$ for some positive integer $r\geq 2$. By bipartiteness of $G_{A}$ we may assume that its eigenvalues have the form $\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{r}>0=\lambda_{r+1}=\ldots=\lambda_{r+k}>-\lambda_{r}\geq\ldots\geq-\lambda_{2}\geq-\lambda_{1}$, so that $\lambda_{+}=\lambda_{r}$ and $\lambda_{-}=-\lambda_{r}$. The earlier used fact that $\lambda_{1}\geq d$ trivially implies that

$$\sum_{i=1}^{r}\lambda_{i}^{2}\geq d^{2}+(r-1)\lambda_{+}^{2}\ \ .$$

(6)

It is well known that the sum of squares $\sum_{i=1}^{m}\lambda_{i}^{2}=trace(A^{2})=md$, where $d$ is the average valency of vertices of $G_{A}$, that is, $md/2$ is the number of edges in the graph $G_{A}$ (cf. Bapat [5]). Combined with the inequality $\lambda_{1}\geq d$ used earlier, we obtain

$$md=2\sum_{i=1}^{r}\lambda_{i}^{2}\geq 2d^{2}+2(r-1)\lambda_{+}^{2}=2d^{2}+(m-k-2)\lambda_{+}^{2}$$

(7)

and evaluation of $\lambda_{+}(A)$ from (7) gives $\lambda_{+}(A)=-\lambda_{-}(A)\leq\sqrt{\frac{d(m-2d)}{m-k-2}}$ which implies the inequality (5) in our statement. ∎

The sum of all eigenvalues of the symmetric matrix $A$ is zero because the trace of $A$ is zero. Hence $\lambda_{min}(A)<0<\lambda_{max}(A)$. Let $\mathscr{A}$ be the $(m+1)\times(m+1)$ adjacency matrix of the graph $G_{\mathscr{A}}$ obtained from $G_{A}$ by adding a vertex connected to a subset of vertices of $G_{A}$. Its adjacency matrix $\mathscr{A}$ has the block form

$$\mathscr{A}=\left(\begin{array}[]{cc}A&e\\ e^{T}&0\end{array}\right),$$

(8)

where $e=(e_{1},\dots,e_{m})^{T}$, $e_{i}\in\{0,1\}$. The maximal eigenvalue $\lambda_{max}(\mathscr{A})$ can be computed by means of the Rayleigh ratio, i.e.

$$\lambda_{max}(\mathscr{A})=\max_{x\in\mathbb{R}^{m},\xi\in\mathbb{R}}\frac{(x^{T},\xi)\left(\begin{array}[]{cc}A&e\\ e^{T}&0\end{array}\right)\left(\begin{array}[]{c}x\\ \xi\end{array}\right)}{|x|^{2}+\xi^{2}}=\max_{x\in\mathbb{R}^{m},\xi\in\mathbb{R}}\frac{x^{T}Ax+2(e^{T}x)\xi}{|x|^{2}+\xi^{2}},$$

where $|x|$ is the Euclidean norm of the vector $x$. Let $\hat{x}$ be an eigenvector for corresponding to the maximal eigenvalue $\lambda_{max}(A)$, that is, $A\hat{x}=\lambda_{max}(A)\hat{x}$. Then

$$\lambda_{max}(\mathscr{A})\geq\max_{\xi\in\mathbb{R}}\frac{\lambda_{max}(A)+2(e^{T}\hat{x})\xi}{1+\xi^{2}}=\lambda_{max}(A)\max_{\xi\in\mathbb{R}}\frac{1+\alpha\xi}{1+\xi^{2}},$$

where $\alpha=2(e^{T}\hat{x})/\lambda_{max}(A)$. Let us introduce the auxiliary function $\psi:\mathbb{R}\to\mathbb{R}$, $\psi(\xi)=(1+\alpha\xi)/(1+\xi^{2})$, where $\alpha\in\mathbb{R}$ is a parameter. Using the first-order necessary condition it is easy to verify that the maximum of the function $\psi$ is attained at $\xi=(-1+\sqrt{1+\alpha^{2}})/\alpha$. As a consequence, we have

$$\max_{\xi}\frac{1+\alpha\xi}{1+\xi^{2}}=\frac{1+\sqrt{1+\alpha^{2}}}{2}>0.$$

Notice that the adjacency matrix contains only nonnegative elements. With regard to the Perron-Frobenius theorem, an eigenvector corresponding to the maximal eigenvalue $\lambda_{max}(A)$ is nonnegative, i.e. $\hat{x}\geq 0$. Consider the vector $e=(1,\dots,1)^{T}$ consisting of ones. It corresponds to the new vertex connected to all the vertices of $G_{A}$. Then $(e^{T}\hat{x})^{2}=(\hat{x}_{1}+\dots+\hat{x}_{m})^{2}\geq|\hat{x}|^{2}=1$ because all $\hat{x}_{i}\geq 0$ are nonnegative. Inserting the parameter $\alpha^{2}=4(e^{T}\hat{x})^{2}/(\lambda_{max}(A))^{2}\geq 4/(\lambda_{max}(A))^{2}$ we obtain $\lambda_{max}(\mathscr{A})\geq\frac{1}{2}(\lambda_{max}(A)+\sqrt{(\lambda_{max}(A))^{2}+4})$, as claimed.