Let $W_1$ and $W_2$ be subspaces of $R^n$, how to prove $(W_1\cap W_2)^{\perp}=W_1^{\perp}+W_2^{\perp}$ and $(W_1^{\perp})^{\perp}=W_1$?
For the first one I have no idea how to get to the other side.
For the second one, let $x\in(W_1^{\perp})^{\perp}$, then $\langle x,y\rangle=0 \ \forall y\in W_1^{\perp}$. Also $\langle y,z\rangle=0 \ \forall z\in W_1$. If we can show that $x, z$ are the same vector, then I think we are done. But I dont know how to do that.
Could someone help?
For the second one, any subspace of $\Bbb R^n$ is a finite dimensional normed vector space, hence any subspace of $\Bbb R^n$ is a Banach subspace. So, in our case $\Bbb R^n = W \oplus W^{\perp}$. Now let $x \in W^{\perp \perp}$, $x$ is in $\Bbb R^n$ so there are $y \in W$ and $z \in W^{\perp}$ such that $x = y+z$.
Then, $z = x - y$, $x \in W^{\perp \perp}$ and $y \in W \subset W^{\perp \perp}$ and $W^{\perp \perp}$ is a subspace, so $z \in W^{\perp \perp}$, hence $z \in W^{\perp \perp} \cap W^{\perp} = \{0\}$, so $x = y \in W$.
Hint for the second one: if for a $y$ holds that $\langle y,z\rangle=0 \ \forall z\in W_1$, then every $z\in W_1$ is perpendicualr to $y$ and $y\in W_1^{\perp}$
