Equality of affine subspaces Let's say we have a vector space $V$ and two linear subspaces $U,W \subseteq V$ with $\dim(U)=\dim(W)$ especially, $U$ is isomorphic to $W$.
Now for any affine subspace by definition 
$$a+U=b+U \Leftrightarrow a-b \in U$$
What can be said about the equality of affine subspaces with different linear subspaces, i.e.
$$ a+W = a+U \Leftrightarrow ??$$
Obviously, this holds when $W = U$. Is $W \cong U$ enough? What does it mean for an affine subspace with different linear subspaces to be equal?
 A: In fact, the vector subspace is determined by the affine subspace. 

If $a+U=b+W$, where $U,W$ are subspaces of $V$, then $U=W$.

Proof. Let $A=a+U$ and $B=b+W$. 
Notice that $a=a+\vec 0\in A$, which implies $a\in B$, i.e., $a=b+\vec w$ for some $\vec w\in W$ and we get
$$b-a\in W.$$
By a very similar argument we can show $a-b\in U$.
$\boxed{U\subseteq W}$: Consider arbitrary $\vec u\in U$. Then we get $a+\vec u\in B$, which means that
$$a+\vec u=b+\vec w$$
for some $\vec w\in W$. From this we get
$$\vec u = \underset{\in W}{\underbrace{(b-a)}}+\underset{\in W}{\underbrace{\vec w}}$$
$\boxed{W\subseteq U}$: Proof of this part is analogous. (Or simply used the first part and symmetry.) $\square$
I will just mention that getting from $a+\vec u=b+\vec w$ to $\vec u = (b-a)+\vec w$ might still need a bit of work. This basically depends on the definition of affine space (and affine subspace) you are using. (But if you only work with $\mathbb R^n$, then it is just the usual addition and subtracting of ordered $n$-tuples.)
A: Perhaps the simplest way is to note that given the affine subspace
$$
\mathfrak{A} = a + U,
$$
then $U$ is uniquely determined as
$$
\{ x - y : x, y \in \mathfrak{A} \}.
$$
In fact, if $x, y \in \mathfrak{A}$, then $x = a + u_{1}$, $y = a + u_{2}$ for some $u_{1}, u_{2} \in U$, so that
$$
x - y = (a + u_{1}) - (a + u_{2}) = u_{1} - u_{2} \in U.
$$
And conversely, if $u \in U$, then $x = a + u, y = a \in \mathfrak{A}$, so that
$$
u = (a + u) - a = x - y.
$$

Allow me to note an easy and well-known generalization.

If $a + U, b + W$ are two affine subspaces, then either $(a + U) \cap (b + W) = \emptyset$,
   or $(a + U) \cap (b + W) = c + (U \cap W)$ for some $c$.

In fact, if $(a + U) \cap (b + W) \ne \emptyset$, let $c \in (a + U) \cap (b + W)$, so that
$$
a + U = c + U, \qquad b + W = c + W.
$$
Now $x \in (a + U) \cap (b + W)$ iff $x \in (c + U) \cap (c + W)$ iff $x = c + u = c + w$ for some $u \in U$ and $w \in W$ iff $x = c + v$, with $v (= u = w) \in U \cap W$.
