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

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.