Actually, that should be d(d+1)/2. But I think my argument still holds: this is a triangular number, corresponding to the number of r’s above the diagonal. Under the rearrangement I propose, you’d have d(d+1)/2 r’s in the first row, and the other d(d+1)/2 r’s in the first column.
Here’s an interesting bound based on Schur’s inequality. Consider the matrix obtained by subtracting the nxn identity matrix from your correlation matrix; in other words, it has zeroes on the diagonal instead of 1’s. Since every vector is an eigenvector of the identity matrix with eigenvalue 1, this new matrix has the same eigenvectors as your original correlation matrix, and the eigenvalues of the new matrix are obtained by subtracting 1 from each of the old eigenvalues.
Now, by Schur’s inequality, we have that
Σ |λ|[sup]2[/sup] <= Σ |a[sub]ij[/sub]|[sup]2[/sup],
where the sum on the left-hand side is over the n eigenvalues, and the sum on the right is over all n[sup]2[/sup] matrix entries. The new matrix will have d(d+1) entries (i.e., twice a triangular number) that are equal to r, and the rest will be equal to zero. Thus, applying Schur’s inequality we have
Σ |λ|[sup]2[/sup] <= d(d+1) r[sup]2[/sup].
But since all the terms on the left-hand side are strictly positive, no individual term can be greater than the right-hand side. Thus, we have that the absolute value of any eigenvalue of our new matrix is bounded above by √(d(d+1)) r.
The block-diagonal matrix you’ve described yields a largest eigenvalue of d*r when transformed in this way, so this constitutes a known lower bound on the largest eigenvalue of all such matrices. There isn’t much “room” between this upper bound and the known lower bound; fractionally, they differ by √(1+1/d), which is relatively small for large d.