Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 8

Let $N=\mathbf{a}{\mathbf{b}}^T$. The orthogonality condition gives $N^2=\mathbf{a}({\mathbf{b}}^T\mathbf{a}){\mathbf{b}}^T=0$: this is the nilpotent part of a transvection when both vectors are nonzero. In two dimensions,Expanding its determinant, the fourth-degree terms cancel:These formulas also cover either vector being zero.

Since $M^T=I-\mathbf{a}{\mathbf{b}}^T=I-N$, nilpotence already gives $(I+N)(I-N)=I$. Explicitly, in index notation,Thus $\boxed{L{M}^T=I}$, or equivalently $M^T=L^{-1}$. Here ${\delta }_{ij}$ is the Kronecker delta.

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.