When does an inner product induce a norm?
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When we consider a vector space $V$ over some field $F$, I know that when the $F=mathbbR$ or $ =mathbbC$, by setting $|x|=leftlanglex,xrightrangle^frac12$ we get a norm. However, since the inner product is a function with its image in $F$, what happens if we consider any $V$ over the rational numbers? For example, if we take $mathbbQ^2$ over $mathbbQ$ with the dot product, then $v=(1,1)$ has norm $sqrt2$, which is not rational. How can one obtain a norm from a given inner product in such cases?
linear-algebra normed-spaces inner-product-space
edited 1 hour ago
zipirovich
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javierochomil
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up vote
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When we consider a vector space $V$ over some field $F$, I know that when the $F=mathbbR$ or $ =mathbbC$, by setting $|x|=leftlanglex,xrightrangle^frac12$ we get a norm. However, since the inner product is a function with its image in $F$, what happens if we consider any $V$ over the rational numbers? For example, if we take $mathbbQ^2$ over $mathbbQ$ with the dot product, then $v=(1,1)$ has norm $sqrt2$, which is not rational. How can one obtain a norm from a given inner product in such cases?
linear-algebra normed-spaces inner-product-space
edited 1 hour ago
zipirovich
10.4k11630
asked 1 hour ago
javierochomil
213
New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
4
down vote
favorite
up vote
4
down vote
favorite
When we consider a vector space $V$ over some field $F$, I know that when the $F=mathbbR$ or $ =mathbbC$, by setting $|x|=leftlanglex,xrightrangle^frac12$ we get a norm. However, since the inner product is a function with its image in $F$, what happens if we consider any $V$ over the rational numbers? For example, if we take $mathbbQ^2$ over $mathbbQ$ with the dot product, then $v=(1,1)$ has norm $sqrt2$, which is not rational. How can one obtain a norm from a given inner product in such cases?
linear-algebra normed-spaces inner-product-space
edited 1 hour ago
zipirovich
10.4k11630
asked 1 hour ago
javierochomil
213
New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When we consider a vector space $V$ over some field $F$, I know that when the $F=mathbbR$ or $ =mathbbC$, by setting $|x|=leftlanglex,xrightrangle^frac12$ we get a norm. However, since the inner product is a function with its image in $F$, what happens if we consider any $V$ over the rational numbers? For example, if we take $mathbbQ^2$ over $mathbbQ$ with the dot product, then $v=(1,1)$ has norm $sqrt2$, which is not rational. How can one obtain a norm from a given inner product in such cases?
linear-algebra normed-spaces inner-product-space
linear-algebra normed-spaces inner-product-space
edited 1 hour ago
zipirovich
10.4k11630
asked 1 hour ago
javierochomil
213
New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
zipirovich
10.4k11630
asked 1 hour ago
javierochomil
213
New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
zipirovich
10.4k11630
edited 1 hour ago
zipirovich
10.4k11630
edited 1 hour ago
zipirovich
10.4k11630
10.4k11630
asked 1 hour ago
javierochomil
213
New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
javierochomil
213
asked 1 hour ago
javierochomil
213
213
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javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
javierochomil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each
ordered field has a Pythagorean extension. If you really want $L^2$-norms
you could always extend your ground field to a Pythagorean extension field.
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each
ordered field has a Pythagorean extension. If you really want $L^2$-norms
you could always extend your ground field to a Pythagorean extension field.
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
add a comment |Â
up vote
4
down vote
An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each
ordered field has a Pythagorean extension. If you really want $L^2$-norms
you could always extend your ground field to a Pythagorean extension field.
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each
ordered field has a Pythagorean extension. If you really want $L^2$-norms
you could always extend your ground field to a Pythagorean extension field.
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
An ordered field where $a^2+b^2$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each
ordered field has a Pythagorean extension. If you really want $L^2$-norms
you could always extend your ground field to a Pythagorean extension field.
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
answered 1 hour ago
Lord Shark the Unknown
92.5k956120
92.5k956120
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
add a comment |Â
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
Thanks, I didn't know about this extensions. It's not that I want a norm like the one I mention, but instead I would like to know how can one obtain a normed space from an inner product space when the field is, like you say, not Pythagorean.
– javierochomil
1 hour ago
add a comment |Â
javierochomil is a new contributor. Be nice, and check out our Code of Conduct.
javierochomil is a new contributor. Be nice, and check out our Code of Conduct.
javierochomil is a new contributor. Be nice, and check out our Code of Conduct.
javierochomil is a new contributor. Be nice, and check out our Code of Conduct.
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