Exercise 1.5.15

Sufficiency: If $u_1=0$ then $S$ is linearly independent. If

for some $k$, say $u_{k+1}=a_1{u}_1+a_2{u}_2+\cdots +a_k{u}_k$, then we have $a_1{u}_1+a_2{u}_2+\cdots +a_k{u}_k-u_{k+1}=0$ is a nontrivial representation. Necessary: If $S$ is linearly dependent, there are some integer $k$ such that there is some nontrivial representation $a_1{u}_1+a_2{u}_2+\cdots +a_k{u}_k+a_{k+1}{u}_{k+1}=0$. Furthermore we may assume that $a_{k+1}\ne 0$ otherwise we may choose less $k$ until that $a_{k+1}\ne =0$. Hence we have $a_{k+1}=-\frac{1}{a_{k+1}}(a_1{u}_1+a_2{u}_2+\cdots +a_k{u}_k)$ and so $a_{k+1}\in \mathrm{span}(\{u_1,u_2,\dots ,u_k\})$.