Prove that $(A \cap B)^\perp ⊃ A^\perp + B^\perp$

Question

Let $A$ and $B$ be subspaces of a vector space V (not necessarily finite-dimensional). For any $C \subset V$, $C^\perp \subset V^∗$ is its annihilator.

1. Prove that $(A \cap B)^\perp ⊃ A^\perp + B^\perp.$

2. Prove that $(A\cap B)^\perp = A^\perp + B^\perp$ if it is given that the following three lemmas are true:

L1. $\dim(A + B) = \dim(A) + \dim(B) – \dim(A \cap B)$

L2. $(A + B)^\perp = A^\perp \cap B^\perp$

L3. $\dim(A^\perp) = \dim(V) – \dim(A)$ for any subspace $A$.

Answer 1

For 1. start with $f \in A^{\perp}$ and $g \in B^{\perp}$. Then $f(a) = 0$ and $g(b) = 0$ for all $a \in A$ and $b \in B$ by definition of $^\perp$. We now want to show that $f + g \in (A \cap B)^{\perp}$, that is $(f + g)(c) = 0$ for all $c \in A \cap B$.

If $c \in A \cap B$, then $c \in A$ and $c \in B$. Can you show that $(f + g)(c) = 0$?

The other statements are similar to prove.