Mixing
I've seen two definitions of subspace; one involving vector spaces and one requiring linear combinations
$begingroup$
The first definition of subspace I was exposed to is: A subspace of $mathbb{R^n}$ is any collection S of vectors in $mathbb{R^n}$ such that:
The zero vector 0 is in S
If u and v are in S, then u + v is in S
If u is in S and c is a scalar, then cu is in S
And you can combine 2 and 3 into one requirement, which is that S be closed under linear combinations.
The second definition of subspace I saw, much later on, was: A subset W of a vector space V is called a subspace of V if W is itself a vector space with the same scalars, addition, and scalar multiplication as V.
I'm a bit confused on the difference between these two definitions. How do I reconcile one with the other? Is there any meaningful difference between the two? As far as I know, vector spaces are also required to be closed under linear combinations, so isn't the second definition saying the same exact thing as the first definition?
One idea I have is that the second definition is a more general form of the first definition, since the first definition is talking about $mathbb{R^n}$ (and $mathbb{R^n}$ is a vector space), while the second definition is talking about all vector spaces (including $mathbb{R^n}$ and more). Is this the correct interpretation? And are there any other differences between the two definitions that I'm missing?
Any help is greatly appreciated.
linear-algebra vector-spaces
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
$endgroup$
add a comment |
$begingroup$
The first definition of subspace I was exposed to is: A subspace of $mathbb{R^n}$ is any collection S of vectors in $mathbb{R^n}$ such that:
The zero vector 0 is in S
If u and v are in S, then u + v is in S
If u is in S and c is a scalar, then cu is in S
And you can combine 2 and 3 into one requirement, which is that S be closed under linear combinations.
The second definition of subspace I saw, much later on, was: A subset W of a vector space V is called a subspace of V if W is itself a vector space with the same scalars, addition, and scalar multiplication as V.
I'm a bit confused on the difference between these two definitions. How do I reconcile one with the other? Is there any meaningful difference between the two? As far as I know, vector spaces are also required to be closed under linear combinations, so isn't the second definition saying the same exact thing as the first definition?
One idea I have is that the second definition is a more general form of the first definition, since the first definition is talking about $mathbb{R^n}$ (and $mathbb{R^n}$ is a vector space), while the second definition is talking about all vector spaces (including $mathbb{R^n}$ and more). Is this the correct interpretation? And are there any other differences between the two definitions that I'm missing?
Any help is greatly appreciated.
linear-algebra vector-spaces
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
$endgroup$
2
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50
add a comment |
$begingroup$
The first definition of subspace I was exposed to is: A subspace of $mathbb{R^n}$ is any collection S of vectors in $mathbb{R^n}$ such that:
The zero vector 0 is in S
If u and v are in S, then u + v is in S
If u is in S and c is a scalar, then cu is in S
And you can combine 2 and 3 into one requirement, which is that S be closed under linear combinations.
The second definition of subspace I saw, much later on, was: A subset W of a vector space V is called a subspace of V if W is itself a vector space with the same scalars, addition, and scalar multiplication as V.
I'm a bit confused on the difference between these two definitions. How do I reconcile one with the other? Is there any meaningful difference between the two? As far as I know, vector spaces are also required to be closed under linear combinations, so isn't the second definition saying the same exact thing as the first definition?
One idea I have is that the second definition is a more general form of the first definition, since the first definition is talking about $mathbb{R^n}$ (and $mathbb{R^n}$ is a vector space), while the second definition is talking about all vector spaces (including $mathbb{R^n}$ and more). Is this the correct interpretation? And are there any other differences between the two definitions that I'm missing?
Any help is greatly appreciated.
linear-algebra vector-spaces
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
$endgroup$
The first definition of subspace I was exposed to is: A subspace of $mathbb{R^n}$ is any collection S of vectors in $mathbb{R^n}$ such that:
The zero vector 0 is in S
If u and v are in S, then u + v is in S
If u is in S and c is a scalar, then cu is in S
And you can combine 2 and 3 into one requirement, which is that S be closed under linear combinations.
The second definition of subspace I saw, much later on, was: A subset W of a vector space V is called a subspace of V if W is itself a vector space with the same scalars, addition, and scalar multiplication as V.
I'm a bit confused on the difference between these two definitions. How do I reconcile one with the other? Is there any meaningful difference between the two? As far as I know, vector spaces are also required to be closed under linear combinations, so isn't the second definition saying the same exact thing as the first definition?
One idea I have is that the second definition is a more general form of the first definition, since the first definition is talking about $mathbb{R^n}$ (and $mathbb{R^n}$ is a vector space), while the second definition is talking about all vector spaces (including $mathbb{R^n}$ and more). Is this the correct interpretation? And are there any other differences between the two definitions that I'm missing?
Any help is greatly appreciated.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
asked Jan 9 at 1:31
James RonaldJames Ronald
1207
1207
2
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50
add a comment |
2
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50
2
2
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50
add a comment |
1 Answer
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$begingroup$
The definitions are equivalent.
A subspace V of a vector space W is a subset of W that is itself a vector space.
It turns out that all we have to do is show the following:
- 0 is in V
- V is closed under addition
- V is closed under scalar multiplication
Since we know that V is a subset of W then we know that all the other properties of a vector space are satisfied. For example, we know that if a and b are in V then a+b = b+a.
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
The definitions are equivalent.
A subspace V of a vector space W is a subset of W that is itself a vector space.
It turns out that all we have to do is show the following:
- 0 is in V
- V is closed under addition
- V is closed under scalar multiplication
Since we know that V is a subset of W then we know that all the other properties of a vector space are satisfied. For example, we know that if a and b are in V then a+b = b+a.
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
$endgroup$
add a comment |
$begingroup$
The definitions are equivalent.
A subspace V of a vector space W is a subset of W that is itself a vector space.
It turns out that all we have to do is show the following:
- 0 is in V
- V is closed under addition
- V is closed under scalar multiplication
Since we know that V is a subset of W then we know that all the other properties of a vector space are satisfied. For example, we know that if a and b are in V then a+b = b+a.
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
$endgroup$
add a comment |
$begingroup$
The definitions are equivalent.
A subspace V of a vector space W is a subset of W that is itself a vector space.
It turns out that all we have to do is show the following:
- 0 is in V
- V is closed under addition
- V is closed under scalar multiplication
Since we know that V is a subset of W then we know that all the other properties of a vector space are satisfied. For example, we know that if a and b are in V then a+b = b+a.
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
$endgroup$
The definitions are equivalent.
A subspace V of a vector space W is a subset of W that is itself a vector space.
It turns out that all we have to do is show the following:
- 0 is in V
- V is closed under addition
- V is closed under scalar multiplication
Since we know that V is a subset of W then we know that all the other properties of a vector space are satisfied. For example, we know that if a and b are in V then a+b = b+a.
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
answered Jan 9 at 1:54
NicNic8NicNic8
4,33531023
4,33531023
add a comment |
add a comment |
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2
$begingroup$
The definitions are equivalent
$endgroup$
– Omnomnomnom
Jan 9 at 1:38
$begingroup$
In the first definition, you can replace $mathbb{R}^n$ with an arbitrary vector space as well.
$endgroup$
– angryavian
Jan 9 at 1:50