Orthogonality and the orthogonal complement

Let $V$ be a real or complex inner-product space. Vectors $x,y\in V$ are orthogonal, written $x\perp y$, when

$⟨x,y⟩=0.$

For a subset $S\subseteq V$ the orthogonal complement of $S$ is

$S^{\perp }:=\{v\in V:⟨v,s⟩=0\text{ for every }s\in S\}\mathrm{.}$

Orthogonality is symmetric. If $⟨x,y⟩=0$, then $⟨y,x⟩=\overline{⟨x,y⟩}=0$, so $x\perp y$ exactly when $y\perp x$; in particular the condition defining $S^{\perp }$ is symmetric in its two arguments.

$S^{\perp }$ is a linear subspace. Let $s\in S$ and scalars $a,b$; then $⟨0,s⟩=0$, and if $u,v\in S^{\perp }$ then $⟨au+bv,s⟩=a⟨u,s⟩+b⟨v,s⟩=0$ by linearity in the first argument, so $au+bv\in S^{\perp }$. Thus $S^{\perp }$ is a linear subspace of $V$ (Linear subspace of a vector space) for every subset $S$, whether or not $S$ is a subspace. Moreover $0\in S^{\perp }$ always, and $v\in V$ lies in $\{0{\}}^{\perp }$ for every $v$, so $\{0{\}}^{\perp }=V$.

Monotonicity. If $S\subseteq T\subseteq V$, then every vector orthogonal to all of $T$ is orthogonal to all of $S$, so $T^{\perp }\subseteq S^{\perp }$.

Nontriviality of orthogonality. By positive definiteness $⟨v,v⟩=\mathrm{\parallel }v{\mathrm{\parallel }}^2=0$ exactly for $v=0$ (The induced length is a norm), so a vector orthogonal to itself is zero, and $\{0{\}}^{\perp }=V$, $V^{\perp }=\{0\}$.