**3. Bounds of** *ρ***(***RDα***(***G***)) of Graphs**

In this section, we find bounds of the spectral radius of generalizes reciprocal distance matrix in terms of parameters associated with the structure of the graph.

Let **e** be the *n*-dimensional vector of ones.

**Theorem 1.** *Let G be a graph with reciprocal distance degree sequence* {*RT*1, *RT*2, ... , *RTn*}*. Then*

$$
\rho(RD\_n(G)) \ge \sqrt{\frac{RT\_1^2 + RT\_2^2 + \dots + RT\_n^2}{n}}.
$$

*The equality holds if and only if G is a reciprocal distance degree regular graph.*

**Proof.** Let **x** = [*x*1, *x*2, ... , *xn*] *<sup>T</sup>* be the unit positive Perron eigenvector of *RDα*(*G*) corresponding to *ρ*(*RDα*(*G*)). We take the unit vector **y** = <sup>√</sup><sup>1</sup> *<sup>n</sup>* **e**. Then, we have

$$\rho(RD\_{\mathfrak{a}}(G)) = \sqrt{\rho^2(RD\_{\mathfrak{a}}(G))} = \sqrt{\mathbf{x}^T(RD\_{\mathfrak{a}}(G))^2 \mathbf{x}} \ge \sqrt{\mathbf{y}^T(RD\_{\mathfrak{a}}(G))^2 \mathbf{y}}.\tag{2}$$

Since (*RDα*(*G*))**y** = <sup>√</sup><sup>1</sup> *<sup>n</sup>* [*RT*1, *RT*2,..., *RTn*] *<sup>T</sup>*, we obtain

$$\mathbf{y}^T (RD\_\mathbf{a}(G))^2 \mathbf{y} = \frac{RT\_1^2 + RT\_2^2 + \dots + RT\_n^2}{n}.$$

Therefore,

$$
\rho(RD\_a(G)) \ge \sqrt{\frac{RT\_1^2 + RT\_2^2 + \dots + RT\_n^2}{n}}.
$$

Now, assume that the equality holds. By Equation (2), we have that **y** is the positive eigenvector corresponding to *ρ*(*RDα*(*G*)). From *RDα*(*G*)**y** = *ρ*(*RDα*(*G*))**y**, we obtain that *RTi* = *ρ*(*RDα*(*G*)), for *i* = 1, 2, ... , *n*. Therefore, graph *G* is a reciprocal distance degree regular graph.

Conversely, if *G* is a reciprocal distance degree regular graph, then *RT*<sup>1</sup> = *RT*<sup>2</sup> = ··· = *RTn* = *k*. From Lemma 2, *k* = *ρ*(*RDα*(*G*)). So

$$
\rho(RD\_a(G)) = k = \sqrt{\frac{nk^2}{n}} = \sqrt{\frac{RT\_1^2 + RT\_2^2 + \dots + RT\_n^2}{n}}.
$$

The equality holds.

**Theorem 2.** *Let G be a graph with reciprocal distance degree sequence* {*RT*1, *RT*2, ... , *RTn*} *and second reciprocal distance degree sequence* {*T*1, *T*2,..., *Tn*}*. Then*

$$\rho(RD\_n(G)) \ge \sqrt{\frac{\left(aRT\_1^2 + (1-a)T\_1\right)^2 + \left(aRT\_2^2 + (1-a)T\_2\right)^2 + \dots + \left(aRT\_n^2 + (1-a)T\_n\right)^2}{\sum\_{i=1}^n RT\_i^2}}.$$

*The equality holds if and only if G is a pseudo reciprocal distance degree regular graph.*

\*\*Proof.\*\* Using  $\mathbf{y} = \frac{1}{\sqrt{\sum\_{i=1}^{n} RT\_i^2}} [RT\_1, RT\_2, \dots, RT\_n]^T$ , the proof is similar to Theorem 1.  $\bigtriangledown\_{\mathbf{y}} = \mathbf{x}$ 

**Remark 1.** *The lower bound given in Theorem 2 improves the bound given in Theorem 1, and the bound given in Theorem 1 improves the bound given in Lemma 3.*

*In fact, from Lemma 1, we have <sup>n</sup>* ∑ *i*=1 *Ti* <sup>=</sup> *<sup>n</sup>* ∑ *i*=1 *RT*<sup>2</sup> *<sup>i</sup>* . *By Cauchy–Schwarz inequality*

$$\begin{split} n \sum\_{i=1}^{n} (aRT\_i^2 + (1-a)T\_i)^2 &\geq \left(\sum\_{i=1}^{n} (aRT\_i^2 + (1-a)T\_i)\right)^2 \\ &= (a \sum\_{i=1}^{n} RT\_i^2 + (1-a) \sum\_{i=1}^{n} T\_i)^2 \\ &= (a \sum\_{i=1}^{n} RT\_i^2 + (1-a) \sum\_{i=1}^{n} RT\_i^2)^2 \\ &= (\sum\_{i=1}^{n} RT\_i^2)^2. \end{split}$$

