[[Vector subspace]]
# Orthogonal complement

Given an [[inner product space]] $(V, \mathbb{K}, \braket{ \cdot | \cdot })$,
the **orthogonal complement** $A^\perp$ of a subset $A \sube V$
is the [[vector subspace]] of vectors orthogonal to those $A$ #m/def/linalg 
$$
\begin{align*}
A^\perp = \{ v \in V : (\forall a \in A) [\braket{ a | v } =0] \}
\end{align*}
$$

> [!check]- Proof of subspace
> Clearly\Span$\vab 0 \in A^\perp$.
> If $v_{1},v_{2} \in A^\perp$ and $\alpha,\beta \in \mathbb{K}$ then $\braket{ a | \alpha v_{1}+\beta v_{2} } = \alpha \braket{ a | v_{1} }+ \beta \braket{ a | v_{2} }=0$ for all $a \in A$,
> and thus $\alpha v_{1} + \beta v_{2} \in A^\perp$.
> Therefore $A^\perp$ is a subspace.
> <span class="QED"/>

## Properties

Let $A \sube V$ be an arbitrary subset. Then

1. $A^\perp$ is [[Topological space|topologically closed]] ^S1
2. $A \cap A^\perp \sube \{ 0 \}$ ^S2
3. $B \sube A \implies A^\perp \sube B^\perp$ ^S3
4. $A \sube (A^\perp)^\perp$ ^S4
5. If $\mathrm{B}_{\epsilon}(v) \sube A$ for some $v \in V$, then $A^\perp = \{ 0 \}$ ^S5
6. $A^\perp = (\Span A)^\perp$ ^S6

> [!check]- Proof of 1–6
> Note that the orthogonal complement of a singleton $\{ v  \}$ can be expressed as a preïmage
> $$
> \begin{align*}
> \{ v \}^\perp = (\bra{v})^{-1} \{ 0 \}
> \end{align*}
> $$
> Since $\{ 0 \}$ is closed in $\mathbb{K}$, and [[the inner product is continuous]],
> it follows $\{ v \}^\perp$ is closed.
> Now for an arbitrary set $A$,
> $$
> \begin{align*}
> A^\perp = \bigcap_{a \in A} \{ a \}^\perp
> \end{align*}
> $$
> which is an intersection of closed sets and is therefore closed,
> proving [[#^S1]].
> 
> Note if $v \in A \cap A^\perp$ then $\braket{ v | v }= 0$, implying $v = 0$ by [[Inner product space#^IP3]],
> proving [[#^S2]].
> 
> Let $B \sube A \sube V$ and $v \in A^\perp$.
> Then $\braket{ v | b }=0$ for any $b \in B \sube A$,
> so $v \in B^\perp$, proving [[#^S3]].
> 
> Let $a \in A$.
> Then by definition $\braket{ a | v }=0$ for any $v \in A^\perp$,
> so $a \in (A^\perp)^\perp$,
> proving [[#^S4]]. 
> 
> Without loss of generality $v=0$, for $\braket{ B_{\epsilon}(v) | a } = 0$ iff $\braket{ \mathrm{B}_{\epsilon}(v) - v | a } = \braket{ \mathrm{B}_{\epsilon}(0) |a  }=0$.
> Now
> $$
> \begin{align*}
> \mathrm{B}_{\epsilon}(0) = \{ v \in V: \braket{ v | v } < \epsilon \}
> \end{align*}
> $$
> and 
> $$
> \braket{ \mathrm{B}_{\epsilon}(0) | y }=0
> \iff
> \braket{ \mathrm{B}_{\epsilon}(0) | \frac{\epsilon y}{2\|y\|} } = 0
> $$
> but since $\frac{\epsilon y}{2\|y\|} \in \mathrm{B}_{\epsilon}(0)$, it follows from [[Inner product space#^IP3]] that $y=0$,
> proving [[#^S5]].
> 
> Let $x \in A^\perp$ and $y \in \Span A$,
> so $y = \sum_{i=1}^n \lambda_{i}a_{i}$ for some $\{ a_{i} \}_{i=1}^n \sube A$.
> Then
> $$
> \begin{align*}
> \braket{ x | y } = \bra{x} \sum_{i=1}^n \lambda_{i} \ket{a_{i}}  = 0
> \end{align*}
> $$
> proving [[#^S6]]. <span class="QED"/>

Let $W \leq V$ be a [[vector subspace]]. Then

1. $V = W \oplus W^\perp$ ([[Direct sum of vector spaces#Internal direct sum]]). ^V1
2. $W=(W^\perp)^\perp$. ^V2

> [!check]- Proof of 1–2
> [[#^V1]] follows directly from [[#^S2]],
> and [[#^V2]] follows directly from [[#^S4]]. <span class="QED"/>

Other properties include

- [[The orthogonal complement of an invariant subspace under a unitary operator is invariant]]

## See also

- [[Orthogonal complement polarity]]

#
---
#state/tidy | #lang/en | #SemBr