17 - Orthogonal Complements

Orthogonal Complement

Let $U\subseteq V$ be a subspace of an inner product space $V$
Orthogonal Complement $U^{\perp }$ of $U$ is defined as

$$U^{\perp }:=\{v\in V\mid ⟨u,v⟩=0\text{ for all }u\in U\}$$

Orthogonal Complement Subspace Property

Let $U\subseteq V$ be a subspace of an inner product space $V$ then

$$U^{\perp }\text{ is a subspace of }V$$

Proof

We have $0\in U^{\perp }$ by definition of $0$
For $v,w\in U^{\perp }$ and $\lambda \in \mathbb{K}$ then for all $u\in U$

$$⟨u,v+\lambda w⟩=⟨u,v⟩+\lambda ⟨u,w⟩=0+\lambda \cdot 0=0$$

Properties of Orthogonal Complement

Let $U\subseteq V$ be a subspace of an inner product space $V$

$$U\cap U^{\perp }=\{0\}$$

$$U\oplus U^{\perp }=V\text{ if }V\text{ is finite dimensional}$$

so

$$\dim U^{\perp }=\dim V-\dim U$$

$$(U+W{)}^{\perp }=U^{\perp }\cap W^{\perp }$$

$$(U\cap W{)}^{\perp }\supseteq U^{\perp }+W^{\perp }$$

with equality if $\dim V<\infty$

$$U\subseteq (U^{\perp }{)}^{\perp }$$

with equality if $V$ is finite dimensional

Isomorphism between Orthogonal Complement and Annihilators

Let $V$ be finite dimensional then

Under $R$-linear isomorphism $\varphi :V\to V^{\prime }$ given by $v\mapsto ⟨v,\cdot ⟩$
Space $U^{\perp }$ maps is0oomorphically to $U^0$ (considered as $\mathbb{R}$ vector spaces)

Proof

Let $v\in U^{\perp }$ then for all $u\in U$

$$⟨u,v⟩=\overline{⟨v,u⟩}=0$$

Hence $⟨v,\cdot ⟩\in U^0$ so $\mathrm{Im}(\varphi {\mid }_{U^{\perp }})\subseteq U^0$ and

$$\dim U^{\perp }=\dim V-\dim U=\dim U^0$$

So

$$\varphi (U^{\perp })=U^0$$

As the kernel of $\varphi$ is trivial then we only need to check equality of dimensions