 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-i/3/a/ii/solution

Using the convention

χ_{T} (x) = det (x I - A)

χ_{T} (x) = (x - 1) det (x - 1 1 x) = (x - 1) (x^{2} + 1) .

Since

x^{2} + 1 > 0

for every

x \in R

, the only eigenvalue over the real numbers is

1

. Solving

(A - I) v = 0

gives

E_{1} = ker (A - I) = span {(0, 0, 1)} .

Thus the only eigenspace over the real numbers is the axis of the rotation.

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