# Linear Algebra - Intersection of Affine Spaces

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.

• 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. – SvanN Dec 24 '18 at 15:44
• If $x\in v_1+W_1$, then $v_1+W_1=x+W_1$ etc. – Angina Seng Dec 24 '18 at 15:53

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