Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 2

Write $P=\mathbf{a}{\mathbf{a}}^T$. Since $\mathbf{a}$ is a unit vector, $P^2=\mathbf{a}({\mathbf{a}}^T\mathbf{a}){\mathbf{a}}^T=P$ and $P^T=P$. Thus this Householder reflection satisfiesGeometrically, the mirror preserves the component tangent to its plane and reverses the normal component; doing it twice restores the original vector.

Expanding the two products of Householder reflections givesThe commuting plane reflections criterion follows: the operator vanishes if the normals are orthogonal, or if they are parallel. Conversely, if $c=\mathbf{a}\cdot \mathbf{b}\ne 0$ and the operator vanishes, applying $\mathbf{b}{\mathbf{a}}^T-\mathbf{a}{\mathbf{b}}^T$ to $\mathbf{a}$ gives $\mathbf{b}-c\mathbf{a}=0$, proving parallelism.

This establishes equality of the reflected directions for every incident ray exactly in those two cases. For one fixed ray the claim needs that qualification: if nonparallel normals are not orthogonal, a ray along $\mathbf{a}\times \mathbf{b}$ has zero normal component for both mirrors and is unchanged in either order. Algebraically it lies in the kernel of the commutator; it is a special direction rather than a pair of commuting reflection operators.

For an arbitrary vector $\mathbf{x}$, expansion in the orthonormal basis gives $x_i=(a_i{a}_j+b_i{b}_j+c_i{c}_j)x_j$, with summation over $j$. Since this holds for every $\mathbf{x}$, the resolution of the identity in an orthonormal basis isLet $P_a=\mathbf{a}{\mathbf{a}}^T$, and similarly for $b,c$. The three orthogonal projection matrices sum to $I$ and have pairwise products zero. ThusAny ray meeting all three ideal perpendicular mirrors therefore returns in the opposite direction, independently of reflection order. This is the geometry of a corner reflector, a kind of retroreflector used in lunar laser ranging.