Proving that the intersection of affine spaces contains only a single vector

Question

Let $V_1$ and $V_2$ be subspaces of $V$, such that $V=V_1 \oplus V_2$. Prove that the intersection of affine spaces $V_1 + a$ and $V_2 + b$ contains a single vector.

I understand the intuition behind this, since $V_1 \cap V_2 = {0}$ and then with two affine spaces formed with vectors from $V$ they're going to intersect at some vector from $V$, but how do I formally prove that?

Answer 1

There is a unique pair $(a_1,a_2)\in V_1\times V_2$ with $a=a_1+a_2$.

There is a unique pair $(b_1,b_2)\in V_1\times V_2$ with $b=b_1+b_2$.

Let $v\in V$ be arbitrary. There is a unique pair $(v_1,v_2)\in V_1\times V_2$ with $v=v_1+v_2$.

Then $v\in V_1+a$ iff $v_2=a_2$ (indeed $V_1+a=V_1+a_1+a_2=V_1+a_2$ since $a_1\in V_1$). Similarly $v\in V_2+b$ iff $v_1=b_1$.

So $v\in (V_1+a)\cap (V_2+b)$ iff we have both $v_1=b_1$ and $v_2=a_2$, in other words $b_1+a_2$ is the unique vector in the intersection.