complement of sum equals intersection of complements I have difficulties with the following exercise:

We have two subspaces of $F^n$, $A_1$ and $A_2$. Prove the following statement:
$$(A_1 + A_2)^{\perp} = A_1^{\perp} \cap A_2^{\perp}$$

To my understanding, the sum of two subspaces is defined as follows:
If $A_1$ and $A_2$ are subspaces of some vector space $V$, then $A_1 + A_2 = \{a_1 + a_2 \mid a_1 \in A_1, a_2 \in A_2\}$
The intersection of two subspaces is simply the set of elements (vectors in the case of subspaces) that are in both $A_1$ and $A_2$.
The orthogonal complement of a subspace $A_1$ is the subspace $A_1^\perp$ that contains the vectors orthogonal to all the vectors in $A_1$.
I'm pretty sure I grasp all of these concepts, but I still am blank when it comes to this proof. I also have trouble visualizing why it would be true, but maybe that's an unnecessary luxury anyway. How does this proof work?
 A: Suppose $v\in A_1^{\perp}\cap A_2^{\perp}$. Then $v\cdot a_1=v \cdot a_2=0$ for all $a_1 \in A_1$ and all $a_2 \in A_2$. Then for all $a_1+a_2 \in A_1+A_2$, $v\cdot (a_1+a_2)=v\cdot a_1+v \cdot a_2=0+0=0$. Thus $v\in (A_1+A_2)^{\perp}$.
Conversely, if $v\not\in A_1^{\perp}\cap A_2^{\perp}$ then without loss of generality let $v\not\in A_1^{\perp}$. Then there is $a_1\in A_1$ with $v\cdot a_1 \neq 0$. We know $0\in A_2$, so $v\cdot (a_1 +0)=v\cdot a_1 \neq 0$. Thus $v \not \in v\in (A_1+A_2)^{\perp}$.
A: This proof works in the way that most proofs of the equality of sets work.  Start with an element $x$ in the first set, and show that it belongs to the second set.  Similarly, show that any element of the second set also belongs to the first set.  That completes the proof.

Here's the easy half of the proof: 
Suppose that $x \in (A_1 + A_2)^\perp$.  Then, for any $y = a_1 + a_2$ with $a_1\in A_1$ and $a_2 \in A_2$, we must have $\langle y,x \rangle = 0$.  
Now, for any $a_1 \in A_1$, we note that $a_1 = a_1 + 0 \in A_1 + A_2$.  So, we must have $\langle a_1,x \rangle = 0$.  Since this holds for any $a_1 \in A_1$, we conclude that $x \in A_1^\perp$.
Similarly, for any $a_2 \in A_2$, we note that $a_2 = 0+a_2 \in A_1 + A_2$.  So, we must have $\langle a_2,x \rangle = 0$.  So, $x \in A_2^\perp$.
Thus, we may conclude that $x$ is in both $A_1^\perp$ and $A_2^\perp$.  So, $x \in A_1^\perp \cap A_2^\perp$.  Thus, $(A_1 + A_2)^\perp \subseteq A_1^\perp \cap A_2^\perp$.
