Why we only need to verify additive identity, and closed under addition and scalar multiplication for...
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In the book Linear Algebra Done Right, it is said that to determine quickly whether a given subset of $V$ is a subspace of $V$, the three conditions, namely additive identity, closed under addition, and closed under scalar multiplication, should be satisfied. The other parts of the definition of a vector space are automatically satisfied.
I think I understand why commutativity, associativity, distributive properties, and multiplicative identity works because their operations are still within the subspace.
But, why don't we need to verify additive inverse, similar to verifying additive identity? Could there be cases where there will be no $v + w = 0$ in the new subspace, $v, w in U$, $U$ is a subspace?
linear-algebra
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
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In the book Linear Algebra Done Right, it is said that to determine quickly whether a given subset of $V$ is a subspace of $V$, the three conditions, namely additive identity, closed under addition, and closed under scalar multiplication, should be satisfied. The other parts of the definition of a vector space are automatically satisfied.
I think I understand why commutativity, associativity, distributive properties, and multiplicative identity works because their operations are still within the subspace.
But, why don't we need to verify additive inverse, similar to verifying additive identity? Could there be cases where there will be no $v + w = 0$ in the new subspace, $v, w in U$, $U$ is a subspace?
linear-algebra
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
$endgroup$
add a comment |
$begingroup$
In the book Linear Algebra Done Right, it is said that to determine quickly whether a given subset of $V$ is a subspace of $V$, the three conditions, namely additive identity, closed under addition, and closed under scalar multiplication, should be satisfied. The other parts of the definition of a vector space are automatically satisfied.
I think I understand why commutativity, associativity, distributive properties, and multiplicative identity works because their operations are still within the subspace.
But, why don't we need to verify additive inverse, similar to verifying additive identity? Could there be cases where there will be no $v + w = 0$ in the new subspace, $v, w in U$, $U$ is a subspace?
linear-algebra
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
$endgroup$
In the book Linear Algebra Done Right, it is said that to determine quickly whether a given subset of $V$ is a subspace of $V$, the three conditions, namely additive identity, closed under addition, and closed under scalar multiplication, should be satisfied. The other parts of the definition of a vector space are automatically satisfied.
I think I understand why commutativity, associativity, distributive properties, and multiplicative identity works because their operations are still within the subspace.
But, why don't we need to verify additive inverse, similar to verifying additive identity? Could there be cases where there will be no $v + w = 0$ in the new subspace, $v, w in U$, $U$ is a subspace?
linear-algebra
linear-algebra
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
asked Dec 9 '18 at 23:05
JOHN JOHN
4279
4279
add a comment |
add a comment |
1 Answer
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You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.
EDIT: Similarly, for the scalar $0$, $0v={bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
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3
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To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
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– GenericMathematician
Dec 9 '18 at 23:15
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I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
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– JOHN
Dec 10 '18 at 22:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.
EDIT: Similarly, for the scalar $0$, $0v={bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
$endgroup$
3
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
add a comment |
$begingroup$
You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.
EDIT: Similarly, for the scalar $0$, $0v={bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
$endgroup$
3
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
add a comment |
$begingroup$
You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.
EDIT: Similarly, for the scalar $0$, $0v={bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
$endgroup$
You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.
EDIT: Similarly, for the scalar $0$, $0v={bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
edited Dec 10 '18 at 0:10
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
answered Dec 9 '18 at 23:12
AnyADAnyAD
2,113812
2,113812
3
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
add a comment |
3
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
3
3
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
To be fair, you will get non-emptiness by verifying that $vec{0}$ is in the set. I think that is what is meant by additive identity.
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:15
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
$begingroup$
I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse.
$endgroup$
– JOHN
Dec 10 '18 at 22:39
add a comment |
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