Affine subspace of two vectors in a field K I have two vectors $v = (v_1,\ v_2)$ and $w = (w_1,\ w_2)$, with $w, v \ \in K^2$ and $v \neq w$. The task is to find the (unique, if I understand the task correctly) affine subspace that contains both vectors.
The definition of a affine subspace $T$ is:
$T = x + U = \big\{ x+u|u \in U \big\}$, where $U \subseteq K^2$ and $x \in K^2$.
The task tells us to find the affine subspace which is a proper subset of K^2 ($T \subsetneq K^2$). The parts I do not understand: unique and proper subset. What if the two vectors are linearly independent, then any subset $U$ that contains them both must be equal to K^2.
Should I in this case construct $T$ by setting $x$ equal to one of the vectors and construct $U$ as the subset containing the other? But in this case the choice wont be unique or will it?
Question: How does one construct a unique (if unique is possible) affine subspace from the two vectors described above.
 A: What is confusing you is that we do not require affine subspaces to be vector spaces themselves, so while it is true that any subspace of the vector space $k^2$ containing two linearly independent vectors must necessarily contain the entirety of $k^2$, the same does not hold for affine subspaces. In your definition, $U$ is a vector subspace of $k^2$, but $T$ is not, because for example, it might not contain the origin.
Start by constructing an affine subspace $T$ which has one of the given vectors, say $v$ as its base point. What is the minimal vector subspace of $U \subseteq k^2$ which contains some $u\in U$ such that $w = v+u$?
A: If you think about affine Spaces U+w they are usual vector spaces U moved by a vector w. You can think about U+w and then try to figure out what U must look like so that v is in U+w.
For uniqueness you can assume that an affine vector space contains both w and v and then show that it must contain U+w. Then try to think about the other inclusion by rewriting the affine space as H+w.
A: I got it, thanks for the tips.
If $w, v \in T = x + U$, where $U \subsetneq K^2$ and $x \in K^2$,then
$w-v \in U$. This can be proven easily.
This implies that span($w-v$) $\subseteq U$. Now if $U$ contains a vector, $v$, with $v \notin $ span($w-v$) then $U=K^2$.
Thus $U =$span($w-v$).
Since both $w$ and $v$ are in $T$ we must have
$w = x + u_2$ and v = $x + u_3$ where $x$ is some vector in $K^2$ and $u_2, u_3 \in U$. This implies that $x = w-u_2$ and $x=v-u_3$. How we chose to write $x$ does not matter, because any vector $x = w-u_2$ can be written as $x=v-u_3$ by setting $u_3=u_2 - (w-v)$ (which is obviously $\in U$).
We then have that $T$ can be wirtten as $T = w-u_2 + U$. Now for any vector, $u\in U$, we have that $u+U=U$. Because a subset is closed under addition.
We can therefore, write $T = w+U$. We can also write $T$ using any of the two forms of $x$ given above, T will always contain the same vectors no matter which form you chose.
Am I right?
