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Is a vector space a subspace of itself?
The 2019 Stack Overflow Developer Survey Results Are Invector space and its subspaceHow to determine vector space?Vector Space vs SubspaceProving a subset is a subspace of a Vector SpaceSubspace of a vector space DefinitionProof these operations is not a vector spaceVector Space definitionAre all vector spaces also a subspace?Understand the definition of a vector subspaceDoes multiplying by the zero vector ALWAYS yield the zero vector?
$begingroup$
We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.
Edit: V1 - V10 are the ten axioms of what makes something a vector space. So for example in $Bbb R^2$, we have that $vecx + vecy = vecy + vec x$, etc..
linear-algebra vector-spaces
asked yesterday
mingming
4556
$endgroup$
|
show 3 more comments
$begingroup$
We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.
Edit: V1 - V10 are the ten axioms of what makes something a vector space. So for example in $Bbb R^2$, we have that $vecx + vecy = vecy + vec x$, etc..
linear-algebra vector-spaces
asked yesterday
mingming
4556
$endgroup$
1
$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
yesterday
3
$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
yesterday
2
$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
yesterday
4
$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
22 hours ago
3
$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
17 hours ago
|
show 3 more comments
$begingroup$
We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.
Edit: V1 - V10 are the ten axioms of what makes something a vector space. So for example in $Bbb R^2$, we have that $vecx + vecy = vecy + vec x$, etc..
linear-algebra vector-spaces
asked yesterday
mingming
4556
$endgroup$
We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.
Edit: V1 - V10 are the ten axioms of what makes something a vector space. So for example in $Bbb R^2$, we have that $vecx + vecy = vecy + vec x$, etc..
linear-algebra vector-spaces
linear-algebra vector-spaces
asked yesterday
mingming
4556
asked yesterday
mingming
4556
edited 9 hours ago
ming
asked yesterday
mingming
4556
asked yesterday
mingming
4556
asked yesterday
mingming
4556
4556
1
$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
yesterday
3
$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
yesterday
2
$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
yesterday
4
$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
22 hours ago
3
$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
17 hours ago
|
show 3 more comments
1
$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
yesterday
3
$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
yesterday
2
$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
yesterday
4
$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
22 hours ago
3
$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
17 hours ago
1
1
$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
yesterday
$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
yesterday
3
3
$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
yesterday
$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
yesterday
2
2
$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
yesterday
$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
yesterday
4
4
$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
22 hours ago
$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
22 hours ago
3
3
$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
17 hours ago
$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
17 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
$endgroup$
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
add a comment |
$begingroup$
Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
$endgroup$
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
add a comment |
$begingroup$
I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
$endgroup$
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
add a comment |
$begingroup$
I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
$endgroup$
I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
answered 23 hours ago
Theo BenditTheo Bendit
20.8k12354
20.8k12354
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
add a comment |
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
1
1
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
$begingroup$
Ohh okay I think I understand. So when we prove that something is a subspace of say $Bbb R^2$, we don't have to prove that $c(vecx + vecy) = cvecx + cvecy$ because we already know that's true inside of $Bbb R^2$ and we know this subspace is a vector space INSIDE of $Bbb R^2$ following all its rules.
$endgroup$
– ming
9 hours ago
1
1
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
But then for questions like "A sequence is a infinite list of real numbers. For example 1, 2, 3, 4, 5 is a sequence, and so is 1, -1, 2, -2, 4, -4. We define addition and scalar multiplication so that...." We actually don't have any known information/facts about this vector space so then we need to prove all ten axioms?
$endgroup$
– ming
9 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
$begingroup$
@ming Exactly. Couldn't put it better myself.
$endgroup$
– Theo Bendit
4 hours ago
add a comment |
$begingroup$
Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
$begingroup$
Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
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Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
answered yesterday
José Carlos SantosJosé Carlos Santos
173k23133242
173k23133242
add a comment |
add a comment |
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1
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How do you define a subspace of a vector space?
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– Brian
yesterday
3
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Is a set a subset of itself?? What’s V1-V10?
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– J. W. Tanner
yesterday
2
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The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
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– Theo Bendit
yesterday
4
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As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
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– Xander Henderson
22 hours ago
3
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@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
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– Theo Bendit
17 hours ago