Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 3 / a (Span and exchange)

Put $U=⟨u_1,\dots ,u_k⟩$. Taking all coefficients zero puts the zero vector in $U$. If $s={\sum }_i{\alpha }_i{u}_i$, $t={\sum }_i{\beta }_i{u}_i$ lie in $U$ and $a,b\in \mathbb{R}$, thenThus $U$ 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.

Suppose ${\sum }_{i=1}^{k+1}{\alpha }_i{u}_i=0$. If ${\alpha }_{k+1}\ne 0$, thencontrary to the hypothesis. Hence ${\alpha }_{k+1}=0$. The assumed linear independence of $u_1,\dots ,u_k$ then gives ${\alpha }_1=\cdots ={\alpha }_k=0$. Therefore $u_1,\dots ,u_{k+1}$ are linearly independent. Adding a vector outside the previous linear span creates no nontrivial relation.

Write $v={\sum }_{i=1}^{k+1}{\alpha }_i{u}_i$. Since $v\notin ⟨u_1,\dots ,u_k⟩$, necessarily ${\alpha }_{k+1}\ne 0$. Rearranging givesAll 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. ThusThis linear span exchange does not require linear independence of the original vectors.