 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/5/ii/a/solution

k

, cancellation of

U^{†} U = I

gives

A^{k} = U D^{k} U^{†}

. The trace is invariant under matrix similarity, since

tr (X Y) = tr (Y X)

, and the determinant is multiplicative. Therefore

tr (A^{k}) = tr (D^{k}) = j = 1 \sum n λ_{j}^{k}, det (A^{k}) = det (D^{k}) = j = 1 \prod n λ_{j}^{k} .

This includes zero eigenvalues without requiring

A

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