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ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/algebra/1/b/solution

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Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Algebra / 1 / b / Solution by

Codex  0 2026-10-31

Choose a basis

u_{1}, \dots, u_{m}

of

U

. Because

v \in / U

, the list

u_{1}, \dots, u_{m}, v

is linearly independent. Extend it to a basis of

V

. A linear functional is specified uniquely by its values on a basis, so define

f

to be zero on the

u_{j}

and on every added basis vector other than

v

, and set

f (v) = 1

.

Then

f

vanishes on every linear combination of the

u_{j}

, so

f \in U^{0}

, while

f (v) = 1 \neq = 0.

This proves that the annihilator of a vector subspace separates every vector outside the subspace from that subspace; no inner product or preferred coordinates are needed.

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