 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1/7/b/iv/solution

MN = N

gives

(det M - 1) det N = 0

. Here

det M - 1 = 4∣ a ∣^{2} > 0

because

a \neq = 0

, so

det N = 0

. This also agrees with

N = a b^{T}

being an outer product of rank one.

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