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Set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$, how do I show that it is not a subspace of $mathbbR^3$.
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So I know that set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$.
Specifically, it is worded in our lecture that it is a " subset of $(mathbbR^3, oplus, odot)$ , where $oplus$ and $odot$ are the usual vector addition and scalar multiplication."
My teacher has stated in our lecture that this set $S$ is not a subspace of $mathbbR^3$.
But from what I can tell $S$ is:
- Closed under addition
- Closed under multiplication
- Contains a zero vector $(0,0,0)$
How is it not a subspace of $mathbbR^3$, what am I missing?
linear-algebra vector-spaces
edited Jul 29 at 1:31
Cornman
2,30021027
asked Jul 28 at 23:16
Wyatt Rose
133
 |Â
show 1 more comment
up vote
2
down vote
favorite
So I know that set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$.
Specifically, it is worded in our lecture that it is a " subset of $(mathbbR^3, oplus, odot)$ , where $oplus$ and $odot$ are the usual vector addition and scalar multiplication."
My teacher has stated in our lecture that this set $S$ is not a subspace of $mathbbR^3$.
But from what I can tell $S$ is:
- Closed under addition
- Closed under multiplication
- Contains a zero vector $(0,0,0)$
How is it not a subspace of $mathbbR^3$, what am I missing?
linear-algebra vector-spaces
edited Jul 29 at 1:31
Cornman
2,30021027
asked Jul 28 at 23:16
Wyatt Rose
133
0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
1
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I know that set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$.
Specifically, it is worded in our lecture that it is a " subset of $(mathbbR^3, oplus, odot)$ , where $oplus$ and $odot$ are the usual vector addition and scalar multiplication."
My teacher has stated in our lecture that this set $S$ is not a subspace of $mathbbR^3$.
But from what I can tell $S$ is:
- Closed under addition
- Closed under multiplication
- Contains a zero vector $(0,0,0)$
How is it not a subspace of $mathbbR^3$, what am I missing?
linear-algebra vector-spaces
edited Jul 29 at 1:31
Cornman
2,30021027
asked Jul 28 at 23:16
Wyatt Rose
133
So I know that set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$.
Specifically, it is worded in our lecture that it is a " subset of $(mathbbR^3, oplus, odot)$ , where $oplus$ and $odot$ are the usual vector addition and scalar multiplication."
My teacher has stated in our lecture that this set $S$ is not a subspace of $mathbbR^3$.
But from what I can tell $S$ is:
- Closed under addition
- Closed under multiplication
- Contains a zero vector $(0,0,0)$
How is it not a subspace of $mathbbR^3$, what am I missing?
linear-algebra vector-spaces
edited Jul 29 at 1:31
Cornman
2,30021027
asked Jul 28 at 23:16
Wyatt Rose
133
edited Jul 29 at 1:31
Cornman
2,30021027
edited Jul 29 at 1:31
Cornman
2,30021027
edited Jul 29 at 1:31
Cornman
2,30021027
2,30021027
asked Jul 28 at 23:16
Wyatt Rose
133
asked Jul 28 at 23:16
Wyatt Rose
133
asked Jul 28 at 23:16
Wyatt Rose
133
133
0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
1
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21
 |Â
show 1 more comment
0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
1
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21
0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
1
1
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.
answered Jul 29 at 0:12
Randall
7,2221825
add a comment |Â
up vote
1
down vote
It is not closed under multiplication.
Take $lambda=sqrt2$ and any vector $v=(a,b,c)in S$. Since $a,b,cin Bbb Z$ then $asqrt2, bsqrt2, csqrt2notin Bbb Z$. Then $lambdacdot v notin S$.
answered Jul 28 at 23:28
Legoman
4,64421033
add a comment |Â
up vote
0
down vote
Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)
First lets look at the definition of a subspace:
All products and sums composed of elements within the subspace also are in the subspace.
All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.
0 is also an element of the subspace.
You remembered rule 1 and 3 however, it's clear S is violating rule 2.
In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.
answered Jul 29 at 1:07
chahochi
166
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.
answered Jul 29 at 0:12
Randall
7,2221825
add a comment |Â
up vote
1
down vote
accepted
$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.
answered Jul 29 at 0:12
Randall
7,2221825
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.
answered Jul 29 at 0:12
Randall
7,2221825
$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.
answered Jul 29 at 0:12
Randall
7,2221825
answered Jul 29 at 0:12
Randall
7,2221825
answered Jul 29 at 0:12
Randall
7,2221825
answered Jul 29 at 0:12
Randall
7,2221825
7,2221825
add a comment |Â
add a comment |Â
up vote
1
down vote
It is not closed under multiplication.
