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:
$a\notin span\{v_1,v_2,v_3...v_n\}$.
It is also known that $v_n\in span\{v_1,v_2...v_{n-1},a\}.$
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?