Orthocomplement equal to Orthocomplement of Linear Span

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\GF$.

Let $S \subseteq V$ be non-empty.

Then:

$S^\bot = \paren {\map \span S}^\bot$

where $\bot$ denotes orthocomplement.

$\paren {\map \span S}^\bot \subseteq S^\bot$

Conversely let $y \in S^\bot$.

Let $x \in \map \span S$.

From the definition of the linear span there exists $x_1, \ldots, x_n \in S$ and $\alpha_1, \ldots, \alpha_n \in \GF$ such that:

$\ds x = \sum_{j \mathop = 1}^n \alpha_j x_j$

We then have:

$\ds \innerprod x y$

$=$

$\ds \innerprod {\sum_{j \mathop = 1}^n \alpha_j x_j} y$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^n \alpha_j \innerprod {x_j} y$

linearity in first argument

$\ds $

$=$

$\ds 0$

since $x_j \in S$ for each $j$, we have $\innerprod {x_j} y = 0$

So $\innerprod x y = 0$ for all $x \in \map \span S$.

So $y \in \paren {\map \span S}^\bot$.

Hence we have $S^\bot \subseteq \paren {\map \span S}^\bot$.

We conclude:

$S^\bot = \paren {\map \span S}^\bot$

$\blacksquare$