 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/6/a/iii/solution

Choose the common orthonormal eigenbasis established in part (ii), and let

U

be the unitary matrix of its columns. Then

U^{†} A U = diag (a_{1}, \dots, a_{n}), U^{†} B U = diag (b_{1}, \dots, b_{n}) .

Consequently

U^{†} (A + i B) U = diag (a_{1} + i b_{1}, \dots, a_{n} + i b_{n}) .

Thus

N

is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives

N N^{†} = N^{†} N = A^{2} + B^{2}

, so

N

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