 Solution

Let $Q\in \mathrm{SO}(3,\mathbb{R})$. Since $Q^{T}Q=I$ and $\det Q=1$,so $\det (Q-I)=0$. Choose a unit vector $e$ with $Qe=e$. Its orthogonal complement is an invariant subspace: for $v\perp e$, $e\cdot Qv=(Q^{T}e)\cdot v=e\cdot v=0$. In an orthonormal basis $(e,e_1,e_2)$, $Q$ therefore has a scalar block $1$ and a two-dimensional orthogonal matrix block of determinant one. Such a block has columns $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. Consequently $Q$ is a rotation in three dimensions about the axis $\mathbb{R}e$ through angle $\theta$, including the identity when $\theta =0$.

To construct the Householder reflections, let $S_0$ reflect in the plane spanned by $e,e_1$ and let $S_{\psi }$ reflect in the plane spanned by $e,\cos \psi e_1+\sin \psi e_2$. They fix $e$, while on $e^{\perp }$ their matrices areTheir product is the planar rotation matrix through $2\psi$. Taking $\psi =\theta /2$ proves $\boxed{Q=S_{\theta /2}{S}_0}$. This gives the composition of two plane reflections explicitly; each is a Householder reflection $I-2n{n}^T$ for the corresponding unit plane normal $n$.