*Moreover, we recall that, n <sup>n</sup>* ∑ *i*=1 *RT*<sup>2</sup> *<sup>i</sup>* ≥ ( *n* ∑ *i*=1 *RTi*)2*. Thus*

$$\sqrt{\frac{\sum\_{i=1}^{n}(\alpha RT\_i^2 + (1-\alpha)T\_i)^2}{\sum\_{i=1}^{n}RT\_i^2}} \ge \sqrt{\frac{\overline{(\sum\_{i=1}^{n}RT\_i^2)^2}{n\sum\_{i=1}^{n}RT\_i^2}} = \sqrt{\frac{\overline{\sum\_{i=1}^{n}RT\_i^2}{n}}{n}}$$

*and*

$$\sqrt{\frac{\sum\_{i=1}^{n} RT\_i^2}{n}} \ge \sqrt{\frac{(\sum\_{i=1}^{n} RT\_i)^2}{n^2}} = \frac{2H(G)}{n}.$$

**Theorem 3.** *Let G be a graph with reciprocal distance degree sequence* {*RT*1, *RT*2, ... , *RTn*} *and second reciprocal distance degree sequence* {*T*1, *T*2,..., *Tn*}*. Then*

$$\min\_{1 \le i \le n} \left\{ \sqrt{(1 - a)T\_i + a(RT\_i)^2} \right\} \le \rho(RD\_a(\mathcal{G})) \le \max\_{1 \le i \le n} \left\{ \sqrt{(1 - a)T\_i + a(RT\_i)^2} \right\}.$$

**Proof.** Let *RDα*(*G*)=(*bi*,*j*). Then (*RDα*(*G*))<sup>2</sup> *<sup>i</sup>*,*<sup>j</sup>* <sup>=</sup> *<sup>n</sup>* ∑ *k*=1 *bi*,*kbk*,*j*, and the row sum of (*RDα*(*G*))<sup>2</sup> should be

$$S\_i((RD\_a(G))^2) = \sum\_{j=1}^n \sum\_{k=1}^n b\_{i,k} b\_{k,j} = \sum\_{k=1}^n (b\_{i,k} \sum\_{j=1}^n b\_{k,j}) = \sum\_{k=1}^n (b\_{i,k}RT\_k) \dots$$

Hence, *Si*((*RDα*(*G*))2)=(1 − *α*)*Ti* + *αRT*<sup>2</sup> *i* .

Now, let **x** be the unit Perron vector corresponding to *ρ*(*RDα*(*G*)). Clearly, *RDα*(*G*)**x** = *ρ*(*RDα*(*G*))**x** and (*RDα*(*G*))2**x** = (*ρ*(*RDα*(*G*)))2**x**. By Lemma 4, we have

$$\min\_{1 \le i \le n} \left\{ (1 - a)T\_i + a(RT\_i)^2 \right\} \le (\rho(RD\_a(G)))^2 \le \max\_{1 \le i \le n} \left\{ (1 - a)T\_i + a(RT\_i)^2 \right\}.$$

Thus

$$\min\_{1 \le i \le n} \left\{ \sqrt{(1 - a)T\_i + a(RT\_i)^2} \right\} \le \rho(RD\_a(G)) \le \max\_{1 \le i \le n} \left\{ \sqrt{(1 - a)T\_i + a(RT\_i)^2} \right\}.$$

**Theorem 4.** *Let G be a graph with n vertices, RTmax and Tmax be the maximum reciprocal distance degree and the maximum second reciprocal distance degree of G, respectively. Then*

$$
\rho(RD\_a(G)) \le \frac{\alpha RT\_{\max} + \sqrt{(\alpha RT\_{\max})^2 + 4(1 - \alpha)T\_{\max}}}{2}.
$$

*The equality holds if and only if G is a reciprocal distance degree regular graph.*

