# Orthogonal Projection onto Closed Linear Span

## Theorem

Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.

Let $E = \set {e_1, \ldots, e_n}$ be an orthonormal subset of $H$.

Let $M = \vee E$, where $\vee E$ is the closed linear span of $E$.

Let $P$ be the orthogonal projection onto $M$.

Then:

$\forall h \in H: P h = \ds \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$

## Proof

Let $h \in H$.

Let:

$\ds u = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$

We have that:

$u \in \map \span E$

and from the definition of closed linear span:

$M = \paren {\map \span E}^-$

We therefore have, by the definition of closure:

$u \in M$

Let $v = h - u$

We want to show that $v \in M^\bot$.

From Intersection of Orthocomplements is Orthocomplement of Closed Linear Span, it suffices to show that:

$v \in E^\bot$

Note that for each $l$ we have:

$\innerprod v {e_l} = \innerprod h {e_l} - \innerprod u {e_l}$

since the inner product is linear in its first argument.

We have:

 $\ds \innerprod u {e_l}$ $=$ $\ds \innerprod {\sum_{k \mathop = 1}^n \innerprod h {e_k} e_k} {e_l}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \innerprod {\innerprod h {e_k} e_k} {e_l}$ linearity of inner product in first argument $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \innerprod h {e_k} \innerprod {e_k} {e_l}$ linearity of inner product in first argument $\ds$ $=$ $\ds \innerprod h {e_l} \innerprod {e_l} {e_l}$ Definition of Orthonormal Subset $\ds$ $=$ $\ds \innerprod h {e_l} \norm {e_l}^2$ Definition of Inner Product Norm $\ds$ $=$ $\ds \innerprod h {e_l}$ since $\norm {e_l} = 1$

so:

$\innerprod v {e_l} = 0$

That is:

$v \in E^\bot$
$v \in M^\bot$

We can therefore decompose each $h \in H$ as:

$h = u + v$

with $u \in M$ and $v \in M^\bot$.

So we have:

 $\ds P h$ $=$ $\ds \map P {u + v}$ $\ds$ $=$ $\ds \map P u + \map P v$ Orthogonal Projection is Linear Transformation $\ds$ $=$ $\ds v$ Kernel of Orthogonal Projection, Fixed Points of Orthogonal Projection $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$

for each $h \in H$.

$\blacksquare$