Show $Q$ with the given addition and scalar multiplication is vector space where $Q = \{v+ H\}$

Question

Let $H$ be a subspace of $V$.

For $c\in V,$ define $E(c) = \{c + h\,|\,h\in H\}$

Let $Q = \{ E(v)\,|\,v\in V\}$.

Define addition in $Q$ by: for $v, w\in V$, $E(v) \bigoplus E(w) = E(v+w)$

Define scalar multiplication in $Q$ by: for $v\in V$ and $\propto \in \mathbb{R}, \propto E(v) = E(\propto v)$.

Show that $Q$ together with this addition and scalar multiplication is a vector space.

My Attempt:

To check that $Q$ is a vector space, one must check each of the 10 axioms of a vector space to see if they hold.

$A_1:$ Let $a, b\in Q$. Then \begin{align*} a + b = E(a) + E(b) \in Q \end{align*} Therefore $Q$ is closed under addition ($A_1$ holds).

I haven't completed the whole solution yet, but you could imagine how long it would take. Is there a shorter and better way of proving?

Answer 1

You can see $Q$ as the subspace of the vector space $V$: Using your definition, we have $Q=\{v+H:v\in V\}$. This is nothing else than the quotient space (that's why the letter 'Q' is used :D) $Q=V/H$ with equivalence classes $[v]=v+H=\{v+h:h\in H\}$.

Let's take the direct sum $V=H\oplus H^\perp$ , then the quotient space is naturally isomorphic to the orthogonal complement of $H$: $$Q=V/H\simeq H^\perp$$.

This shows already, that $Q$ is isomorphic to a subspace of $V$ and thus a subspace itself.

See https://en.wikipedia.org/wiki/Quotient_space_(linear_algebra) for more details about Quotients of vector spaces.

Edit: If you really want to calculate the vector space axioms you can use the fact that the $+$ and $\cdot$ operation are inherited from $V$ and thus you only need to verify that $Q$ is a subspace. For this you only need to verify two axioms

1. $Q\neq \emptyset$
2. $\forall [v],[w]\in Q,\lambda\in\mathbb{R}$ we have $[v]+\lambda [w]\in Q$

and that should be easy.