**Proof.** Since *RDα*(*G*) = *αRT*(*G*)+(1 − *α*)*RD*(*G*), 0 ≤ *α* ≤ 1, it can be obtained by simple calculation

$$S\_i(RD\_\pi(G)) = RT\_{i\prime}$$

$$S\_i((RT(G))^2) = S\_i(RT(G)RD(G)) = RT\_{i\prime}^2$$

$$S\_i((RD(G))^2) = S\_i(RD(G)RT(G)) = T\_i.$$

Then

$$\begin{aligned} S\_l(\left(RD\_{\mathfrak{a}}(G)\right)^2) &= S\_l(\operatorname{a}^2(RT(G))^2 + \mathfrak{a}(1-\mathfrak{a})RT(G)RD(G) \\ &+ \mathfrak{a}(1-\mathfrak{a})RD(G)RT(G) + (1-\mathfrak{a})^2RD(G)^2) \\ &= S\_l(\operatorname{a}RT(G)(\operatorname{a}RT(G) + (1-\mathfrak{a})RD(G))) \\ &+ \mathfrak{a}(1-\mathfrak{a})S\_l(\operatorname{RD}(G)RT(G)) + (1-\mathfrak{a})^2S\_l(\operatorname{RD}(G)^2) \\ &= \operatorname{a}RT\_lS\_l(\operatorname{RD}\_{\mathfrak{a}}(G)) + (1-\mathfrak{a})T\_l \\ &\leq \operatorname{a}RT\_{\operatorname{max}}S\_l(\operatorname{RD}\_{\mathfrak{a}}(G)) + (1-\mathfrak{a})T\_{\operatorname{max}}. \end{aligned}$$

that is,

$$S\_i(\left(R D\_\mathfrak{a}(G)\right)^2 - \mathfrak{a} R T\_{\max} R D\_\mathfrak{a}(G)) \le (1 - \mathfrak{a}) T\_{\max}.$$

By Lemma 4,

$$
\rho^2(RD\_a(G)) - \alpha RT\_{\max} \rho(RD\_a(G)) - (1 - a)T\_{\max} \le 0.
$$

For any vertex *vi*, when the inequality is equal, *RTi* = *RTmax*, *Ti* = *Tmax*. That is, *G* is a reciprocal distance degree regular graph.

On the contrary, when *G* is a reciprocal distance degree regular graph, the inequality is equal.

**Theorem 5.** *Let G be a graph with n vertices, RTmin and Tmin be the minimum reciprocal distance degree and the minmum second reciprocal distance degree of G, respectively. Then*

$$
\rho(RD\_a(G)) \ge \frac{\alpha RT\_{\min} + \sqrt{(\alpha RT\_{\min})^2 + 4(1-\alpha)T\_{\min}}}{2}.
$$

*The equality holds if and only if G is a reciprocal distance degree regular graph.*

**Proof.** The method is the same as Theorem 4.

**Theorem 6.** *Let G be a graph with reciprocal distance degree sequence* {*RT*1, *RT*2, ... , *RTn*} *and second reciprocal distance degree sequence* {*T*1, *T*2,..., *Tn*}*. Then*

$$\rho(RD\_n(G)) \le \max\_{1 \le i,j \le n} \left\{ \frac{a(RT\_i + RT\_j) + \sqrt{a^2(RT\_i - RT\_j)^2 + 4(1 - a)^2 \frac{T\_i T\_j}{RT\_i RT\_j}}}{2} \right\}.\tag{3}$$

*The equality holds if and only if G is a reciprocal distance degree regular graph.*

**Proof.** Let **x** = (*x*1, *x*2, ... , *xn*) be the eigenvector corresponding to the eigenvalue *ρ*(*G*) of the matrix *RT*(*G*)−1*RDα*(*G*)*RT*(*G*), *xs* = max{*xi*|*i* = 1, 2, ... , *n*}, *xt* = max{*xi*|*xi* = *xs*, *i* = 1, 2, . . . , *n*}.

Through simple calculation, the value of the (*i*, *j*)-th element of *RT*(*G*)−1*RDα*(*G*)*RT*(*G*)is

$$ \begin{cases} \alpha RT\_{i\prime} & \text{if } i = j\_{\prime\prime} \\ (1 - \alpha) \frac{RT\_i}{RT\_i} \frac{1}{d\_{ij}} \prime & \text{if } i \neq j\_{\prime\prime} \end{cases} $$

