 Solution

The supplied matrix obeysSeparate the even and odd powers in the matrix exponential to obtainIt fixes the $z$ component and is a counterclockwise active rotation viewed from $+z$. This sign differs from the clockwise convention in b, because the source fixes the sign of both $J_3$ and the exponential. Replacing $\theta$ by $-\theta$ yields the clockwise rotation.

With $(J_k{)}_{ij}=-i{ϵ}_{kij}$, the generators of rotations obeyThe commutator measures the leading difference between performing two infinitesimal rotations in opposite orders. For example, with $X=-i\alpha J_1$ and $Y=-i\beta J_2$,The failure to commute is itself an infinitesimal rotation about the perpendicular axis; the Lie algebra closes on the same three generators of rotations.

To pass from this local statement to the full group property, put $A_k=-i{J}_k$. These are real antisymmetric matrices, with $[A_i,A_j]={ϵ}_{ijk}{A}_k$. The Baker--Campbell--Hausdorff formula shows locally that the logarithm of a product of their exponentials remains a linear combination of the $A_k$, because all nested commutators do. Globally, each exponential has $R^{T}R=I$ and $\det R=e^{\mathrm{tr}A}=1$. If $R_1,R_2$ have these properties, thenEvery real three-dimensional orthogonal matrix of determinant $+1$ has an eigenvector with eigenvalue $+1$: its nonreal eigenvalues come in conjugate pairs, and the product of all three is $+1$. On the perpendicular plane it acts as a two-dimensional rotation. It therefore has axis-angle form and can be written $e^{-i\theta {\hat{n}}^k{J}_k}$. Thus the product is again a rotation. The identity is $\theta =0$, the inverse reverses $\theta$, and associativity comes from matrix multiplication. This proves that all these rotations form the special orthogonal group $SO(3)$. The commutator supplies local closure, while the orthogonality and axis-angle argument supplies the necessary global step.