Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 6C (Algebra and Geometry)

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

In suffix notation, $x_i^{\prime }=a{x}_i+b{ϵ}_{ikj}{n}_k{x}_j$, so the matrix entries areHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The diagonal determines $a=2$. Comparing the off-diagonal entries withgives $b\mathbf{n}=(1,2,2)$. As $\mathbf{n}$ is a unit vector and $b>0$, $b=3$ and $\mathbf{n}=(1,2,2)/3$. ThusIn this example the transverse scaling is $\sqrt{13}$ and the axial scaling is two; the rotation angle is $\arctan (3/2)$.

Let $P\mathbf{x}=(\mathbf{n}\cdot \mathbf{x})\mathbf{n}$ be the axial orthogonal projection and $N\mathbf{x}=\mathbf{n}\times \mathbf{x}$. Then $NP=PN=0$, while the vector triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular plane,Thus the inverse of an axis-angle linear map isor, in the requested vector form,The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the matrix inverse without choosing a coordinate frame.