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Let V be a vector space over a field F and let W be a subspace of V . Show that if B ={$w_1,...,w_m$} is a basis for W, and $v_1,...,v_k$ $\in$ V are such that $B \cup$ {$v_1,...,v_k$} is a basis for V , then{$[v_1],[v_2],...,[v_k]$} is a basis for the quotient space V/W.

Consider $a_1[v_1]+....+a_k[v_k]=0$ By linearity, we get $[a_1v_1+....+a_kv_k]=0$. By definition of the equivalence class, $a_1v_1+.....+a_kv_k\in W$. Hence $a_1v_1+....+a_kv_k=B_1w_1+....+B_mw_m$. Then $a_1v_1+....+a_kv_k-B_1w_1-....-B_mw_m=0$. As $a_1,v_1....,v_k,w_1,....,w_m$ is a basis for $V$, we get from linear independence, $a_1=a_2=…..=w_1=…=w_m=0$.

In particular, $a_1=a_2=…..=a_k=0$.

As $dim(V/W)=dim(V)-dim(W)=n+k-k=k$, it suffices to show linear independence.

Is the above proof correct?

Note: The below proof below proved $dim(V/W)=dim(V)-dim(W)$, where $V/W$ is finite dimensional.

it suffices to consider the map $\Pi : V\rightarrow V/W$ given by $\Pi(v)=[v]$. This is well defined, because the operations are well defined, and similarly, linear. It is clearly surjective and so by the Rank Nullity theorem, $dim(V)=dim(V/W)+dim(ker\Pi)$, hence it suffices to show that $dim(Ker\Pi)=dim(W)$. Observe that if $v\in Ker\Pi$ then $\Pi(a)=[a]=[0]$ and consequently, $a=a-0\in W$. Further, if $v\in W$ then $\Pi(v)=[v]=[0]$, since $v=v-0\in W$ so $v\in ker(\Pi)$. Therefore, $dim(Ker\Pi)=dim(W)$. Hence we conclude that $dim(V/W)=dim(V)-dim(W)$.

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    $\begingroup$ But how did you prove that $\dim(V/W) = \dim V - \dim W$ in the first place? If you use that fact you are correct and done, but you have to make sure you haven't been circular i.e. that this thing you proved above was not used in the proof of $\dim(V / W) = \dim V - \dim W$ To get around this, showing that $[v_i]$ span $V/W$ should be doable easily. $\endgroup$ Commented Nov 14, 2019 at 6:29
  • $\begingroup$ @астонвіллаолофмэллбэрг This was already proved $\endgroup$ Commented Nov 14, 2019 at 6:31
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    $\begingroup$ Then you are done. Just asking : how was it proved? (In brief) $\endgroup$ Commented Nov 14, 2019 at 6:31
  • $\begingroup$ @астонвіллаолофмэллбэрг I just wrote the proof. Is it correct? $\endgroup$ Commented Nov 14, 2019 at 6:42
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    $\begingroup$ You are welcome! For completeness I have written up the spanning part as well. +1 for your question, I wish others put effort into their questions like you do. $\endgroup$ Commented Nov 14, 2019 at 7:05

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I am convinced by your proof and the fact that it is not circular.

If you wanted to avoid using $\dim(V / W) = \dim V - \dim W$ then you could have shown that $[v_i]$ spanned $V / W$ as follows : if $[v] \in V$, we write $v = \sum c_iv_i + \sum b_kw_k$ since $B \cup \{v_i\}$ spans $V$. But then, by linearity we get $[v] = \sum c_i [v_i] + \sum b_k[w_k] = \sum c_{i}[v_i]$ since each $w_k \in W$ so $[w_k] = [0]$. Thus spanning is clear.

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