Quotient spaces in linear algebra I'm having a bit of difficulty understanding what a quotient space is to a vector space $V$. I will present the part I'm finding trouble with below. 
Let $V$ be a vector space and let $U$ be a sub space of $V$. We define an equivalence relation by setting $v\approx w$ if $v-w\in U$, the way I understand this is it means squishing the size of the sub-set $U$ to zero? The set of equivalence classes of this relation is denoted $V/U$, it is itself a vector space. The map from $V$ to $V/U$ is the projection $\pi$, such that $\pi:V\rightarrow V/U$. 
$v$ is a class representative of $[v]$ the equivalence class. Lastly the dim$(V/U)=$dim$V-$dim$U$.
It is this last section I'm having trouble with. What is $[v]$, it is seemingly called the same thing as $V/U$ namely a equivalence class. So here are my questions;
1) what does it mean that $v$ and $w$ are linked by an equivalence relation?
2) what is the difference between $[v]$ and $V/U$? 
 A: (1) An equivalence relation is a relation $\sim$ which satisfies the following conditions:
$$x \sim x \qquad \mbox{ for all } x \tag{reflexive}$$
$$x \sim y \iff y \sim x \tag{symmetric}$$
$$x \sim y \mbox{ and } y \sim z \implies x \sim z \tag{transitive}$$
If you replace $\sim$ with $=$, these three properties hold, so we might informally say that an equivalence relation is a relation which is 'like' equality.  
In your case, $x \sim y \iff x-y \in U$, so saying $v$ and $w$ are linked by the equivalence relation $\sim$ means (geometrically) that if we consider the hyperplane $U+v$ of all vectors which have the form $u+v$ for $u \in U$, then $w$ is in this hyperplane.  
(2) The equivalence class $[v]$ is the set of all vectors $w$ such that $v \sim w$.  In fact, $V/U$ is not an equivalence class: it is the vector space of all equivalence classes $[v]$, where $v \in V$.  To see why $V/U$ is a vector space define addition and scalar multiplication by
$$[v] + [w] = [v+w]$$
$$k[v] = [kv]$$
It is not entirely obvious that these operations are well defined: if $v' \in [v]$ (that is, if $v' \sim v$) and $w' \in [w]$, then $[v'] = [v]$ and $[w'] = [w]$, so we would need to show that $[v'+w'] = [v+w]$ and that $[kv'] = [kv]$ (that is, that the value of $[v] + [w]$ and $k[v]$ does not depend on the representative we choose for $[v]$ and $[w]$).  However, we can prove that if $v \sim v'$ and $w' \sim w$, then
$$kv' \sim kv$$
and 
$$v'+w' \in v + w$$
(I'll leave the details to you.  Just work using the definition of $\sim$).  From this, we have that $[v'+w'] = [v+w]$ and $[kv'] = [kv]$ by transitivity.  
A: On $V$ define the relation $v_1 \sim v_2$ if $v_1 - v_2 \in U.$ Then it is an equivalence relation on $V.$ An equivalence class (under this equivalence relation) is denoted by $[v]$ for some $v \in V.$ By definition, $[v] = \{w \in V |  v-w \in U \}.$ The set of all equivalence classes is denoted by $V/U.$ Then $V/U$ has a natural vector space structure induced from the vector space structure of $V$ itself.
$V/U = \{[v] | v \in V \}$ where we have the assumption (by construction) that $[v_1] = [v_2] \Leftrightarrow v_1 \sim v_2 \Leftrightarrow v _1 - v_2 \in U.$ The map $\pi : V \rightarrow V/U$ is defined by $v \mapsto [v].$ This is a linear transformation.
A: 1) Not a term, I have heard of, but in this case I think it simply means: $v \approx w$ or equivalently $v-w\in U$.
You can view an equivalence relation as a kind of "weak equality", since it behaves like equality, it is reflexive, transitive and symmetric, but $v \approx w$ follows from $v = w$, but not the other way around. Of course: $v \approx w \Leftrightarrow [v]=[w]$.
2) $[v]\in V/U$.  The set $[v]$ contains all $w\in V$ with $w\approx v$. The set $V/U$ contains all equivalence classes, that is sets of the form $[v]$ for every $v\in V$.
Specifially:
$$[v] = \{ w\in V : w\approx v\}$$
for all $v\in V$ and:
$$V/U = \{ S \subseteq V : \exists v\in V : S=[v]\} = \{[v] : v\in V \}$$
