I have a question about how to prove a certain property of orthogonal complements of vector subspaces. Given $\mathrm E$, a vector space over a commutative field k, define: $$\phi\text{ : } \mathrm E \times \mathrm E^* \to k\text{ , where } \mathrm E^* \text{ is the dual space of E, and } \\ \phi(x,\omega) = \omega(x)$$
we can now define $\mathrm F^ \bot = \{\omega \in \mathrm E^* \mid \phi(x,\omega)=0 , \forall x\in \mathrm F \}$, where $\mathrm F $ is a vector subspace of $\mathrm E $
The first couple of properties such as proving that $\phi$ is in fact a mixed tensor, that $\mathrm F^ \bot $ is also a subspace and $ \dim F + \dim F^\bot = \dim E$ are simple enough, but I've gotten to one that I don't know how to go about, which is: $$ {(F+G)}^\bot = F^\bot\cap G^\bot $$
Where, of course, $F$ and $G$ are vector subspaces. Also for anyone wondering this is in the context of duality in projective spaces, because proving these properties is fundamental to the principle of duality, or at least the way I'm learning it. Thanks in advance.
Try this: $\omega \in F^\bot \cap G^\bot $ implies $\omega \in F^\bot $ and $\omega \in G^\bot$. That is, $\omega (x) = 0$ for every $x \in F$ and $\omega (y) = 0$ for every $y\in G$. Hence, $\omega (x+y) = 0$ for all $x+y \in F + G$. So $\omega \in (F+G)^\bot$.
So far we have proved one inclusion: $F^\bot \cap G^\bot \subset (F+G)^\bot$. For the other inclusion, namely $(F+G)^\bot \subset F^\bot \cap G^\bot$, just do the same: start with the definition of $\omega$ being in $(F+G)^\bot$ and see what happens. (If you get stuck, ask me again.)