Because

$$RT(G)^{-1}RD\_{\mathfrak{a}}(G)RT(G)\mathbf{x} = \rho(RD\_{\mathfrak{a}}(G))\mathbf{x},\tag{4}$$

row *s* and *t* in Equation (4) are

$$\rho \left( R D\_{\mathfrak{a}} (G) \right) \mathfrak{x}\_{s} = \mathfrak{a} R T\_{s} \mathfrak{x}\_{s} + (1 - \mathfrak{a}) \sum\_{i=1}^{n} \frac{R T\_{i}}{R T\_{s}} \frac{\mathfrak{x}\_{i}}{d\_{si}},\tag{5}$$

$$\rho(RD\_{\mathfrak{a}}(G))\mathbf{x}\_{t} = \mathfrak{a}RT\_{t}\mathbf{x}\_{t} + (1-\mathfrak{a})\sum\_{i=1}^{n}\frac{RT\_{i}}{RT\_{t}}\frac{\mathbf{x}\_{i}}{d\_{ti}}.\tag{6}$$

After shifting the item of Equations (5) and (6), we can get

$$\begin{split} (\rho(RD\_{\mathfrak{a}}(G) - aRT\_{\mathfrak{s}})) \mathbf{x}\_{\mathfrak{s}} &= (1 - a) \sum\_{i=1}^{n} \frac{RT\_{i}}{RT\_{\mathfrak{s}}} \frac{\mathbf{x}\_{i}}{d\_{\mathfrak{s}i}} \\ &\leq (1 - a) \frac{\mathbf{x}\_{t}}{RT\_{\mathfrak{s}}} \sum\_{i=1}^{n} RT\_{i} \frac{1}{d\_{\mathfrak{s}i}} \\ &= (1 - a) \frac{T\_{\mathfrak{s}}}{RT\_{\mathfrak{s}}} \mathbf{x}\_{t'} \\ (\rho(RD\_{\mathfrak{a}}(G) - aRT\_{t})) \mathbf{x}\_{t} &= (1 - a) \sum\_{i=1}^{n} \frac{RT\_{i}}{RT\_{\mathfrak{s}}} \frac{\mathbf{x}\_{i}}{d\_{\mathfrak{s}i}} \end{split} \tag{7}$$

$$\begin{split} \mathcal{X} &= \mathcal{X} \sum\_{i=1}^{n} RT\_{t} \, d\_{ti} \\ &\leq (1 - \alpha) \frac{\mathcal{X}\_{\text{s}}}{RT\_{t}} \sum\_{i=1}^{n} RT\_{i} \frac{1}{d\_{ti}} \\ &= (1 - \alpha) \frac{T\_{t}}{RT\_{t}} \mathcal{X}\_{\text{s}}. \end{split} \tag{8}$$

Multiply Equation (7) and (8) to simplify (*ρ*(*RDα*(*G*) − *αRTs*)(*ρ*(*RDα*(*G*) − *αRTt*)*xsxt* ≤ (1 − *α*)<sup>2</sup> *TsTt RTsRTt xtxs*. Then

$$\left(\rho (RD\_{\mathfrak{a}}(G))^2 - a(RT\_{\mathfrak{s}} + RT\_{\mathfrak{t}})\rho (RD\_{\mathfrak{a}}(G)) + a^2 RT\_{\mathfrak{s}}RT\_{\mathfrak{t}} - (1 - a)^2 \frac{T\_{\mathfrak{s}}T\_{\mathfrak{t}}}{RT\_{\mathfrak{s}}RT\_{\mathfrak{t}}} \le 0.1\right)$$

$$\rho (RD\_{\mathfrak{a}}(G)) \le \frac{a(RT\_{\mathfrak{s}} + RT\_{\mathfrak{t}}) + \sqrt{a^2 (RT\_{\mathfrak{s}} - RT\_{\mathfrak{t}})^2 + 4(1 - a)^2 \frac{T\_{\mathfrak{s}}T\_{\mathfrak{t}}}{RT\_{\mathfrak{s}}RT\_{\mathfrak{t}}}}$$

Hence

$$\rho(RD\_{\mathfrak{a}}(G)) \le \max\_{1 \le i,j \le n} \left\{ \frac{a(RT\_i + RT\_j) + \sqrt{a^2(RT\_i - RT\_j)^2 + 4(1 - \alpha)^2 \frac{T\_i T\_j}{RT\_i RT\_j}}}{2} \right\}.$$

Suppose *G* is a *k*-reciprocal distance regular graph, *RTi* = *k*, *Ti* = *k*2, *i* = 1, 2, ... , *n*. According to Lemma 2, *ρ*(*RDα*(*G*)) = *k*, so Equation (3) holds. On the contrary, if inequality (3) is equal, *x*<sup>1</sup> = *x*<sup>2</sup> = ··· = *xn* can be obtained from (7) and (8), that is, *ρ*(*RDα*(*G*)) =

*αRT*<sup>1</sup> + (1 − *α*) *<sup>T</sup>*<sup>1</sup> *RT*<sup>1</sup> <sup>=</sup> *<sup>α</sup>RT*<sup>2</sup> + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*) *<sup>T</sup>*<sup>2</sup> *RT*<sup>2</sup> <sup>=</sup> ··· <sup>=</sup> *<sup>α</sup>RTn* + (<sup>1</sup> <sup>−</sup> *<sup>α</sup>*) *Tn RTn* , which means that *G* is a reciprocal distance degree regular graph.

