Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 7

The trace is unchanged by transposition and is cyclic under a product. With $S^T=S$ and $A^T=-A$,Thus $\boxed{\mathrm{tr}(AS)=0}$. In components this is the vanishing contraction of a symmetric tensor with an antisymmetric tensor.

The Levi-Civita symbol vanishes when two indices coincide, so

The antisymmetric contribution is the negative of twice the cross-product matrix for $\mathbf{a}$:The cross-product matrix spectrum is $0,\pm i|\mathbf{a}|$, so $M$ has eigenvalues $1,1\pm 2i|\mathbf{a}|$. Multiplication givesDirect expansion of the displayed matrix yields the same real positive determinant.

Use index contraction:The last term vanishes because $a_k{a}_{\ell }$ is symmetric in the contracted indices while the Levi-Civita symbol is antisymmetric. Hence $\boxed{MN=N}$.

Taking determinants in $MN=N$ gives $(\det M-1)\det N=0$. Here $\det M-1=4|\mathbf{a}{|}^2>0$ because $\mathbf{a}\ne 0$, so $\boxed{\det N=0}$. This also agrees with $N=\mathbf{a}{\mathbf{b}}^T$ being an outer product of rank one.