 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-2/17g/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

A Hermitian two-by-two matrix has the form

A = (α \overset{z}{ˉ} z δ), α, δ \in R, z \in C .

Addition and real scalar multiplication preserve this condition, and the two diagonal entries plus the real and imaginary parts of

z

give four independent real coordinates. Thus $S$ is a real four-dimensional vector space. The trace product is real because

\overline{tr (A B)} = tr (B A) = tr (A B)

; this also proves symmetry of

b

Directly,

tr (A^{2}) = α^{2} + δ^{2} + 2∣ z ∣^{2}

and

(tr A)^{2} = α^{2} + 2 α δ + δ^{2}

. Hence

b (A, A) = ∣ z ∣^{2} - α δ = - det A .

Since

tr I = 2

and

tr (A I) = tr A

b (A, I) = - \frac{1}{2} tr A .

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