Proving Statement about Independent Sets and Spans

Question

I am having trouble proving the following statement:

Suppose that $V$ is a vector space and $S \subset V$. If $S$ is a linearly independent set and $S \cup \{v\}$ is linearly dependent, then $v \in \text{span}(S)$.

My thinking:

let $S = \{v_1,...,v_n\}$. Since $S$ is a linearly independent set, we have that $$\alpha_1 v_1 + ... + \alpha_n v_n = \mathbf{0} \Rightarrow \alpha_1 = ... = \alpha_n = 0$$ Where $\alpha_1,...,\alpha_n$ are scalars.

The aim is to show that there exist scalars $\beta_1,...,\beta_n$ such that we can write $v = \beta_1 v_1 + ... + \beta_n v_n$

I want to use the fact that the equation $$\gamma_1 v_1 + ... + \gamma_n v_n + \gamma v = \textbf{0}$$ $\gamma_1,...,\gamma_n, \gamma$ are scalars; has a solution other than $\gamma_1 = ... = \gamma_n = \gamma = 0$. I am unsure where to go from here.

Answer 1

Solve for $v$

$$\gamma_1 v_1 + ... + \gamma_n v_n + \gamma v = \textbf{0}$$

If $\gamma$ is $0$ then by linear independence of $v_1, v_2,..., v_n$ all the coefficients will be $0$ and as a result $v=0$ which is in the span.

Otherwise, you divide by $\gamma$ to get $$v= -(\gamma_1 v_1 + ... + \gamma_n v_n )/\gamma$$

Thus, in any case, $v$ is the span of $v_1, v_2, ...,v_n$