**Theorem 7.** *Let G be a graph with reciprocal distance degree sequence* {*RT*1, *RT*2, ... , *RTn*} *and second reciprocal distance degree sequence* {*T*1, *T*2,..., *Tn*}*. Then*

$$\rho(RD\_{\mathfrak{a}}(G)) \ge \min\_{1 \le i,j \le \mathfrak{a}} \left\{ \frac{a(RT\_i + RT\_j) + \sqrt{a^2(RT\_i - RT\_j)^2 + 4(1 - \mathfrak{a})^2 \frac{T\_i T\_j}{RT\_i RT\_j}}}{2} \right\}.$$

*The equality holds if and only if G is a reciprocal distance degree regular graph.*

**Proof.** The method is the same as Theorem 6.

**Theorem 8.** *Let G be a graph of order n and* 0 ≤ *α* < 1*, then*

*i*=1

$$\rho(RD\_{\mathfrak{a}}(G)) \le \frac{2\mathfrak{a}H(G)}{n} + \sqrt{\frac{n-1}{n}(\|RD\_{\mathfrak{a}}(G)\|\_F^2 - \frac{(2\mathfrak{a}H(G))^2}{n})}.$$

*The equality holds if and only if G* = *Kn.*

**Proof.** We recall that *<sup>n</sup>* ∑ *i*=1 *<sup>λ</sup>i*(*RDα*(*G*)) = *<sup>α</sup> <sup>n</sup>* ∑ *i*=1 *RTi* <sup>=</sup> <sup>2</sup>*αH*(*G*), and *<sup>n</sup>* ∑ *i*=1 *λi*(*RDα*(*G*))<sup>2</sup> = *RDα*(*G*) <sup>2</sup> *<sup>F</sup>*. Clearly, *<sup>n</sup>* ∑ (*λi*(*RDα*(*G*)) <sup>−</sup> <sup>2</sup>*αH*(*G*) *<sup>n</sup>* ) = 0.

By Lemma 8,

$$\rho(RD\_{\mathfrak{a}}(G)) - \frac{2aH(G)}{n} \le \sqrt{\frac{n-1}{n} \sum\_{i=1}^{n} (\lambda\_i(RD\_{\mathfrak{a}}(G)) - \frac{2aH(G)}{n})^2},\tag{9}$$

with equality holds if and only if

$$\lambda\_2(RD\_a(G)) - \frac{2aH(G)}{n} = \dots = \lambda\_n(RD\_a(G)) - \frac{2aH(G)}{n} = -\frac{\rho(RD\_a(G)) - \frac{2aH(G)}{n}}{n-1}.\tag{10}$$

Since

$$\begin{split} \sum\_{i=1}^{n} (\lambda\_i(RD\_a(G)) - \frac{2aH(G)}{n})^2 &= \sum\_{i=1}^{n} (\lambda\_i(RD\_a(G)))^2 - \frac{4aH(G)}{n} \sum\_{i=1}^{n} \lambda\_i(RD\_a(G)) + n(\frac{2aH(G)}{n})^2 \\ &= \|RD\_a(G)\|\_F^2 - 2\frac{(2aH(G))^2}{n} + \frac{(2aH(G))^2}{n} \\ &= \|RD\_a(G)\|\_F^2 - \frac{(2aH(G))^2}{n} .\end{split}$$

The upper bound (9) is equivalent to

$$\rho(RD\_a(G)) \le \frac{2aH(G)}{n} + \sqrt{\frac{n-1}{n}(\|RD\_a(G)\|\_F^2 - \frac{(2aH(G))^2}{n})} \tag{11}$$

with the necessary and sufficient condition for the equality given in (10).

Now, suppose that the equality holds. Therefore, the equality condition for (11) can be given in (10), and we obtain that *G* has only two distinct generalized reciprocal distance eigenvalues. Hence, from Lemma 7, *G* = *Kn*.

Conversely, from Lemma 7 the generalized reciprocal distance eigenvalues of *Kn* are *ρ*(*RDα*(*Kn*) = *n* − 1 and *λi*(*RDα*(*G*)) = *αn* − 1, for *i* = 2, 3, ... , *n*. Then, the equality holds.
