Proving a set of equivalence classes is a basis for quotient space $V/W$

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.

Note that\begin{align}a_1[v_1]+\cdots+a_k[v_k]=0&\iff[a_1v_1+\cdots+a_kv_k]=[0]\\&\iff a_1v_1+\cdots+a_kv_k\in W\\&\iff(\exists b_1,\ldots,b_m):a_1v_1+\cdots+a_nv_n-b_1w_1-\cdots-b_mw_m=0.\end{align}Since $$\mathscr{B}\cup\{v_1,\ldots,v_k\}$$ is a basis of $$V$$, this implies that all coefficients are $$0$$.

• Could you elaborate on how you got to the third line from the second? – sangstar Nov 11 '18 at 17:39
• In order to do that, all I did was to use two facts: that $a_1v_1+\cdots+a_nv_n\in W$ and that $\mathscr B$ is a basis of $W$ (which implies that every element of $W$ can be written as a linear combination of the elements of $\mathscr B$). – José Carlos Santos Nov 11 '18 at 17:41
• Right, and a couple things: firstly, we know that, for instance, $[a_1v_1] \implies a_1v_1 \in W \ \text{because} \implies a_1v_1 \ R_w \ a_1v_1 \implies a_1 v_1 - a_1v_1 = 0 \in W$? If not, I'm not sure how you went from line 1 to 2. Secondly, why does $B$ being a basis for $W$ imply the third line exactly? It doesn't come together like that for me. – sangstar Nov 11 '18 at 17:49
• First of all, $[v]=[v^\star]\iff v-v^\star\in W$. In particular,$$[a_1v_1+\cdots+a_nv_n]=[0]\iff a_1v_1+\cdots+a_nv_n\in W.$$On the other hand,$$a_1v_1+\cdots+a_nv_n\in W\implies(\exists b_1,\ldots,b_m):a_1v_1+\cdots+a_nv_n=b_1w_1+\cdots+b_mw_m$$because $\{w_1,\ldots,w_m\}=\mathscr B$, which is a basis of $W$. – José Carlos Santos Nov 11 '18 at 17:59
• I agree with $[v] = [v^*] \iff v - v^* \in W$. However, how does this imply that $[a_1v_1 + ... + a_nv_n] = [0] \iff a_1v_1 + ...+a_nv_n \in W$? – sangstar Nov 11 '18 at 18:05

For $$a_1,\ldots,a_k \in F$$ such that $$a_1 [v_1] + a_2 [v_2] + \cdots + a_k [v_k] = [0]$$, we have $$[a_1 v_1 + \cdots + a_k v_k] = [0]$$, i.e. $$a_1v_1 + \cdots + a_kv_k \in W$$. Hence, as $$\{w_1, \ldots, w_m\}$$ is a basis of $$W$$, there are $$b_1,\ldots, b_m \in F$$ such that $$a_1 v_1 + \cdots + a_k v_k = b_1 w_1 + \cdots + b_m w_m$$, equivalently, $$a_1 v_1 + \cdots + a_k v_k + (-b_1)w_1 + \cdots + (-b_m)w_m = 0$$. By the linear independence of $$\{w_1,\ldots, w_m, v_1,\ldots, v_k\}$$, we get that $$a_1 = \cdots = a_k = 0$$ $$(= b_1 = \cdots = b_m)$$.

Denote $$V_1=\bigl\langle v_1,\dots,v_k\bigr\rangle$$. If $$\;\{w_1,\dots,w_m\}\cup\{v_1,\dots,v_k\}\;$$ is a basis for $$V$$, it means $$W\cap V_1=\{0\}$$.

Now $$\; a_1[v_1]+\cdots+a_k[v_k]=[a_1v_1+\cdots+a_kv_k]=[0]$$ means that $$\;a_1v_1+\cdots+a_kv_k\in W$$, hence it is $$0$$ since $$W\cap V_1=\{0\}$$. However, these vectors are linearly independent since they're a subset of a basis of $$V$$. So $$\;a_1=\cdots=a_k=0$$, which prove the equivalence classes $$\;[v_1], \cdots,[v_k]\;$$ are linearly independent.