1090: Gram-Schmidt Orthonormalization of Countable Subset of Vectors Space with Inner Product

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definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of inner product on real or complex vectors space.

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Target Context

• The reader will have a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$ with any inner product, $⟨\bullet ,\bullet ⟩:V\times V\to F$
$S$: $\in \{\text{ the countable subsets of }V\}$, $=\{s_1,s_2,...\}$
$\ast \tilde{S}$: $\in \{\text{ the countable subsets of }V\}$, $=\{b_1,b_2,...\}$, as specified below
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Conditions:
$b_n$ is defined iteratively:
$\{\}$ has been already determined from $\{\}$
Suppose that $\{b_1,...,b_{n-1}\}$ has been already determined from $\{s_1,...,s_{n^{\prime }}\}$
When there is the smallest $\{s_1,...,s_{n^{″}}\}$ such that $\{b_1,...,b_{n-1},s_{n^{″}}\}$ is linearly independent, $b_n:=(s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert$, and $\{b_1,...,b_n\}$ has been determined from $\{s_1,...,s_{n^{″}}\}$
Otherwise, the iteration is finished, and $\tilde{S}:=\{b_1,...,b_{n-1}\}$
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"countable" does not mean that it is infinite: any finite set is countable.

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2: Note

Often, $S$ is supposed to be linearly independent, but this definition does not require $S$ to be linearly independent.

Let us see that $\tilde{S}$ is orthonormal, iteratively.

For $\{b_1\}$, $⟨b_1,b_1⟩=⟨s_{n^{″}}/\Vert s_{n^{″}}\Vert ,s_{n^{″}}/\Vert s_{n^{″}}\Vert ⟩=1/\Vert s_{n^{″}}{\Vert }^2⟨s_{n^{″}},s_{n^{″}}⟩=1/\Vert s_{n^{″}}{\Vert }^2\Vert s_{n^{″}}{\Vert }^2=1$.

So, $\{b_1\}$ is orthonormal.

For $\{b_1,b_2\}$, $⟨b_1,b_2⟩=⟨b_1,(s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert ⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert ⟨b_1,(s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j)⟩=1/\Vert s_{n^{″}}-⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_1,s_{n^{″}}⟩-\overline{⟨s_{n^{″}},b_1⟩}⟨b_1,b_1⟩)=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_1,s_{n^{″}}⟩-\overline{⟨s_{n^{″}},b_1⟩}1)=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_1,s_{n^{″}}⟩-\overline{⟨s_{n^{″}},b_1⟩})=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_1,s_{n^{″}}⟩-⟨b_1,s_{n^{″}}⟩)=0$.

$⟨b_2,b_2⟩=⟨(s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert ,(s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j\Vert ⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j{\Vert }^2⟨s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j,s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j{\Vert }^2\Vert s_{n^{″}}-\sum _{j=1}^1⟨s_{n^{″}},b_j⟩b_j{\Vert }^2=1$.

So, $\{b_1,b_2\}$ is orthonormal.

Let us suppose that $\{b_1,...,b_{n-1}\}$ is orthonormal.

For $\{b_1,...,b_n\}$, for each $1\le l\le n-1$, $⟨b_l,b_n⟩=⟨b_l,(s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert ⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert ⟨b_l,s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_l,s_{n^{″}}⟩-\sum _{j=1}^{n-1}\overline{⟨s_{n^{″}},b_j⟩}⟨b_l,b_j⟩)=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_l,s_{n^{″}}⟩-\sum _{j=1}^{n-1}\overline{⟨s_{n^{″}},b_j⟩}{\delta }_{l,j})=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_l,s_{n^{″}}⟩-\overline{⟨s_{n^{″}},b_l⟩})=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert (⟨b_l,s_{n^{″}}⟩-⟨b_l,s_{n^{″}}⟩)=0$.

$⟨b_n,b_n⟩=⟨(s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert ,(s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j)/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j\Vert ⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j{\Vert }^2⟨s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j,s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j⟩=1/\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j{\Vert }^2\Vert s_{n^{″}}-\sum _{j=1}^{n-1}⟨s_{n^{″}},b_j⟩b_j{\Vert }^2=1$.

So, $\{b_1,...,b_n\}$ is orthonormal.

Then, let $b_l,b_m\in \tilde{S}$ be any such that $l\ne m$.

Let $l<m$ without loss of generality.

So, $b_l,b_m\in \{b_1,...,b_m\}$.

As $\{b_1,...,b_m\}$ is orthonormal, $⟨b_l,b_m⟩=0$.

$⟨b_l,b_l⟩=1$.

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References

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