Why ${mathbf{0}}$ has dimension zero?
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According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).
Then why does ${mathbf{0}}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
linear-algebra
asked Feb 5 '14 at 13:40
Qingtian
3131411
add a comment |
up vote
24
down vote
favorite
According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).
Then why does ${mathbf{0}}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
linear-algebra
asked Feb 5 '14 at 13:40
Qingtian
3131411
13
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
1
it is a definition
– Semsem
Feb 5 '14 at 13:57
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42
add a comment |
up vote
24
down vote
favorite
up vote
24
down vote
favorite
According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).
Then why does ${mathbf{0}}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
linear-algebra
asked Feb 5 '14 at 13:40
Qingtian
3131411
According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).
Then why does ${mathbf{0}}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
linear-algebra
linear-algebra
asked Feb 5 '14 at 13:40
Qingtian
3131411
asked Feb 5 '14 at 13:40
Qingtian
3131411
edited Nov 22 at 23:14
asked Feb 5 '14 at 13:40
Qingtian
3131411
asked Feb 5 '14 at 13:40
Qingtian
3131411
asked Feb 5 '14 at 13:40
Qingtian
3131411
3131411
13
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
1
it is a definition
– Semsem
Feb 5 '14 at 13:57
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42
add a comment |
13
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
1
it is a definition
– Semsem
Feb 5 '14 at 13:57
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42
13
13
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
1
1
it is a definition
– Semsem
Feb 5 '14 at 13:57
it is a definition
– Semsem
Feb 5 '14 at 13:57
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42
add a comment |
2 Answers
2
active
oldest
votes
up vote
29
down vote
accepted
A vector by itself doesn't have a dimension. A subspace has a dimension. Why ${mathbf{0}}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in ${mathbf{0}}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in ${mathbf{0}}$ is the empty set that has zero elements.
answered Feb 5 '14 at 14:19
egreg
177k1484198
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
|
show 2 more comments
up vote
5
down vote
The sum over the empty set is the additive identity... in this case zero.
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
29
down vote
accepted
A vector by itself doesn't have a dimension. A subspace has a dimension. Why ${mathbf{0}}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in ${mathbf{0}}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in ${mathbf{0}}$ is the empty set that has zero elements.
answered Feb 5 '14 at 14:19
egreg
177k1484198
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
|
show 2 more comments
up vote
29
down vote
accepted
A vector by itself doesn't have a dimension. A subspace has a dimension. Why ${mathbf{0}}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in ${mathbf{0}}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in ${mathbf{0}}$ is the empty set that has zero elements.
answered Feb 5 '14 at 14:19
egreg
177k1484198
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
|
show 2 more comments
up vote
29
down vote
accepted
up vote
29
down vote
accepted
A vector by itself doesn't have a dimension. A subspace has a dimension. Why ${mathbf{0}}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in ${mathbf{0}}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in ${mathbf{0}}$ is the empty set that has zero elements.
answered Feb 5 '14 at 14:19
egreg
177k1484198
A vector by itself doesn't have a dimension. A subspace has a dimension. Why ${mathbf{0}}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.
What's the largest linearly independent set in ${mathbf{0}}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in ${mathbf{0}}$ is the empty set that has zero elements.
answered Feb 5 '14 at 14:19
egreg
177k1484198
edited Aug 6 '16 at 20:11
answered Feb 5 '14 at 14:19
egreg
177k1484198
answered Feb 5 '14 at 14:19
egreg
177k1484198
answered Feb 5 '14 at 14:19
egreg
177k1484198
177k1484198
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
|
show 2 more comments
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
1
1
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
Very nice. Thanks for pointing out my confusion!
– Qingtian
Feb 5 '14 at 18:05
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
I think it is just a matter of definition! And the reason for such a definition as you said is for consistency with other cases. However, I think that the part that you are saying ${} in {mathbf{0}}$ is not true. Am I right? :)
– H. R.
Aug 6 '16 at 13:26
1
1
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@H.R. Where am I saying that? The only sets of vectors (included) in it...
– egreg
Aug 6 '16 at 14:18
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
@egreg: Specifically, I mean this phrase "The only sets of vectors in it are the empty set and the whole set"
– H. R.
Aug 6 '16 at 14:33
2
2
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
@H.R. Done as requested
– egreg
Aug 6 '16 at 20:12
|
show 2 more comments
up vote
5
down vote
The sum over the empty set is the additive identity... in this case zero.
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
add a comment |
up vote
5
down vote
The sum over the empty set is the additive identity... in this case zero.
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
add a comment |
up vote
5
down vote
up vote
5
down vote
The sum over the empty set is the additive identity... in this case zero.
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
The sum over the empty set is the additive identity... in this case zero.
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
answered Feb 5 '14 at 13:42
JP McCarthy
5,54912440
5,54912440
add a comment |
add a comment |
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13
The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space.
– Michael Joyce
Feb 5 '14 at 13:45
1
it is a definition
– Semsem
Feb 5 '14 at 13:57
@MichaelJoyce see this math.stackexchange.com/questions/1315457/…
– user75086
Jun 7 '15 at 8:55
Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different.
– jdods
Aug 6 '16 at 13:42