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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.

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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 $0$.

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  • $\begingroup$ Could you elaborate on how you got to the third line from the second? $\endgroup$ – sangstar Nov 11 '18 at 17:39
  • $\begingroup$ 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 $\mathscr B$). $\endgroup$ – José Carlos Santos Nov 11 '18 at 17:41
  • $\begingroup$ 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. $\endgroup$ – sangstar Nov 11 '18 at 17:49
  • $\begingroup$ 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 $W$. $\endgroup$ – José Carlos Santos Nov 11 '18 at 17:59
  • $\begingroup$ I agree with $[v] = [v^*] \iff v - v^* \in W$. However, how does this imply that $[a_1v_1 + ... + a_nv_n] = [0] \iff a_1v_1 + ...+a_nv_n \in W$? $\endgroup$ – sangstar Nov 11 '18 at 18:05
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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 $a_1 = \cdots = a_k = 0$ $(= b_1 = \cdots = b_m)$.

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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.

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