I am tasked with the following:
Let $V$ be a vector space over a field $F$ and let $W$ be a subspace of $V$. Show that if $\mathscr{B}=\{w_1,\dots,w_m\}$ is a basis for $W$, and $v_1,\dots,v_k\in V$ are such that $\mathscr{B}\:\cup\{v_1,\dots,v_k\}$ is a basis for $V$, then $\{[v_1],[v_2],\dots,[v_k]\}$ is a basis for the quotient space $V/W$.
I tackle this first, by noting that the dimension of the set $\left\{[v_1],[v_2,],...,[v_k]\right\}$ is equal in dimension to $\dim V/W$, as $\dim V/W = \dim V - \dim W = (k + \dim W) - \dim W = k$.
I then attempted to prove that $\left\{[v_1],[v_2,],...,[v_k]\right\}$ is linearly independent, which is where I fell short. I postulate that, for linear independence,
$a_1 [v_1] + a_2[v_2] + \ ... \ + a_k [v_k] = [0]$
With the only solution $a_1=a_2= \dots =a_k = 0$.
Due to the quotient space as a vector space definition, I know I can do this..
$\implies [a_1 v_1] + [a_2v_2] + \ ... \ + [a_k v_k] = [0]$
$\implies a_1=a_2= \ ...\ =a_k = 0$. Or,
$\implies [v_i] = [v_j], \text{for some $i,j$ such that $1 \le i,j \le k$}$.
$\implies v_i R_w v_j$
As far as I know in my class, the only equivalence relation we've been dealing with in this class regarding the quotient space is:
$\implies v_i R_w v_j \implies v_i - v_j \in W$
However, I have no idea how this is a contradiction, and it must be for linear independence.