Exercise 6.2.14

We prove the first equality first. If $x\in {(W_1+W_2)}^{\text{⊥}}$, we have $x$ is orthogonal to $u+v$ for all $u\in W_1$ and $v\in W_2$. This means that $x$ is orthogonal to $u=u+0$ for all $u$ and so $x$ is an element in $W_1^{\text{⊥}}$. Similarly, $x$ is also an element in $W_2^{\text{⊥}}$. So we have

$${(W_1+W_2)}^{\text{⊥}}\subset W_1^{\text{⊥}}.$$

Conversely, if $x\in W_1^{\text{⊥}}\cap W_2^{\text{⊥}}$, then we have

$$⟨x,u⟩=⟨x,v⟩=0$$

for all $u\in W_1$ and $v\in W_2$. This means

$$⟨x,u+v⟩=⟨x,u⟩+⟨x,v⟩=0$$

for all element $u+v\in W_1+W_2$. And so

$${(W_1+W_2)}^{\text{⊥}}\supset W_1^{\text{⊥}}.$$

For the second equality, we have, by Exercise 6.2.13(c),

$${(W_1\cap W_2)}^{\text{⊥}}={({(W_1^{\text{⊥}})}^{\text{⊥}}\cap {(W_2^{\text{⊥}})}^{\text{⊥}})}^{\text{⊥}}$$

$$={({(W_1^{\text{⊥}}+W_2^{\text{⊥}})}^{\text{⊥}})}^{\text{⊥}}=W_1^{\text{⊥}}+W_2^{\text{⊥}}.$$