What is it called when each element of one vector is greater than each element of another?
$begingroup$
I'm currently writing a term paper and while writing some proofs I've used an operator I don't know the name of.
Let us assume we have $x_1, x_2 in mathcal{R}^n$. Now lets define an operator '$succeq$' such that $x_1 succeq x_2$ implies that $x_1(i) geq x_2(i),,forall,, i in {1, .. n}$. Essentially, the operator implies that each element of one vector is greater than each element of the other. My question: is there a specific term for this operator? I strongly feel as though I've come across papers that have used this operator in proofs but I'm having a hard time finding them.
So far I've come across the concept of 'majorization' (http://mathworld.wolfram.com/Majorization.html) that seems to come close to what I want. Any help/information would be appreciated.
vector-spaces operator-theory operations-research
asked Jan 2 at 21:34
dar_devildar_devil
236
$endgroup$
add a comment |
$begingroup$
I'm currently writing a term paper and while writing some proofs I've used an operator I don't know the name of.
Let us assume we have $x_1, x_2 in mathcal{R}^n$. Now lets define an operator '$succeq$' such that $x_1 succeq x_2$ implies that $x_1(i) geq x_2(i),,forall,, i in {1, .. n}$. Essentially, the operator implies that each element of one vector is greater than each element of the other. My question: is there a specific term for this operator? I strongly feel as though I've come across papers that have used this operator in proofs but I'm having a hard time finding them.
So far I've come across the concept of 'majorization' (http://mathworld.wolfram.com/Majorization.html) that seems to come close to what I want. Any help/information would be appreciated.
vector-spaces operator-theory operations-research
asked Jan 2 at 21:34
dar_devildar_devil
236
$endgroup$
add a comment |
$begingroup$
I'm currently writing a term paper and while writing some proofs I've used an operator I don't know the name of.
Let us assume we have $x_1, x_2 in mathcal{R}^n$. Now lets define an operator '$succeq$' such that $x_1 succeq x_2$ implies that $x_1(i) geq x_2(i),,forall,, i in {1, .. n}$. Essentially, the operator implies that each element of one vector is greater than each element of the other. My question: is there a specific term for this operator? I strongly feel as though I've come across papers that have used this operator in proofs but I'm having a hard time finding them.
So far I've come across the concept of 'majorization' (http://mathworld.wolfram.com/Majorization.html) that seems to come close to what I want. Any help/information would be appreciated.
vector-spaces operator-theory operations-research
asked Jan 2 at 21:34
dar_devildar_devil
236
$endgroup$
I'm currently writing a term paper and while writing some proofs I've used an operator I don't know the name of.
Let us assume we have $x_1, x_2 in mathcal{R}^n$. Now lets define an operator '$succeq$' such that $x_1 succeq x_2$ implies that $x_1(i) geq x_2(i),,forall,, i in {1, .. n}$. Essentially, the operator implies that each element of one vector is greater than each element of the other. My question: is there a specific term for this operator? I strongly feel as though I've come across papers that have used this operator in proofs but I'm having a hard time finding them.
So far I've come across the concept of 'majorization' (http://mathworld.wolfram.com/Majorization.html) that seems to come close to what I want. Any help/information would be appreciated.
vector-spaces operator-theory operations-research
vector-spaces operator-theory operations-research
asked Jan 2 at 21:34
dar_devildar_devil
236
asked Jan 2 at 21:34
dar_devildar_devil
236
asked Jan 2 at 21:34
dar_devildar_devil
236
asked Jan 2 at 21:34
dar_devildar_devil
236
asked Jan 2 at 21:34
dar_devildar_devil
236
236
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't really like a description of these as "vectors": I'd rather call them functions $mathbb Nto mathbb R$, or possibly with a domain consisting of some segment of $mathbb N$.
Anyhow, in that context, I thought I had seen this called "the dominance order" on functions, which is a partial order you get when you say $fleq g$ if $f(x)leq g(x)$ for all $x$.
For example here or here.
However, while searching, I see that people use this term for lots of other partial orders that seem unrelated. Still, it seems like a pretty sensible name. You could say that one function(/vector) dominates another if it is greater at each point.
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
add a comment |
$begingroup$
I've encountered the notation $x_1 succeq x_2$ in the book Convex Optimization by Boyd and Vandenberghe. If $K$ is a cone, then they define the "generalized inequality" $x succeq_K y$ to mean that $x - y in K$. In your case, $K$ is the nonnegative orthant.
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
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active
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votes
$begingroup$
I don't really like a description of these as "vectors": I'd rather call them functions $mathbb Nto mathbb R$, or possibly with a domain consisting of some segment of $mathbb N$.
