 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/6a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

A^{†} = A

, and unitary when

U^{†} U = U U^{†} = I

A u = λ u

with

u \neq = 0

, then

λ u^{†} u = u^{†} A u = (u^{†} A u)^{*} = λ^{*} u^{†} u,

so every eigenvalue is real. If

A u = λ u

and

A v = μ v

, then

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

U = (u_{1} \dots u_{n})

satisfies

U^{†} U = I

and is unitary. With

D = diag (λ_{1}, \dots, λ_{n}),

A u_{j} = λ_{j} u_{j}

say

A U = U D

, hence

A = U D U^{†}

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian:

U = (1)

and

D = (i)

give

U D U^{†} = (i)

. It is Hermitian exactly when all diagonal entries of

D

are real.

For the displayed matrix, one valid decomposition is

U = i / 2 0 1/ 2 010 - i / 2 0 1/ 2, D = diag (5, 2, - 1) .

The columns are orthonormal eigenvectors, so direct multiplication gives

U D U^{†} = A

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