I'm self-studying quantum mechanics with Sakurai's book (Modern Quantum Mechanics, 2nd edition) and came across the following in reference to the operator $\textbf{S}^2$:
As will be shown in Chapter 3, for spins higher than $\frac{1}{2}$, $\textbf{S}^2$ is no longer a multiple of the identity operator; however, $[\textbf{S}^2, S_i] = 0$ still holds (for $i = x, y, z$). (page 28)
The square of the total spin commuting with the components, I'm comfortable with. But the first part just confuses me: for a system with spin $s$, is it not true that
$$\textbf{S}^2|\cdot\rangle = \hbar^2s(s+1)|\cdot\rangle$$
whether or not $s = \frac{1}{2}$? Or do I have a fundamental misunderstanding of the situation? (I have read through Chapter 3, but apparently I kept missing the part where the book addresses this.)