Question

Suppose v_1,v_2,........,v_{m }is a linearly independent in V and w∈V. Show that v_1,v_2,........,v_{m }is linearly independent if and only if w∉span(v_1,v_2,........,v_{m )}

Answer 1

Suppose that $w\in \text{Span}(v_1,...,v_m)$, then we have $w=\sum \alpha_i v_i$ and thus $\sum \alpha_iv_i-w=0$ is a non-trivial (as at least the coefficient before $w$ is $-1\neq 0$ ) linear combination of $(v_1,...,v_m,w)$ that gives zero and thus $(v_1,...,v_m,w)$ is not linearly independent.

Now let's suppose that $(v_1,...,v_m,w)$ is linearly dependent, then there exists a non-trivial linear combination such that $\sum \alpha_i v_i +bw=0$. We can not have $b=0$ as in this case we would have $\sum \alpha_i v_i=0$ for sum non-trivial combination $(\alpha_i)$. Therefore we have $w=\sum(-\alpha_i/b)v_i \in \text{Span}(v_1,...,v_m)$.

$\neg((v_1,...,v_m,w) \text{ are linearly independent}) \leftrightarrow \neg (w\notin \text{Span}(v_1,...,v_m))$ and therefore we have (by contraposition)

$((v_1,...,v_m,w) \text{ are linearly independent}) \leftrightarrow (w\notin \text{Span}(v_1,...,v_m))$