Question about proof from Friedberg's Linear Algebra

Theorem 1.7 Let $$S$$ be a linearly independent subset of a vector space $$V$$, and let $$v$$ be a vector in $$V$$ that is not in $$S$$. Then $$S \cup \{v\}$$ is linearly dependent if and only if $$v\in$$ span($$S$$).

The following is the first part of the proof:

Proof . If $$S \cup \{v\}$$ is linearly dependent, then there are vectors $$u_1,u_2,...,u_n$$ in $$S \cup \{v\}$$ such that $$a_1u_1 + a_2u_2 + \cdots + a_nu_n = 0$$ for some nonzero scalars $$a_1,a_2, \cdots , a_n$$. Because $$S$$ is linearly independent, one of the $$u_{i}$$'s, say $$u_1$$, equals $$v$$. Thus $$a_1v + a_2u_2 + \cdots + a_nu_n = 0$$

$$v = a_{1}^{-1}(-a_2u_2 - \cdots -a_nu_n) = -(a_{1}^{-1}a_2)u_2 - \cdots -(a_{1}^{-1}a_n)u_n.$$

Since $$v$$ is a linear combination of $$u_2, \cdots , u_n$$, which are in $$S$$, we have $$v \in$$ span(S).

I am confused by the statement, "because $$S$$ is linearly independent, one of the $$u_{i}$$'s, say $$u_1$$, equals $$v$$." Why exactly does it follow from $$S$$ being linearly independent that one of the $$u_{i}$$'s is equal to $$v$$?

• If none of the $u_i$'s were equal to $v$, then the statement would exactly (literally) contradict the fact that $S$ is linearly independant. Jun 13 '18 at 0:48

The zero vector can be expressed as a non-trivial combination of vectors from $S\cup \{u\}$ (by definition, since this set is lin. dip.). This gives
$$0=a_1u_1+a_2u_2+\dots +a_nu_n$$ and not all $a_i$ zero. This is true only if one of the $u_i$s is $v$ because the above equation has only trivial solution (all $a_i$s zero) if considered for vectors from $S$ only (S being linearly independent).
Hence without loss of generality (relabelling the vectors if your prefer) we can set/claim $u_1=v$.
$$u_1,u_2,...,u_n \in S \cup \{v\}$$ and $$a_1u_1 + a_2u_2 + \cdots + a_nu_n = 0$$
along with linear independence of $S$ implies that either all coefficients are $0$ or one of $u_i$ is $v$. Since the coefficients are not all $0$, one of $u_i$ is $v$