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.
Best Answer
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 $(x+W_1)\cap (x+W_2)=x+(W_1\cap W_2)$.