Show that $(U \cap V)^\perp = U^\perp + V^\perp$. Let $U$ and $V$ be subspaces of $\mathbb R^n$. Show that $(U \cap V)^\perp = U^\perp + V^\perp$.
I know that in order for $V$ to be an orthogonal complement it must fulfil $x^*v=0$ for all $v$ in $V$.
I also know that  $U \cap V$ means all elements lying in both $U$ and $V$.
I am unsure how to use this to prove the equation above.
 A: Let $x\in A^\perp+B^\perp$. That is, $x=y+z$ with $y\in A^\perp$ and $B\in  B^\perp$. Then, for any $v\in A\cap B$, we have $y^*v=0$ (since $y\in A^\perp$ and $v\in A$) and $z^*v=0$ (since $z\in B^\perp$ and $v\in B$). Then
$$
x^*v=(y^*+z^*)v=y^*v+z^*v=0+0=0.
$$
So $$\tag{1}A^\perp+B^\perp\subset (A\cap B)^\perp.$$ 
Now take $x\in (A^\perp+B^\perp)^\perp$. Then $x\in A^{\perp\perp}=A$, and $x\in B^{\perp\perp}=B$. So $x\in A\cap B$. That is, 
$$
(A^\perp+B^\perp)^\perp\subset A\cap B.
$$
Taking orthogonals, 
$$\tag{2}
(A\cap B)^\perp\subset (A^\perp+B^\perp)^{\perp\perp}=A^\perp+B^\perp.
$$
Now, from (1) and (2), 
$$
(A\cap B)^\perp= (A^\perp+B^\perp)^{\perp\perp}=A^\perp+B^\perp.
$$
A: You could first prove the easier equality $$(C+D)^{\perp}=C^{\perp}\cap D^{\perp}$$
by proving the inclusions $(C+D)^{\perp}\subset C^{\perp}\cap D^{\perp}$ and $C^{\perp}\cap D^{\perp}\subset (C+D)^{\perp}$.
For $C=A^{\perp}$ and $D=B^{\perp}$, this equality gives
$$(A^{\perp}+B^{\perp})^{\perp}=A\cap B.$$
Taking orthogonals of both sides then gives
$$A^{\perp}+B^{\perp}=(A\cap B)^{\perp}.$$
