Condition for valid correlation matrix with constant pairwise-correlation

I was reading a paper about variance reduction techniques, where they want to create a valid correlation matrix of size $n \times n$, but with the same pairwise-correlation coefficient, therefore, every off-diagonal entries are set to the same number $\alpha$:

$$R_n = (r_{ij}) =\begin{pmatrix} 1 & \alpha & \cdots & \alpha \\ \alpha & 1 & \cdots & \alpha \\ \vdots & \vdots & \ddots & \vdots \\ \alpha & \alpha & \cdots & 1 \end{pmatrix}$$

Then, they say that this matrix must be positive-definite to be a valid correlation matrix which I understand perfeclty, but they say that this matrix is therefore constrained by the following relation:

$$0 \leq \sum_{i=1}^{n} \sum_{j=1}^{n} r_{ij} \leq n^2$$

This relation, however, is totally new for me and I couldn't find where it came from. So my question is: Is this relation correct? Is there some kind of proof for it? Any help would be greatly appreaciated.

PS: English is not my first language, so apologies for my grammar.

• Hint: The expression simplifies to $$\sum_{i=1}^n\sum_{j=1}^n r_{ij} = n(1-\alpha) + n^2\alpha = n^2(\alpha + (1-\alpha)/n).$$This reduces the relation to $$0 \le \alpha + (1-\alpha)/n \le 1.$$ – whuber Feb 16 '18 at 19:21

The matrix will be positive semi-definite if and only if $-1/(n-1) \le \alpha \le 1$, as shown in answers at Bound for the correlation of three random variables .
The upper bound of $n^2$ in $0 \leq \sum_{i=1}^{n} \sum_{j=1}^{n} r_{ij} \leq n^2$ is achieved using $\alpha = 1.$
The lower bound of $0$ is achieved using $\alpha = -1/(n-1)$, because there are $n(n-1)$ occurrences of $\alpha$ in the double sum, together with n occurrences of $1$ (the diagonal elements). Therefore the double sum = $n(n-1)*(-1)/(n-1) + n*1 = 0$.