description/proof of that countable-dimensional vectors space with inner product has orthonormal basis
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of inner product on real or complex vectors space.
- The reader knows a definition of basis of module.
- The reader knows a definition of orthonormal subset of vectors space with inner product.
- The reader knows a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
- The reader admits the proposition that any vectors space has a basis.
- The reader admits the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent.
Target Context
- The reader will have a description and a proof of the proposition that any countable-dimensional vectors space with any inner product has an orthonormal basis.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the countable-dimensional } F \text{ vectors spaces }\}\), with any inner product
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Statements:
\(\exists B \in \{\text{ the orthonormal bases for } V\}\)
//
\(V\) does not need to be finite-dimensional, but at least this proposition requires \(V\) to be countable-dimensional (this proposition does not claim anything about any uncountable-dimensional vectors space).
2: Proof
Whole Strategy: Step 1: take any basis, \(B = \{b_1, b_2, ...\}\); Step 2: take the Gram-Schmidt orthonormalization of \(B\), \(\widetilde{B}\); Step 3: see that \(\widetilde{B}\) is an orthonormal basis.
Step 1:
There is a basis, \(B = \{b_1, b_2, ...\}\), by the proposition that any vectors space has a basis.
The basis is countable because this proposition requires so.
Step 2:
Let us take the Gram-Schmidt orthonormalization of \(B\), \(\widetilde{B} = \{\widetilde{b}_1, \widetilde{b}_2, ...\}\).
Step 3:
\(\widetilde{B}\) is indeed orthonormal, as has been checked in Note for the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
\(\widetilde{B}\) is linearly independent, by the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent.
Let us see that \(\widetilde{B}\) spans \(V\).
Let \(v \in V\) be any.
As is mentioned in Note for the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product, as \(B\) is linearly independent, each \(\{\widetilde{b}_1, ..., \widetilde{b}_n\}\) is determined from \(\{b_1, ..., b_n\}\), and each \(b_j\) is a linear combination of \(\{\widetilde{b}_1, ..., \widetilde{b}_n\}\).
As \(B\) spans \(V\), \(v = v^1 b_1 + ... + v^n b_n\).
But as \(b_j\) is a linear combination of \(\{\widetilde{b}_1, ..., \widetilde{b}_n\}\), \(v\) is a linear combination of \(\{\widetilde{b}_1, ..., \widetilde{b}_n\}\).
So, \(\widetilde{B}\) spans \(V\).
So, \(\widetilde{B}\) is a basis for \(V\).
So, \(\widetilde{B}\) is an orthonormal basis for \(V\).
