Exercise 1.5.14

Sufficiency: It’s natural that $0$ is linearly dependent. If $v$ is a linear combination of $u_1,u_2,\dots ,u_n$ , say $v=a_1{u}_1+a_2{u}_2+\cdots a_n{u}_n$, then $v-a_1{u}_1-a_2{u}_2-\cdots -a_n{u}_n=0$ implies $S$ is linearly dependent. Necessity: If $S$ is linearly dependent and $S\ne \{0\}$ we have some nontrivial representation $a_0{u}_0+a_1{u}_1+\cdots +a_n{u}_n=0$ with at least one of the coefficients is nonzero, say $a_0\ne 0$ without loss the generality. Then we can let $v=u_0=-\frac{1}{a_0}(a_1{u}_1+a_2{u}_2+\cdots +a_n{u}_n)$.