Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 3 / a / iii

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- Solution iii

Solution

Write

v = \sum_{i = 1}^{k + 1} α_{i} u_{i}

. Since

v \in / ⟨ u_{1}, \dots, u_{k} ⟩

, necessarily

α_{k + 1} \neq = 0

. Rearranging gives

u_{k + 1} = \frac{1}{α _{k + 1}} v - \sum_{i = 1}^{k} \frac{α _{i}}{α _{k + 1}} u_{i} \in ⟨ u_{1}, \dots, u_{k}, v ⟩ .

All generators of

⟨ u_{1}, \dots, u_{k + 1} ⟩

therefore belong to

⟨ u_{1}, \dots, u_{k}, v ⟩

, proving one set inclusion. Conversely,

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

all belong to the linear subspace

⟨ u_{1}, \dots, u_{k + 1} ⟩

, proving the other set inclusion. Thus

⟨ u_{1}, \dots, u_{k}, v ⟩ = ⟨ u_{1}, \dots, u_{k + 1} ⟩ .

This linear span exchange does not require linear independence of the original vectors.

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