Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 4

Decompose ${\mathbb{R}}^n=\mathrm{span}\{a\}\oplus a^{\perp }$. The rank-one matrix update $a{a}^T$ acts as multiplication by $\Vert a{\Vert }^2$ along $a$ and as zero on its orthogonal complement. ThusThe perpendicular eigenvalue has multiplicity $n-1$ and is absent when $n=1$. Consequently $A^{+}$ is always an invertible matrix, while

Put $\beta =\Vert b\Vert >0$ and choose an oriented orthonormal basis with third vector $b/\beta$. The Levi-Civita symbol term has action ${ϵ}_{ijk}{b}_k{x}_j=x\times b$, the negative of $b\times x$. Thus this axisymmetric linear map has block formThe transverse eigenvalues $1\pm i\beta$ are not real. The only real eigenvalue and its eigenvectors are thereforeThe determinants are $\det B^{+}=(1+{\beta }^2{)}^2>0$ and $\det B^{-}=(1-{\beta }^2)(1+{\beta }^2)$, proving that $B^{+}$ is always invertible, and $B^{-}$ is invertible exactly when $\Vert b\Vert \ne 1$. The complex transverse pair is important for this determinant even though the question asks for real eigenvalues.