Orthogonal Real Subspaces

Definition

Two subspaces $S, T \subset \R ^n$ are orthogonal if

\[ x^\tp y = 0 \text{ for all } x \in S, y \in T. \]

For any set $S \subset \R ^n$ (not necessarily a subspace), the orthogonal complement of $S$ is the set

\[ S^\perp = \Set{x \in \R ^n}{x^\top y = 0 \text{ for all } y \in S}. \]

$S^\perp $ is the set of all vectors which are orthogonal to every vector in $S$.

Orthgonal complement is a subspace

Notice that $S^\perp $ is always a subspace. If $x \in S^\perp $, then $x^\top y = 0$ for all $y \in S$. So then $(\alpha x)^\top y = \alpha (x^\top y) = 0$ for all $\alpha \in \R $ and $y \in S$. We conclude $\alpha x \in S^\perp $ for all $\alpha \in \R $. In other words, $S^\perp $ is closed under scalar multiplication. If $x, z \in S^\perp $, then $(x+z)^\top y = x^\top y + z^\top y = 0 + 0 = 0$. We conclude that $x + z \in S^\perp $ for all $x, z \in S^\perp $. In other words, $S^\perp $ is closed under vector addition. Consequently, $S^\perp $ is a subspace.