 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/6/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

A^{†} = \overline{A}^{T} = A

. If

A v = λ v

with

v \neq = 0

, then

v^{†} A v = λ v^{†} v

is real because its conjugate equals

v^{†} A^{†} v = v^{†} A v

. Since

v^{†} v > 0

λ

is real. If

A w = μ w

with

μ \neq = λ

, then

μ v^{†} w = v^{†} A w = (A v)^{†} w = λ v^{†} w,

v^{†} w = 0

The converse is false. For example

(1012)

has the two real eigenvalues one and two but is not a Hermitian matrix. It can be embedded as a block in a

3 \times 3

example if desired.

For Hermitian

A, B

(A B)^{†} = B^{†} A^{†} = B A

. Thus

A B

is Hermitian exactly when

A B = B A

, equivalently when their commutator vanishes. Also

(A B + B A)^{†} = A B + B A

, so a value that works for every such pair is

c = 1.

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