# Prove that, with vector addition and scalar multiplication well-defined, $V/W$ becomes a vector space over $k$.

Let $V$ be a vector space over a field $k$ and let $W$ be a subspace of $V$. Prove that, with vector addition and scalar multiplication well-defined, $V/W$ = {$v+W | w\in W$} becomes a vector space over $k$.

I know that vector addition being well-defined means that $(v+w) - (v'+w') = w_1+w_2 \in W$ where $v-v'= w_1 \in W$ and $w-w' = w_2 \in W$

And I know that scalar multiplication being well-defined means that $ru-ru' = r(u-u') = ru \in W$ where $u-u'=W$

I know how to prove the additive identity (it is $0$) and the additive inverse (it is $-v+w$) but I'm getting confused on how to prove the commutativity, associativity, and the distributive properties. Would i just use an element in $V$ to prove those or do I need to use an element from $V/W$?

• You have to lift the elements of $V/W$ in $V$, use the corresponding properties in $V$ and get back down to $V/W$. Feb 24 '16 at 1:02

your approach on vector addition and scalar multiplication being well-defined are spotted on!

To prove linearity,

Let $\alpha: -> V/W$ be the function $v \in V$ the equivalence class $\alpha(v) \in V/W$ to which $v$ belongs. Then we know $\alpha$ is a linear map

To prove it, we could say the following:

$\alpha(av+bw)$ is the equivalent to which $av+bw$ belongs. This is by definition of the sum of the classes containing $av, bw$, and it means

= $a$ times the class of $v$ + $b$ times the class of $w$,

That is $\alpha(av+bw)=a\alpha(v)+b\alpha(w)$ and we are done