The Orthogonal Subspace S⟂ of an Inner Product Space

Definition: Let $H$ be an inner product space and let $S \subseteq H$. The corresponding Orthogonal Set is defined to be the set $S^{\perp} = \{ h \in H : h \perp s, \: \forall s \in S \}$.

The following proposition tells us that given any nonempty subset $S$ of an inner product space $H$, the corresponding orthogonal set $S^{\perp}$ is always a subspace of $H$.

Proposition 1: Let $H$ be an inner product space and let $S \subseteq H$ be nonempty. Then $S^{\perp}$ is a subspace of $H$.

• Proof: First observe that $\langle 0, s \rangle = 0$ for all $s \in S$ and so $0 \in S^{\perp}$.

• Let $h_1, h_2 \in S^{\perp}$. Then $\langle h_1, s \rangle = 0$ and $\langle h_2, s \rangle = 0$ for all $s \in S$. So for every $s \in S$:

(1)

\begin{align} \quad \langle h_1 + h_2, s \rangle = \langle h_1, s \rangle + \langle h_2, s \rangle = 0 + 0 = 0 \end{align}

• Thus $S^{\perp}$ is closed under addition.

• Now let $h \in S^{\perp}$ and let $\lambda \in \mathbb{R}$. Then $\langle h, s \rangle = 0$ for all $s \in S$ and so for every $s \in S$:

(2)

\begin{align} \quad \langle \lambda h, s \rangle = \lambda \langle h, s \rangle = \lambda \cdot 0 = 0 \end{align}

• Thus $S^{\perp}$ is closed under scalar multiplication. We conclude that $S^{\perp}$ is subspace of $H$. $\blacksquare$