Solution
= Solution
{c}
Put $U=\langle u_1,\ldots,u_k\rangle$. 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 $a,b\in\mathbb R$, then
$$
as+bt=\sum_{i=1}^k(a\alpha_i+b\beta_i)u_i\in U.
$$
Thus $U$ is closed under <vector addition> and <scalar multiplication>, contains the <zero vector>, and is therefore \b[a <linear subspace> of $V$]. This is precisely why the <linear span> is a <linear subspace>.