Take $lambda=sqrt2$ and any vector $v=(a,b,c)in S$. Since $a,b,cin Bbb Z$ then $asqrt2, bsqrt2, csqrt2notin Bbb Z$. Then $lambdacdot v notin S$.
answered Jul 28 at 23:28
Legoman
4,64421033
add a comment |Â
up vote
1
down vote
It is not closed under multiplication.
Take $lambda=sqrt2$ and any vector $v=(a,b,c)in S$. Since $a,b,cin Bbb Z$ then $asqrt2, bsqrt2, csqrt2notin Bbb Z$. Then $lambdacdot v notin S$.
answered Jul 28 at 23:28
Legoman
4,64421033
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is not closed under multiplication.
Take $lambda=sqrt2$ and any vector $v=(a,b,c)in S$. Since $a,b,cin Bbb Z$ then $asqrt2, bsqrt2, csqrt2notin Bbb Z$. Then $lambdacdot v notin S$.
answered Jul 28 at 23:28
Legoman
4,64421033
It is not closed under multiplication.
Take $lambda=sqrt2$ and any vector $v=(a,b,c)in S$. Since $a,b,cin Bbb Z$ then $asqrt2, bsqrt2, csqrt2notin Bbb Z$. Then $lambdacdot v notin S$.
answered Jul 28 at 23:28
Legoman
4,64421033
answered Jul 28 at 23:28
Legoman
4,64421033
answered Jul 28 at 23:28
Legoman
4,64421033
answered Jul 28 at 23:28
Legoman
4,64421033
4,64421033
add a comment |Â
add a comment |Â
up vote
0
down vote
Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)
First lets look at the definition of a subspace:
All products and sums composed of elements within the subspace also are in the subspace.
All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.
0 is also an element of the subspace.
You remembered rule 1 and 3 however, it's clear S is violating rule 2.
In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.
answered Jul 29 at 1:07
chahochi
166
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
add a comment |Â
up vote
0
down vote
Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)
First lets look at the definition of a subspace:
All products and sums composed of elements within the subspace also are in the subspace.
All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.
0 is also an element of the subspace.
You remembered rule 1 and 3 however, it's clear S is violating rule 2.
In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.
answered Jul 29 at 1:07
chahochi
166
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)
First lets look at the definition of a subspace:
All products and sums composed of elements within the subspace also are in the subspace.
All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.
0 is also an element of the subspace.
You remembered rule 1 and 3 however, it's clear S is violating rule 2.
In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.
answered Jul 29 at 1:07
chahochi
166
Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)
First lets look at the definition of a subspace:
All products and sums composed of elements within the subspace also are in the subspace.
All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.
0 is also an element of the subspace.
You remembered rule 1 and 3 however, it's clear S is violating rule 2.
In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.
answered Jul 29 at 1:07
chahochi
166
edited Jul 29 at 1:34
answered Jul 29 at 1:07
chahochi
166
answered Jul 29 at 1:07
chahochi
166
answered Jul 29 at 1:07
chahochi
166
166
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
add a comment |Â
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
But point 2 isn't the real problem. OP is working over the reals, so s/he definitely has inverses of all non-zero scalars. The problem is that the scalar multiplication is not closed on the potential-subspace $S$.
– Randall
Jul 29 at 1:21
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
@Randall Oh you're right. I misread the question. I'll edit my answer.
– chahochi
Jul 29 at 1:25
add a comment |Â
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0.5 times any vector is once again in the set?
– Randall
Jul 28 at 23:17
You need to check the definition of a vector subspace, and check each of the things your set needs to satisfy. There is at least one which it clearly doesn't satisfy.
– Matt
Jul 28 at 23:19
this is called an integer lattice. For a linear subspace, any real mulitple of a vector must also be in the subspace
– Will Jagy
Jul 28 at 23:19
No, 0.5 times any vector in S would not be within the set anymore. But being closed under multiplication means that the end result would need to be within R3, not still within S. Am I misunderstanding that? Does the result need to stay within S as well?
– Wyatt Rose
Jul 28 at 23:20
1
Yes you are misunderstanding that.
– Randall
Jul 28 at 23:21