Solution
= Solution
{c}
Suppose $\sum_{i=1}^{k+1}\alpha_i u_i=0$. If $\alpha_{k+1}\ne0$, then
$$
u_{k+1}=-\sum_{i=1}^k\frac{\alpha_i}{\alpha_{k+1}}u_i\in\langle u_1,\ldots,u_k\rangle,
$$
contrary to the hypothesis. Hence $\alpha_{k+1}=0$. The assumed <linear independence> of $u_1,\ldots,u_k$ then gives $\alpha_1=\cdots=\alpha_k=0$. Therefore \b[$u_1,\ldots,u_{k+1}$ are <linearly independent>]. Adding a <vector> outside the previous <linear span> creates no nontrivial relation.