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I have a question regarding affine set.

Say given a vector space $V$,and two subspace say $U_1,U_2$ of $V$, and also given $x_1,x_2 \in V$. Also let $L_1$ = $x_1 +U_1$ and $L_2$ = $x_2 +U_2$

How do we show : $L_1 \subseteq L_2$ iff these two conditions $U_1 \subseteq U_2$ and $x_1 - x_2 =0$ are true.

Also I am little bit confused about what the $x_1,x_2$ could be, could they be vector like $x_1= [4,3,7,1]$, or $x_1,x_2$ has to be some constant number ?

I am new to affine space , so I am very confused.

Thank you

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  • $\begingroup$ There is no such thing as a "constant number" in a vector space. Elements of a vector space are called vectors, so yes, $x_1$ and $x_2$ are vectors. After the choice of a basis (and assuming the space is finite dimensional), they may be written as a column with some coefficients inside it. $\endgroup$
    – Suzet
    Jun 28, 2018 at 6:37
  • $\begingroup$ but could you also comment on how to solve the problem? thank you $\endgroup$
    – john_w
    Jun 28, 2018 at 7:03

2 Answers 2

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Your statement is wrong. If $U_1=U_2,x_2=0$ and $x_1 \in U_2\setminus \{0\}$ then $L_1 \subset L_2$ but $x_1 \neq x_2$. The correct statement is $L_1 \subset L_2$ iff $U_1 \subset U_2$ and $x_1-x_2 \in U_2$.

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As Kavi Rama Murthy stated it, the correct statement is $L_1 \subset L_2$ if and only if $U_1 \subset U_2$ and $x_1-x_2 \in U_2$.

For the direct implication, assume that $L_1 \subset L_2$. Then $x_1\in L_2=x_2+U_2$, which exactly means that $x_1-x_2 \in U_2$. Then, let us consider an element $z\in U_1$. We know by definition that $x_1+z$ belongs to $L_1\subset L_2=x_2+U_2$. It follows that $z\in (x_2-x_1)+U_2=U_2$ because we already know that $x_1-x_2 \in U_2$.

For the converse, assume that $U_1 \subset U_2$ and $x_1-x_2 \in U_2$. Let us consider any element of $L_1$. It can be written (uniquely) as $x_1+z$ for some $z\in U_1$. Then, $z\in U_2$ and our element $x_1+z$ may be written as $x_2 + (x_1-x_2+z)$. Because $x_1-x_2 \in U_2$, we know that this is an element of $x_2+U_2=L_2$, which is exactly what we wanted.

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  • $\begingroup$ hi Suzet, sorry I am just wondering could you explain your first paragraph where you said $x_1 \in L_2 ...$, how does $L_1 \subset L_2$ implies $x_1 \in L_2 $? $\endgroup$
    – john_w
    Jul 2, 2018 at 16:32
  • $\begingroup$ This is because $x_1=x_1+0 \in L_1$. Remember that $0$ is an element that lies in every vector spaces, so in particular in $U_1$. $\endgroup$
    – Suzet
    Jul 2, 2018 at 22:08

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