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Rényi entropy is defined as:

\begin{aligned} S_n(\alpha)&:=\frac{1}{1-\alpha}\log\left({\sum_{a_{1\to n}}}\sum_{\mu=1}^{N}\left(\frac{p_{a_1a_2\cdots a_n}}{N}\right)^\alpha\right) \\ &=\frac{1}{1-\alpha}\log\left(\frac{1}{\mathcal{D}^{2\alpha(n-1)}}{{{\sum_{a_{1\to n}}}N_{a_1a_2\cdots a_n}{\prod_{i=1}^n d_{a_i}^\alpha}}}\right) \\ &=\frac{1}{1-\alpha}\left[\log\left({{{\sum_{a_{1\to n}}}N_{a_1a_2\cdots a_n}{\prod_{i=1}^n d_{a_i}^\alpha}}}\right)-\alpha(n-1)\log \mathcal{D}^2\right]\\ &=\frac{\alpha(n-1)\log \mathcal{D}^2-\log t_n(\alpha)}{\alpha-1} \end{aligned}

We are using

\begin{aligned} t_n(\alpha)&:={{{\sum_{a_{1\to n}}}N_{a_1a_2\cdots a_n}{\prod_{i=1}^n d_{a_i}^\alpha}}}\\ &= {\sum_{a_{1\to n}}}{\sum_{x_{1\to n-1}}}N_{x_0a_1}^{x_1}N_{x_1a_2}^{x_2}\cdots N_{x_{n-1}a_n}^{x_n}{\prod_{i=1}^n d_{a_i}^\alpha},\quad x_0=x_n=1\\ &= {\sum_{x_{1\to n-1}}}\prod_{i=1}^n\left({\sum_{a_{1\to n}}}d_{a_i}^\alpha N_{x_{i-1}a_i}^{x_i}\right)\end{aligned}

Define

$$T_{bc}(\alpha):=\sum_ad_a^\alpha N_{ba}^c$$

Then

$$t_n(\alpha)={\sum_{x_{1\to n-1}}}\prod_{i=1}^nT_{x_{i-1}x_i}(\alpha)\left[\bm T^n(\alpha)\right]_{x_0x_n}=\left[\bm T^n(\alpha)\right]_{11} \label{t11}$$

It’s obvious that $$T$$ is an elementwise positive matrix. Moreover, $$T$$ is normal:

\begin{aligned} \label{key} T_{ab}T_{cb}&=\sum_x\sum_y d_x^\alpha d_y^\alpha \sum_b N_{xa}^b N_{yc}^b\\ &=\sum_x\sum_y d_x^\alpha d_y^\alpha \sum_b N_{x\bar b}^{\bar a} N_{y\bar b}^{\bar c}\\ &=\sum_x\sum_y d_x^\alpha d_y^\alpha \sum_b N_{\bar y b}^{c}N_{\bar xb}^{a}\\ &=T_{ba}T_{bc}\end{aligned}

Thus, the eigenvectors correspond to distinct eigenvalues must be orthogonal. Assume the eigenvalues and orthonormal eigenvectors are $$\bm x_i$$ with eigenvalue $$\lambda_i(\alpha)$$. The eigenvector $$\bm x_m:=d_i\bm e_i/\mathcal{D}$$ has the maximum eigenvalue because its all components are positive. The eigenvalue is

$$\lambda_{\max}(\alpha)=\sum_a d_a^{\alpha+1}, \quad\lambda_{\max}(1)=\mathcal{D}^2$$

We decompose $$\bm e_1$$ into $$\bm x_i$$ to calculate $$[\bm{T}^n]_{11}$$:

\begin{aligned} \bm e_1&=\sum_i c_i\bm x_i,\quad c_i=\bm e_1\cdot \bm x_i\\ [\bm{T}_\alpha^n]_{11}&=\bm{e}_1^T\bm{T}_\alpha^n \bm{e}_1=\sum_i c_i^2\lambda_i^n\\ &=c_m^2\lambda_{\max}^n\left[1+\sum_{i\neq m}\frac{c_i^2}{c_m^2}\left( \frac{\lambda_i}{\lambda_{\max}}\right) ^n\right]\\ &=\frac{\lambda_{\max}^n}{\mathcal{D}^2}\left[1+\Theta(k^n)\right],\quad k:=\frac{\lambda_{\mathrm{sub}}}{\lambda_{\max}}<1\end{aligned}
\begin{aligned} S_n(\alpha)&=\frac{\alpha(n-1)\log \mathcal{D}^2-\log [\bm{T}_\alpha^n]_{11}}{\alpha-1}\\ &=n\left[\frac{\alpha\log \lambda_{\max}(1)-\log \lambda_{\max}(\alpha)}{\alpha-1}\right]-\log \mathcal{D}^2+\Theta(k^n)\label{al}\end{aligned}

In the $$\alpha=1$$ case, there is no other nontrivial eigenvector except $$\bm x_m$$ because $$\bm T=\bm x_m\bm x_m^\mathsf{T}$$. So the $$\Theta(k^n)$$ term in [al] reduce to zero.

\begin{aligned} S_{n}(1)&=n\frac{\dd}{\dd\alpha}\Big|_{\alpha=1}\Big[\alpha\log \lambda_{\max}(1)-\log \lambda_{\max}(\alpha)\Big] -\log \mathcal{D}^2\\ &=n\Big[\log \mathcal{D}^2-\sum_a \frac{d_a^2\log d_a}{\mathcal{D}^2}\Big]-\log \mathcal{D}^2\\ &=(n-1)\log \mathcal{D}^2-n\sum_a p_a\log d_a\end{aligned}