Properties of annihilators

Question

Let $S$ and $T$ be two subspaces of $V$ and denote the annihilator of a set $A \subset V$ by $A^\perp$. Then how do I prove the following propositions:

1. $\left(S^\perp\right)^\perp =S$.
2. $(S+T)^\perp = S^\perp \cap T^\perp$

Answer 1

HINT

(1) Let $a \in A = S^\perp$. Then, $a \cdot s = 0 \forall s \in S$. So, $\{a\}^\perp \supset S$. $$ A^\perp = \bigcap_{a \in A} \{a\}^\perp \supseteq S. $$ Can you prove the inclusion the other way using a similar technique?

(2) Similar approach. Let $x \in S+T$, then $\exist x_s \in S, x_t \in T$ such that $x = x_s+x_t$. Thus, if $a \in (S+T)^\perp$, then $$ 0 = a \cdot x = a \cdot (x_s + x_t) = a\cdot x_s + a \cdot x_t, \quad \forall x \in S+T. $$ Can you show this implies $a\cdot x_s = 0 = a \cdot x_t$?

Then, reverse the inclusion argument as before...