Let $p,q$ be arbitrary primes.

Let $N = p * q$.

Let $I$ be the $N * N$ identity matrix.

Let $R$ be the $N * N$ matrix defined as follows: $R[x_0 * p + y_0, x_1 * p + y_1]=1$ if and only if $x_0+1 \equiv x_1 (\mod q)$ and $y_0 + 1 \equiv y_1 (\mod p)$.

Let $A = \begin{pmatrix} \frac12I & \frac12R \\\\ \frac12R & \frac12I \end{pmatrix}$.

The largest eigenvalue of $A$ is 1.0, associated with the all 1 vector.

Question: how can I show that the second largest (absolute value of) eigenvalue is < 1?

I'm not particularly concerned with the bound. For example, $\lambda < 1 - 2^{p*q}$ is perfectly fine. I just need to show that it's < 1.

Context: derandomization.