Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 17

By the definition of matrix multiplication and trace, with all indices summed from $1$ to $D$,The entries are scalars, so their multiplication commutes. Reordering them and renaming the summed indices givesEquivalently, in Einstein notation, $P_{ij}{Q}_{jk}{R}_{ki}=Q_{jk}{R}_{ki}{P}_{ij}$. This proves the cyclic property of the trace.

The outer product of the column vector with its matrix transpose is the $2\times 2$ matrix

The reversed product is the dot product, a scalar represented by a $1\times 1$ matrix:

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

Using Einstein notation and the Kronecker delta,The unit vector condition is essential in the last step.

The matrix issoHere $\Vert \mathbf{p}{\Vert }^2=2$, so $P=I-2\hat{\mathbf{p}}{\hat{\mathbf{p}}}^T$ with $\hat{\mathbf{p}}=\mathbf{p}/\sqrt{2}$. This is a Householder transformation: reflection in the plane $x_1+x_2=0$. The component parallel to the normal vector $\mathbf{p}$ changes sign, while the perpendicular component is fixed.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.

From part (f), $PQ=\mathrm{diag}(-1,-1,1)$. To obtain $PQR=-I$, choose the Householder transformation that reverses only the third coordinate:Indeed $\mathbf{r}{\mathbf{r}}^T=\mathrm{diag}(0,0,2)$, so $R_{ij}={\delta }_{ij}-r_i{r}_j$, and direct matrix multiplication gives $PQR=-I$. The alternative vector $-\mathbf{r}$ produces the same matrix. Geometrically $R$ reflects in the plane $x_3=0$, completing the reversal of all three coordinates.