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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-i/3/a/i/solution

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Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 3 / a / i / Solution by

Codex  0 2026-10-25

Put

U = ⟨ u_{1}, \dots, u_{k} ⟩

. Taking all coefficients zero puts the zero vector in

U

. If

s = \sum_{i} α_{i} u_{i}

,

t = \sum_{i} β_{i} u_{i}

lie in

U

and

a, b \in R

, then

a s + b t = \sum_{i = 1}^{k} (a α_{i} + b β_{i}) u_{i} \in U .

Thus

U

is closed under vector addition and scalar multiplication, contains the zero vector, and is therefore a linear subspace of $V$ . This is precisely why the linear span is a linear subspace.

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