Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 7B (Algebra and Geometry)

Let $[v{]}_{\times }$ denote the cross product matrix, with $[v{]}_{\times }x=v\times x$. ThenThe two eigenvalue calculations use an orthonormal basis adapted to $u,v$.

If $u\cdot v=0$, use the positively oriented orthonormal basis $(u,v\times u,v)$. The cross product identities give $Au=u+v\times u$, $A(v\times u)=-u$ and $Av=0$. In this basis,Its characteristic polynomial is $\lambda ({\lambda }^2-\lambda +1)$, so

If $v=\pm u$, then $Au=u$. On the perpendicular plane, $u{u}^T$ vanishes and $[v{]}_{\times }$ is a signed right-angle rotation, whose square is minus the identity. Hence its complex eigenvalues are $\pm i$, and

A trace-zero Hermitian matrix has the unique formSet $x_3=h$, $x_1=\mathrm{Re}w$ and $x_2=-\mathrm{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real Pauli matrix representation $R(x)=x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$.

For a unitary matrix, $U^{-1}=U^{†}$. Thusand invariance of the matrix trace under similarity transformations gives $\mathrm{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.

Finally,The determinant is invariant under similarity transformations, so $\det R(y)=\det R(x)$ implies