Anyhow, in that context, I thought I had seen this called "the dominance order" on functions, which is a partial order you get when you say $fleq g$ if $f(x)leq g(x)$ for all $x$.
For example here or here.
However, while searching, I see that people use this term for lots of other partial orders that seem unrelated. Still, it seems like a pretty sensible name. You could say that one function(/vector) dominates another if it is greater at each point.
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
add a comment |
$begingroup$
I don't really like a description of these as "vectors": I'd rather call them functions $mathbb Nto mathbb R$, or possibly with a domain consisting of some segment of $mathbb N$.
Anyhow, in that context, I thought I had seen this called "the dominance order" on functions, which is a partial order you get when you say $fleq g$ if $f(x)leq g(x)$ for all $x$.
For example here or here.
However, while searching, I see that people use this term for lots of other partial orders that seem unrelated. Still, it seems like a pretty sensible name. You could say that one function(/vector) dominates another if it is greater at each point.
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
add a comment |
$begingroup$
I don't really like a description of these as "vectors": I'd rather call them functions $mathbb Nto mathbb R$, or possibly with a domain consisting of some segment of $mathbb N$.
Anyhow, in that context, I thought I had seen this called "the dominance order" on functions, which is a partial order you get when you say $fleq g$ if $f(x)leq g(x)$ for all $x$.
For example here or here.
However, while searching, I see that people use this term for lots of other partial orders that seem unrelated. Still, it seems like a pretty sensible name. You could say that one function(/vector) dominates another if it is greater at each point.
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
$endgroup$
I don't really like a description of these as "vectors": I'd rather call them functions $mathbb Nto mathbb R$, or possibly with a domain consisting of some segment of $mathbb N$.
Anyhow, in that context, I thought I had seen this called "the dominance order" on functions, which is a partial order you get when you say $fleq g$ if $f(x)leq g(x)$ for all $x$.
For example here or here.
However, while searching, I see that people use this term for lots of other partial orders that seem unrelated. Still, it seems like a pretty sensible name. You could say that one function(/vector) dominates another if it is greater at each point.
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
answered Jan 2 at 21:48
rschwiebrschwieb
107k12102251
107k12102251
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
add a comment |
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Thank you for your answer! Out of curiosity, may I ask why you don't like a description of those as 'vectors'? I want to make sure I avoid inaccuracies or faux pax.
$endgroup$
– dar_devil
Jan 2 at 21:57
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
Actually, each function from the natural numbers to the reals is a vector.
$endgroup$
– Michael Hoppe
Jan 2 at 22:00
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@MichaelHoppe Nobody’s disputing that. I just said I didn’t like that description (in this case). Some folks tend to cling to tightly to vectors as “lists of numbers” and I prefer the function aspect of it.
$endgroup$
– rschwieb
Jan 2 at 22:58
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@dar_devil Just a preference in this context. Are you talking in the context of a vector space after all?
$endgroup$
– rschwieb
Jan 3 at 1:18
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
$begingroup$
@rschwieb That makes sense. Yes, I am talking in the context of vector spaces. I do think of vectors as "lists of numbers" rather than from the function aspect. I'm in a field that is more engineering than math and maybe that's why.
$endgroup$
– dar_devil
Jan 7 at 20:01
add a comment |
$begingroup$
I've encountered the notation $x_1 succeq x_2$ in the book Convex Optimization by Boyd and Vandenberghe. If $K$ is a cone, then they define the "generalized inequality" $x succeq_K y$ to mean that $x - y in K$. In your case, $K$ is the nonnegative orthant.
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
$endgroup$
add a comment |
$begingroup$
I've encountered the notation $x_1 succeq x_2$ in the book Convex Optimization by Boyd and Vandenberghe. If $K$ is a cone, then they define the "generalized inequality" $x succeq_K y$ to mean that $x - y in K$. In your case, $K$ is the nonnegative orthant.
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
$endgroup$
add a comment |
$begingroup$
I've encountered the notation $x_1 succeq x_2$ in the book Convex Optimization by Boyd and Vandenberghe. If $K$ is a cone, then they define the "generalized inequality" $x succeq_K y$ to mean that $x - y in K$. In your case, $K$ is the nonnegative orthant.
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
$endgroup$
I've encountered the notation $x_1 succeq x_2$ in the book Convex Optimization by Boyd and Vandenberghe. If $K$ is a cone, then they define the "generalized inequality" $x succeq_K y$ to mean that $x - y in K$. In your case, $K$ is the nonnegative orthant.
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
answered Jan 2 at 21:56
littleOlittleO
29.9k646110
29.9k646110
add a comment |
add a comment |
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