Why does the quotient space V/W not equal the vectorspace V?

Question

Let's say $V=\Re^3$ and $W$ is a plane through the origin.

The way I understand the quotient space $V/W$ is that it's formed by taking every vector $\vec{v}^{\,} \in V$ and adding it to the subspace $W$.

Why doesn't this make $V/W = V$?

Answer 1

The quotient space is the set $$V/W = \{[v]\mid v \in V\},$$ where $[v] = \{ v + w \mid w \in W \}$. Now, it does not really make sense to ask if $V$ and $V/W$ are equal since their elements do not belong to the same kind of set (one consists of vectors, the other of classes of vectors). The best you could hope for would be that the natural map $V \to V/W$ given by $v \mapsto [v]$ is a bijection, so that the two spaces are "the same size" (that they are isomorphic as vector spaces), with the elements in one set corresponding naturally to elements in the other. Now in your example, this is not the case because you could take two different vectors $w_1, w_2 \in W$ and such two would satisfy $[w_1] = [w_2]$, so that the map is not injective. It will in fact be surjective, and it would have also been injective if $W$ were the trivial space. More generally, as long as we're dealing with finite dimensional spaces, $$\dim(V/W) = \dim(V) - \dim(W).$$

(Edit, just before someone complains: The above formula should convince you that as vector spaces, the two sets are really different. However, you could cook up a bijective correspondence between the two spaces so that in some (explicit) sense, they have the same number of elements but this correspondence will not be linear and therefore somewhat uninteresting from the viewpoint that the spaces are not just arbitrary sets but actually vector spaces.)

Answer 2

The sets $V/W$ and $V$ consist of completelly different elements. It is why $V/W \neq V$. $V$ consists of vectors but $V/W$ consists of a set of vectors constructed with some rules. The rules does not matter in this context.