 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/8e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

H : V \times V \to C

such that

H (v, w) = \overline{H (w, v)} .

Use the convention that

H

is linear in its first argument and conjugate-linear in its second. Its matrix in the basis

(v_{1}, \dots, v_{n})

A_{ij} = H (v_{i}, v_{j}) .

v = \sum_{i} x_{i} v_{i}

and

w = \sum_{j} y_{j} v_{j}

, then

H (v, w) = x^{T} A \overline{y} .

Since

w_{i} = \sum_{j} p_{ij} v_{j}

, the coordinate row of each new basis vector is a row of

P

. Direct substitution gives

H (w_{i}, w_{k}) = \sum_{j, l} p_{ij} A_{j l} \overline{p_{k l}},

so the new matrix is

P A \overline{P}^{T} .

The subspace

N = V^{⊥}

is the radical of

H

. In coordinates it is the kernel of

A

dim N = n - rank A .

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