 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/5/a/solution

The matrix sending the standard basis vectors to the given columns is

U = \frac{1}{6} 31 1 + i - i 3 i - 1 + i 0 - 2 1 + i .

With the Hermitian inner product, the squared column norms are

3/6 + 1/6 + 2/6 = 1

3/6 + 1/6 + 2/6 = 1

, and

4/6 + 2/6 = 1

. The off-diagonal products are

u_{1}^{†} u_{2} u_{1}^{†} u_{3} u_{2}^{†} u_{3} = \frac{- 3 i + i + ( 1 - i ) ( - 1 + i )}{6} = 0, = \frac{- 2 + ( 1 - i ) ( 1 + i )}{6} = 0, = \frac{2 i + ( - 1 - i ) ( 1 + i )}{6} = 0.

Thus the columns form an orthonormal basis, so

U^{†} U = I

and

U

is a unitary matrix. Here the operation maps the old basis vectors to the new vectors; the inverse coordinate change of basis uses

U^{†}

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