Proof that operations of $V/W$ are well defined?

Question

I have a rather challenging question on my assignment and I have put in my best effort for now. I think I just need a tiny nudge to set me in the right direction to finish this proof. If you could have a look, that would be great!

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Background on Cosets and Operations Defined on $V/W$

Let $W$ be a subspace of a vector space $V$ over $\mathbb{F}$. For some fixed $\mathbf{v} \in V$, a coset of $W$ is defined to be the set

$$ \{\mathbf{v}\} + W = \{ \mathbf{v} + \mathbf{w} \,\,|\,\, \mathbf{w} \in W \}. $$ We usually denote this as $\mathbf{v} + W$, though.

The set of all cosets of $W$ is denoted by $$ V/W=\{ \mathbf{v} + W \,\,|\,\, \mathbb{v} \in V \}. $$

Addition and scalar multiplication are defined on $V/W$ by $$ (\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1} + \mathbf{v_2}) + W, \quad \text{and} \\ k(\mathbf{v} + W) = k\mathbf{v} + W, \quad k \in \mathbb{F}. $$

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The Question

I am required to prove that these operations are well-defined, i.e. if $\mathbf{v_1} + W = \mathbf{v_1'} + W$ and $\mathbf{v_2} + W = \mathbf{v_2'} + W$, then $$ (\mathbf{v_1} + W) + (\mathbf{v_2} + W) = (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W), \quad \text{and}\\ k(\mathbf{v_1} + W) = k(\mathbf{v_1'} + W), \quad \forall k \in \mathbb{F}. $$

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My Problem

I am having great difficulty proving the first part. So far I have this. $$ \begin{align*} \mathbf{v_1} + W = \mathbf{v_1'} + W &\implies \mathbf{v_1} - \mathbf{v_1'} = \mathbf{w_1} \in W \implies \mathbf{v_1} = \mathbf{v_1'} + \mathbf{w_1}, \quad \text{and}\\ \mathbf{v_2} + W = \mathbf{v_2'} + W &\implies \mathbf{v_2} - \mathbf{v_2'} = \mathbf{w_2} \in W \implies \mathbf{v_2} = \mathbf{v_2'} + \mathbf{w_2}. \end{align*} $$

Then, $$ \begin{align*} (\mathbf{v_1} + W) + (\mathbf{v_2} + W) &= (\mathbf{v_1} + \mathbf{v_2}) + W\\ &=\left( (\mathbf{v_1'} + \mathbf{w_1}) + (\mathbf{v_2'} + \mathbf{w_2}) \right), \quad \mathbf{w_1}, \mathbf{w_2} \in W\\ &\quad \,\, \text{Let } \mathbf{w_1} + \mathbf{w_1} = \mathbf{w_3} \in W.\\ &=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + \mathbf{w_3} \right) + W\\ &=\left( (\mathbf{v_1'} + \mathbf{v_2'}) + W \right) + (\mathbf{w_3} + W)\\ &= (\mathbf{v_1'} + W) + (\mathbf{v_2'} + W) + (\mathbf{w_3} + W), \quad \mathbf{w_3} \in W. \end{align*} $$

As you can see, this is frustratingly close to the result I wanted to prove. The only thing in the way is the extra $+ (\mathbf{w_3} + W)$. Is there some way I could make that disappear though? Or is my whole proof just wrong?

The only thing I know about the coset $\mathbf{w} + W$ where $\mathbf{w} \in W$ is that it is a subspace of the vector space $V$. Could I somehow use this fact?

Answer 1

Your proof is correct and can be finished by noting that if $w \in W$ then $w + W = W$. The reason is that

$$ w + W = \{ w + w' \, | \, w' \in W \} $$

and since $W$ is a subspace, it is closed under addition so we have $w + W \subseteq W$. On the other hand, if $w' \in W$ then $w' = w + (w' - w)$ (where $w' - w \in W$ again because $W$ is a subspace) so we also have $W \subseteq w + W$ and hence $w + W = W$.

Answer 2

$w_3 + W = W$!

Proof : $x \in w_3 + W \implies x = w_3 + w$ for some $w \in W $, which means $x \in W$ since $W$ is a subspace.

On the other hand, $x \in W \implies x = w_3 + (x - w_3) \in w_3 + W$, since $W$ is a subspace, so $x-W_3 \in W$.

Hence, continuing from your line of thought: $$ (v_1'+W) + (v_2' + W) +(w_3'+W) = (v_1'+W) + (v_2'+W) + W \\ = (v_1' + W) + (v_2' + W + W) = (v_1' + W) + (v_2' + W) $$

Which completes the proof.

Use a similar logic to prove that $kW = W$ for any non-zero scalar $k$. This will be helpful for your second part.

Answer 3

Suppose $v_{1} + W = v_{1}^{'} + W$ and $v_{2} + W = v_{2}^{'} + W.$ Then, as you note, we have $v_{1}-v_{1}^{'} \in W$ and $v_{2}-v_{2}^{'} \in W.$ As $W$ is a subspace, it is closed under addition, so $(v_{1}-v_{1}^{'}) + (v_{2}-v_{2}^{'}) \in W.$

Now, rearrange the terms to obtain $(v_{1}+v_{2}) - (v_{1}^{'} + v_{2}^{'}) \in W.$ Then, we have $(v_{1}+v_{2}) + W = (v_{1}^{'} + v_{2}^{'}) + W.$ Thus, from the definition of addition, it follows that $$(v_{1} + W) + (v_{2} + W) = (v_{1}^{'} + W) + (v_{2}^{'} + W),$$ which shows that addition is well defined.

Similarly, we show that scalar multiplication is well defined. Namely, $a(v_{1} - v_{1}^{'}) \in W,$ since it is closed under scalar multiplication. Then, we have $av_{1} - av_{1}^{'} \in W,$ which implies $av_{1} + W = av_{1}^{'} + W.$ Thus, from the definition of scalar multiplication, it follows that $$a(v_{1} + W) = a(v_{1}^{'} + W),$$ which shows that scalar multiplication is well defined.