User:Anghel/Sandbox

Theorem
Let $\struct {V, \innerprod \cdot \cdot}$ be a inner product space.

Let $S = \set {v_n: n \in \N_{>0} }$ be a linearly independent subset of $V$.

Then there exists an orthonormal subset $E = \set {e_n: n \in \N_{>0} }$ of $V$ such that:


 * $\forall k \in \N: \span \set {v_1, \ldots , v_k} = \span \set {e_1 , \ldots ,e_k}$

where $\span$ denotes linear span.

Proof
For all $k \in \N_{>0}$, define $u_k, e_k \in V$ inductively as:

where $\norm \cdot$ denotes the inner product norm on $V$.

We prove the theorem by induction for $k \in \N_{>0}$.

Base Case for k = 1
For $k=1$, the sum in the definition of $u_1$ is empty, so $u_1 = v_1$.

Let $\mathbf 0$ denote the zero vector of $V$, which by definition is the identity of $V$.

From Subset of Module Containing Identity is Linearly Dependent, it follows that $\mathbf0 \notin E$, so $v_1 \ne \mathbf 0$.

By, it follows that $\norm {v_1} \ne 0$.

It follows that $e_1 = \dfrac {1}{\norm {v_1} }$ is well-defined.

Hence, $\set {e_1}$ is an orthonormal subset.

Induction Hypothesis
Suppose that for $k \in \N_{>1}$, the set $\set {e_1, \ldots ,e_{k-1} }$ is a linear independent orthonormal subset such that:


 * $\span \set {v_1, \ldots , v_{k-1} } = \span \set {e_1 , \ldots ,e_{k-1} }$

This is our induction hypothesis.

Base Case for k = 1
qed}}