Prove that $\{v_1,v_2...v_n\}$ is linearly dependent if $a\notin \text{span}\{v_1,v_2,v_3...v_n\}$ and $v_n\in \text{span}\{v_1,v_2...v_{n-1},a\}$. I ran into the next problem and got really confused:

Let $\{ v_1, v_2,v_3... v_n \}$ be a set of vectors in the vector space $V$, and let $a\in V$ in such a way that:
$a\notin \text{span}\{v_1,v_2,v_3...v_n\}$.
It is also known that $v_n\in \text{span}\{v_1,v_2...v_{n-1},a\}.$
Prove that the set $\{v_1,v_2,v_3...v_n\}$ is linearly dependant.

So what I have in mind is that $a$ can't be represented as a linear combination of vectors in $\{v_1,v_2,...v_n\}$, so $a\ne \gamma_1v_1+\gamma_2v_2+...+\gamma_nv_n$.
But, on the other hand, $v_n = \beta_1v_1+\beta_2v_2+...+\beta_{n-1}v_{n-1}+\beta_na$.
Is it alright to manipulate this thing to become:
$-\beta_na=\beta_1v_1+\beta_2v_2+...+\beta_{n-1}v_{n-1}-v_n$?
And now I can divide by $(-\beta_n)$ to represent $a$ as a linear combination of the others while I give the new divided scalars a new notation "$\gamma_i$"
such that:
$a=\gamma_1v_1+\gamma_2v_2+...+\gamma_nv_n $ and of course that could not happen by the intel they gave us above. Therefore the scalar of $a$ back then, the $\beta_n$ must have been equal to zero so I wont be able to divide and get to that conclusion.
Therefore I could represent $v_n$ by the span of $\{v_1,v_2...v_{n-1}\}$ and this is why  $\{v_1,v_2,v_3...v_n\}$ is linearly dependent.
Is it the right way to look at it?
 A: $\text{ Since } v_n \in \text{ span } \{v_1,v_2,\cdots,v_{n-1},a\} \to v_n = c_1v_1 + c_2v_2 +\cdots c_{n-1}v_{n-1} + c_na$. $\text{ But }  a \notin \text{span } \{v_1,v_2,\cdots,v_n\}, c_n = 0 \to 0 = c_1v_1+c_2v_2+\cdots + c_{n-1}v_{n-1} + (-1)v_n \to v_1,v_2,\cdots, v_n \text{ are linearly dependent }$.
A: Looks like you have it, though you might want to focus on writing cleanly and simply. You can write your steps very clearly:
By hypothesis, $v_n\in\text{span}(v_1,\dots,v_{n-1},a)$ which means by definition $v_n = \beta a + \sum \beta_iv_i$.
We can rewrite this as $-\beta a = -v_n + \sum \beta_iv_i$.
Next we want to show $\beta=0$. Suppose $\beta\neq 0$. Then we can divide by $-\beta$ and we get $a = -\frac{1}{\beta}v_n + \sum \frac{\beta_i}{\beta}v_i$. But this contradicts the hypothesis that $a$ is not in the span of the $v_i$'s. In my definitions, this includes the case of $a=0$ (think through that).
Therefore we have $0 = -v_n + \sum \beta_i v_i$
Since at least one of coefficients is not zero, this shows the $v$'s are linearly independent.
I think the important thing here is that you clearly state each step and make sure each line follows from the previous one. You have the right ideas but I think you might want to work on clearly/cleanly writing them down and connecting them. Make sure you feel comfortable going from one to the other. This is the key to knowing you have a correct proof.
