Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 6C (Vectors and Matrices)

The distinguished unit vector is $n=(1,1,1)/\sqrt{3}$. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts givesso $A_{ij}=\alpha {\delta }_{ij}+\beta n_i{n}_j+\gamma {\epsilon }_{ijk}{n}_k$.

On the line spanned by $n$, $A$ multiplies by $L=\alpha +\beta =a+b+c$. On $n^{\perp }$, it is the composition of a rotation with a dilation by $r=\sqrt{{\alpha }^2+{\gamma }^2}$. Consequently every area in that plane is multiplied by $r^2={\alpha }^2+{\gamma }^2$.

For a plane reflection, the $n$ direction is reversed while $n^{\perp }$ is fixed. Thus $L=-1$, $\alpha =1$, and $\gamma =0$, equivalently

The map is a rotation about the $n$ axis exactly whenor explicitlyThese conditions make the axis fixed and the perpendicular-plane action length preserving; they are also necessary.

Let $N=n{n}^T$ and $K_{ij}={\epsilon }_{ijk}{n}_k$. Since $K^2=N-I$, inversion separately on $\mathbb{R}n$ and $n^{\perp }$ givesprovided $(\alpha +\beta )({\alpha }^2+{\gamma }^2)\ne 0$.

The line/plane decomposition givesSubstituting the values in part (a) and simplifying yields