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Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$. $v_1, v_2 \in V$ and $(v_1 + W_1) \cap(v_2 + W_2) \neq \emptyset$.

Prove that $(v_1 + W_1) \cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 \in V$ so that $(v_1 + W_1) \cap(v_2 + W_2) = v_3 + W_3 $.

I have found this previous question but I couldn't figure out the next steps of proving this.

We know that $\exists x \in (v_1 + W_1) \cap(v_2 + W_2) $.

I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 \in W_1, w_2 \in W_2 , w_1 = w_2$.

Would appreciate some points and guidelines about how to approach this.

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    $\begingroup$ It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection. $\endgroup$
    – SvanN
    Dec 24, 2018 at 15:44
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    $\begingroup$ If $x\in v_1+W_1$, then $v_1+W_1=x+W_1$ etc. $\endgroup$ Dec 24, 2018 at 15:53

1 Answer 1

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The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that $(x+W_1)\cap (x+W_2)=x+(W_1\cap W_2)$.

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