Why are all vectors with exactly one nonzero component not a subspace of $mathbb{R}^3$?
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In order for a set of vectors to be part of a subspace there can't be any redundant vectors included in the set, i.e the set should contain only the minimum amount of vectors necessary which can be scaled and combined in different ways to span the entire subspace. Correct?
Might it be the case then that if my set includes ALL vectors with exactly one nonzero component that there will clearly be redundancy in my set?
The zero vector is included and the set is both closed under scalar multiplication and addition so I don't see any other explanation.
linear-algebra vector-spaces
asked Nov 16 at 8:33
Mango164
313
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up vote
-1
down vote
favorite
In order for a set of vectors to be part of a subspace there can't be any redundant vectors included in the set, i.e the set should contain only the minimum amount of vectors necessary which can be scaled and combined in different ways to span the entire subspace. Correct?
Might it be the case then that if my set includes ALL vectors with exactly one nonzero component that there will clearly be redundancy in my set?
The zero vector is included and the set is both closed under scalar multiplication and addition so I don't see any other explanation.
linear-algebra vector-spaces
asked Nov 16 at 8:33
Mango164
313
3
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
1
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In order for a set of vectors to be part of a subspace there can't be any redundant vectors included in the set, i.e the set should contain only the minimum amount of vectors necessary which can be scaled and combined in different ways to span the entire subspace. Correct?
Might it be the case then that if my set includes ALL vectors with exactly one nonzero component that there will clearly be redundancy in my set?
The zero vector is included and the set is both closed under scalar multiplication and addition so I don't see any other explanation.
linear-algebra vector-spaces
asked Nov 16 at 8:33
Mango164
313
In order for a set of vectors to be part of a subspace there can't be any redundant vectors included in the set, i.e the set should contain only the minimum amount of vectors necessary which can be scaled and combined in different ways to span the entire subspace. Correct?
Might it be the case then that if my set includes ALL vectors with exactly one nonzero component that there will clearly be redundancy in my set?
The zero vector is included and the set is both closed under scalar multiplication and addition so I don't see any other explanation.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 16 at 8:33
Mango164
313
asked Nov 16 at 8:33
Mango164
313
asked Nov 16 at 8:33
Mango164
313
asked Nov 16 at 8:33
Mango164
313
asked Nov 16 at 8:33
Mango164
313
313
3
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
1
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11
add a comment |
3
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
1
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11
3
3
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
1
1
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11
add a comment |
1 Answer
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The set is not closed under addition. Consider the vectors $vec{v} = (1,0,0)$ and $vec{w} = (0,1,0)$. The sum is $vec{v}+vec{w} = (1,1,0)$, i.e. a vector with two nonzero components.
Also I don't think the zero vector is included, since it has exactly zero nonzero components.
answered Nov 16 at 8:36
Eevee Trainer
1,459216
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The set is not closed under addition. Consider the vectors $vec{v} = (1,0,0)$ and $vec{w} = (0,1,0)$. The sum is $vec{v}+vec{w} = (1,1,0)$, i.e. a vector with two nonzero components.
Also I don't think the zero vector is included, since it has exactly zero nonzero components.
answered Nov 16 at 8:36
Eevee Trainer
1,459216
add a comment |
up vote
2
down vote
accepted
The set is not closed under addition. Consider the vectors $vec{v} = (1,0,0)$ and $vec{w} = (0,1,0)$. The sum is $vec{v}+vec{w} = (1,1,0)$, i.e. a vector with two nonzero components.
Also I don't think the zero vector is included, since it has exactly zero nonzero components.
answered Nov 16 at 8:36
Eevee Trainer
1,459216
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The set is not closed under addition. Consider the vectors $vec{v} = (1,0,0)$ and $vec{w} = (0,1,0)$. The sum is $vec{v}+vec{w} = (1,1,0)$, i.e. a vector with two nonzero components.
Also I don't think the zero vector is included, since it has exactly zero nonzero components.
answered Nov 16 at 8:36
Eevee Trainer
1,459216
The set is not closed under addition. Consider the vectors $vec{v} = (1,0,0)$ and $vec{w} = (0,1,0)$. The sum is $vec{v}+vec{w} = (1,1,0)$, i.e. a vector with two nonzero components.
Also I don't think the zero vector is included, since it has exactly zero nonzero components.
answered Nov 16 at 8:36
Eevee Trainer
1,459216
answered Nov 16 at 8:36
Eevee Trainer
1,459216
answered Nov 16 at 8:36
Eevee Trainer
1,459216
answered Nov 16 at 8:36
Eevee Trainer
1,459216
1,459216
add a comment |
add a comment |
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3
Closed under addition, you said? (1,0,0) + (0,0,1) = ?
– Ivan Neretin
Nov 16 at 8:36
1
It''s hard to be certain whether the first sentence is correct since it's very unclear what it means. But it sounds like you're confusing "part of a subspace" with "basis for a subspace"...
– David C. Ullrich
Nov 16 at 15:11