 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/6/ii/a/solution

For

B x_{i} = λ_{i} x_{i}

λ_{j} x_{i}^{†} x_{j} = x_{i}^{†} B x_{j} = (B x_{i})^{†} x_{j} = λ_{i} x_{i}^{†} x_{j} .

Distinct eigenvalues therefore imply

x_{i}^{†} x_{j} = 0

for

i \neq = j

. Together with the unit normalizations,

x_{i}^{†} x_{j} = δ_{ij} .

The

n

mutually orthogonal nonzero vectors are linearly independent and span the

n

-dimensional space, so they form an orthonormal basis. Here

δ_{ij}

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