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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/6/a/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 6 / a / Solution by

Codex  0 2026-10-18

Let

A = A^{†}

be a Hermitian matrix and

A v = λ v

with

v \neq = 0

v^{†} A v

equals its own complex conjugate, because

(v^{†} A v)^{*} = v^{†} A^{†} v = v^{†} A v .

Since

v^{†} A v = λ v^{†} v

and the squared norm

v^{†} v

is positive,

λ \in R

.

Now take

A u = λ u

and

A v = μv

. The inner product satisfies

u^{†} A v = μ u^{†} v, u^{†} A v = (A u)^{†} v = \overline{λ} u^{†} v = λ u^{†} v .

Subtracting gives

(μ - λ) u^{†} v = 0

. Thus distinct eigenvalues imply

u^{†} v = 0

, proving orthogonality of the corresponding eigenvectors. Within a degenerate eigenspace, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same eigenspace need not be orthogonal.

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