= Solution
An <irreducible spin representation> is a nonzero <Hilbert space> with no proper nonzero closed <invariant subspace> under all three <angular momentum operators>. Equivalently, the associated <unitary representation> of <SU(2)> cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a <spin> $j\in\{0,\tfrac12,1,\tfrac32,\ldots\}$ and have <dimension> $2j+1$.
Choose an <orthonormal basis> $|j,m_s\rangle$, with $m_s=-j,-j+1,\ldots,j$. The <angular momentum commutation relations> imply $[\mathbf S^2,S_i]=0$, $[S_3,S_\pm]=\pm\hbar S_\pm$, $[S_+,S_-]=2\hbar S_3$, and $S_+^\dagger=S_-$. The complete action is
$$
\boxed{\begin{aligned}
\mathbf S^2|j,m_s\rangle&=\hbar^2j(j+1)|j,m_s\rangle,\\
S_3|j,m_s\rangle&=\hbar m_s|j,m_s\rangle,\\
S_\pm|j,m_s\rangle&=\hbar\sqrt{(j\mp m_s)(j\pm m_s+1)}\,|j,m_s\pm1\rangle.
\end{aligned}}
$$
In particular, the <spin raising operator> annihilates $|j,j\rangle$ and the <spin lowering operator> annihilates $|j,-j\rangle$. The coefficient follows by taking the squared <norm> and using $S_\mp S_\pm=\mathbf S^2-S_3^2\mp\hbar S_3$. Thus the <ladder operators> connect every member of the <basis>, while terminating at its two endpoints.
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