 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/6/b/solution

Let

A v_{i} = λ_{i} v_{i}

and

A v_{j} = λ_{j} v_{j}

λ_{j} v_{i}^{†} v_{j} = v_{i}^{†} A v_{j} = - (A v_{i})^{†} v_{j} = - \overline{λ}_{i} v_{i}^{†} v_{j} = λ_{i} v_{i}^{†} v_{j},

where the last equality uses part (a). Distinct eigenvalues therefore imply

v_{i}^{†} v_{j} = 0 for i \neq = j

. Normalize each eigenvector to length one. Since there are

n

Put them in the columns of

U

. Then

U

A U = U D

, so

A = U D U^{†}, D = diag (λ_{1}, \dots, λ_{n}) .

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