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Can we define inner product on every vector space? [duplicate]
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Is there a vector space that cannot be an inner product space?
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Can we define inner product on every vector space?
I don't know any example of any vector space that do not have any inner product .
Help me
functional-analysis
asked Jul 25 at 12:28
yourmath
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marked as duplicate by Rhys Steele, Chappers, Gerry Myerson, José Carlos Santos, mechanodroid Jul 25 at 13:16
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up vote
1
down vote
favorite
This question already has an answer here:
Is there a vector space that cannot be an inner product space?
2 answers
Can we define inner product on every vector space?
I don't know any example of any vector space that do not have any inner product .
Help me
functional-analysis
asked Jul 25 at 12:28
yourmath
1,7951617
marked as duplicate by Rhys Steele, Chappers, Gerry Myerson, José Carlos Santos, mechanodroid Jul 25 at 13:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Is there a vector space that cannot be an inner product space?
2 answers
Can we define inner product on every vector space?
I don't know any example of any vector space that do not have any inner product .
Help me
functional-analysis
asked Jul 25 at 12:28
yourmath
1,7951617
This question already has an answer here:
Is there a vector space that cannot be an inner product space?
2 answers
Can we define inner product on every vector space?
I don't know any example of any vector space that do not have any inner product .
Help me
This question already has an answer here:
Is there a vector space that cannot be an inner product space?
2 answers
functional-analysis
asked Jul 25 at 12:28
yourmath
1,7951617
asked Jul 25 at 12:28
yourmath
1,7951617
asked Jul 25 at 12:28
yourmath
1,7951617
asked Jul 25 at 12:28
yourmath
1,7951617
1,7951617
marked as duplicate by Rhys Steele, Chappers, Gerry Myerson, José Carlos Santos, mechanodroid Jul 25 at 13:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Rhys Steele, Chappers, Gerry Myerson, José Carlos Santos, mechanodroid Jul 25 at 13:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44
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$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44
$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44
$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44
add a comment |Â
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If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.
But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.
But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
add a comment |Â
up vote
1
down vote
If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.
But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.
But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
If you have a finite dimensional vector space just choose a basis and then define an inner product by assuming that basis is orthonormal. Another way to say the same thing: choosing a basis of a finite dimensional vector space over a field $K$ establishes an isomorphism with $K^n$ where there's a natural inner product.
But I think your question misses an important point. We don't find inner products on vector spaces at random. They come to us because they provide useful information that comes essentially from the source of the vector space itself. You've tagged your question "functional analysis". There you regularly encounter inner products that help you do functional analysis - for example, for Fourier analysis.
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
answered Jul 25 at 12:36
Ethan Bolker
35.7k54199
35.7k54199
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
add a comment |Â
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
1
1
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
My question is- Can we define inner product on any vector space?
– yourmath
Jul 25 at 12:38
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
See this essential duplicate @RhysSteele found: math.stackexchange.com/questions/247425/…
– Ethan Bolker
Jul 25 at 12:43
1
1
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
Even in an infinite-dimensional vector space (over $mathbb R$ or $mathbb C$), your method works fine. You need the axiom of choice to ensure that there is a basis, but then every basis gives rise to an inner product, in which that basis is orthonormal.
– Andreas Blass
Jul 25 at 13:05
add a comment |Â
$L^p$ spaces with $p$ different of two are the typical example .
– Gustave
Jul 25 at 13:44