 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/7b/c/solution

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

v_{1} = P v, v_{0} = (I - P) v .

Then

v = v_{0} + v_{1}

, and idempotence gives

P v_{0} = P (I - P) v = (P - P^{2}) v = 0, P v_{1} = P^{2} v = P v = v_{1} .

Moreover,

(v_{0}, v_{1}) = ((I - P) v, P v) = (v, (I - P) P v) = 0,

because both

P

and

I - P

are Hermitian. This is the orthogonal decomposition into the kernel and image of the projection.

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