Vector Space is a Free Module 
Is this argument correct? They assume $B$ doesn’t span $V$. Then ${B}^{‘} = B \cup (v) $. And they say $\alpha v + \sum_{i=1}^r \alpha_ib_i$ = 0 implies that $\alpha\neq0$ otherwise every scalar becomes zero. My question here is that there is typo here isn’t it? $B^‘$ must be linearly dependent right if $\alpha\neq0$? Another question is what is the problem if all scalars are zero? That will show that $B^‘$ is linearly independent and will belong in $\mathscr F$ and that will contradict the maximality of $B$. Isn’t this argument sounding circular? Or am I missing something? 

$v$ is by assumption (that $B$ is not a basis) not a linear combination of any finite subset of $B$. In particular, $v \notin \text{span}(B)$. So it's right, that $B' = B \cup \{v\}$ is linear independent (because if it was linear dependent, then there would be an $(\alpha,\alpha_1,...,\alpha_r) \neq 0$ and a sum $\alpha v+\sum_{i=1}^r \alpha_i b_i =0$ which would imply $v \in \text{span}(B)$ (as in the text)). Thus $B' \in \mathscr{F}$ which is a contradiction to the maximality.
