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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.

HaakonA
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  • This may answer your question, though it's not exactly a duplicate. https://math.stackexchange.com/questions/2262258/what-does-it-mean-to-be-affinely-independent-and-why-is-it-important-to-learn/2262272#2262272 – Ethan Bolker Jun 30 '21 at 15:48

3 Answers3

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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 $w = v+u$?

Noah Solomon
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  • Thank you for your reply. The problem I have is the following. Lets say I construct $T=v+U$. Now for $w$ to be in $T$, the smallest subspace I can come up with is $U={0,w,\lambda w}$, with $\lambda \in K$. I figure it must be like this since a subspace is closed under addition and multiplication. Thus $T$ will now contain $−w$, which would not be in $T$ if I started from $T=w+U$. Furthermore, both vectors can not be in $U$ because then you run the risk of spanning $K^2$ and it becomes difficult to select a unique base vector so that $T$ is a proper subspace. – HaakonA Jun 30 '21 at 05:41
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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.

Dude1662
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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?

HaakonA
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  • It makes it more readable if you mark the first paragraph in some way to show that you do a uniqueness proof. And mark the second paragraph to show you have an existence proof. – Dude1662 Jun 30 '21 at 22:32
  • Your proof is correct, but there were some errors writing everything down. – Dude1662 Jun 30 '21 at 22:33
  • For example: "Now if U contains a vector, v, with v∉ span(w−v) then U=K2. Thus U=span(w−v)." Here you use the variable v twice. – Dude1662 Jun 30 '21 at 22:35
  • For the uniqueness part: You have only shown that U is span(w-v). You still have to show that if there are x and y, such that x+U contains v and w and y+U contains v and w, x+U=y+U – Dude1662 Jun 30 '21 at 22:42
  • "How we chose to write x does not matter, because any vector x=w−u2 can be written as x=v−u3 by setting u3=u2−(w−v) (which is obviously ∈U)." This sentence is unneccessary. You have already shown that $x=w-u_2$. This is all you need for the rest of the proof.(to show existence of such an affine space containing v and w but not the whole space.) – Dude1662 Jun 30 '21 at 22:45