 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/8/a/solution

Let

N = a b^{T}

. The orthogonality condition gives

N^{2} = a (b^{T} a) b^{T} = 0

: this is the nilpotent part of a transvection when both vectors are nonzero. In two dimensions,

tr L = 2 + a_{1} b_{1} + a_{2} b_{2} = 2 .

Expanding its determinant, the fourth-degree terms cancel:

det L = (1 + a_{1} b_{1}) (1 + a_{2} b_{2}) - a_{1} b_{2} a_{2} b_{1} = 1 + a \cdot b = 1 .

These formulas also cover either vector being zero.

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