= Solution
Put $s=\gamma^{-1/2}$. Then
$$
\alpha=s^2,
\qquad
\beta=(1-s)^2,
\qquad
\sqrt\beta=1-s.
$$
For an eigenvalue $\lambda$, the corresponding $2\times2$ block has characteristic polynomial
$$
z^2-(1+\beta-\alpha\lambda)z+\beta.
$$
For $\lambda=1$,
$$
1+\beta-\alpha=2(1-s)=2\sqrt\beta,
$$
so the polynomial is
$$
(z-\sqrt\beta)^2.
$$
For $\lambda=\gamma$, it is
$$
z^2-\beta z+\beta.
$$
Its discriminant is $\beta(\beta-4)<0$, so its conjugate roots have product $\beta$ and modulus $\sqrt\beta$. Both blocks therefore have the same <spectral radius>, and
$$
\boxed{\rho(M)=\sqrt\beta=1-\frac1{\sqrt\gamma}.}
$$
This is the <heavy-ball rate for a two-eigenvalue diagonal quadratic>.
By contrast, the steepest-descent factor is
$$
\frac{\gamma-1}{\gamma+1}
=1-\frac{2}{\gamma}+O(\gamma^{-2}).
$$
The heavy-ball factor is $1-\gamma^{-1/2}$, which is substantially smaller for $\gamma\gg1$: it gives an iteration scale of order $\sqrt\gamma$ rather than order $\gamma$.
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