 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/5/c/iii/solution

Taking determinants and using invariance under transpose,

det (I + A) = det ((I + A)^{T}) = det (I - A), det N = 1.

This determinant identity actually holds in every dimension. In dimension three, skew symmetry additionally gives

det A = det A^{T} = det (- A) = (- 1)^{3} det A = - det A,

det A = 0

. Hence there is a real nonzero vector

x

in the kernel:

A x = 0, x \neq = 0.

The eigenvalue relation then gives

N x = x

. Thus the Cayley transform fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.

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