Help understanding proof for vector subspace (Hoffman and Kunze)
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In Hoffman and Kunze, following proof is provided for this theorem:
Theorem: A non-empty subset W of V is a subspace of V iff for each pair of vectors a,b in W and each scalar c in F the vector ca + b is again in W.
Proof.
Suppose that W is a non-empty subset of V such that ca + b belongs to W for all vectors a, b in W and all scalars c in F.Since W is non-empty, there is a vector p in W, and hence (-1)p+p= 0 is in W. Then if a is any vector in W and c any scalar, the vector ca = ca + 0 is in W. In particular, (-1)a = -a is in W. Finally, if a and b are in W, then a + b = 1a + b is in W.Thus W is a subspace of V.
Conversely, if W is a subspace of V, a and b are in W, and c is a scalar, certainly ca + b is in W.
What I don't get is:
How (-1)p+p= 0 is concluded to be in W?
How proving a + b = 1a + b is in W helps?
Please help in solving my above queries.
Thanks.
linear-algebra proof-explanation
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
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add a comment |
$begingroup$
In Hoffman and Kunze, following proof is provided for this theorem:
Theorem: A non-empty subset W of V is a subspace of V iff for each pair of vectors a,b in W and each scalar c in F the vector ca + b is again in W.
Proof.
Suppose that W is a non-empty subset of V such that ca + b belongs to W for all vectors a, b in W and all scalars c in F.Since W is non-empty, there is a vector p in W, and hence (-1)p+p= 0 is in W. Then if a is any vector in W and c any scalar, the vector ca = ca + 0 is in W. In particular, (-1)a = -a is in W. Finally, if a and b are in W, then a + b = 1a + b is in W.Thus W is a subspace of V.
Conversely, if W is a subspace of V, a and b are in W, and c is a scalar, certainly ca + b is in W.
What I don't get is:
How (-1)p+p= 0 is concluded to be in W?
How proving a + b = 1a + b is in W helps?
Please help in solving my above queries.
Thanks.
linear-algebra proof-explanation
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
$endgroup$
add a comment |
$begingroup$
In Hoffman and Kunze, following proof is provided for this theorem:
Theorem: A non-empty subset W of V is a subspace of V iff for each pair of vectors a,b in W and each scalar c in F the vector ca + b is again in W.
Proof.
Suppose that W is a non-empty subset of V such that ca + b belongs to W for all vectors a, b in W and all scalars c in F.Since W is non-empty, there is a vector p in W, and hence (-1)p+p= 0 is in W. Then if a is any vector in W and c any scalar, the vector ca = ca + 0 is in W. In particular, (-1)a = -a is in W. Finally, if a and b are in W, then a + b = 1a + b is in W.Thus W is a subspace of V.
Conversely, if W is a subspace of V, a and b are in W, and c is a scalar, certainly ca + b is in W.
What I don't get is:
How (-1)p+p= 0 is concluded to be in W?
How proving a + b = 1a + b is in W helps?
Please help in solving my above queries.
Thanks.
linear-algebra proof-explanation
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
$endgroup$
In Hoffman and Kunze, following proof is provided for this theorem:
Theorem: A non-empty subset W of V is a subspace of V iff for each pair of vectors a,b in W and each scalar c in F the vector ca + b is again in W.
Proof.
Suppose that W is a non-empty subset of V such that ca + b belongs to W for all vectors a, b in W and all scalars c in F.Since W is non-empty, there is a vector p in W, and hence (-1)p+p= 0 is in W. Then if a is any vector in W and c any scalar, the vector ca = ca + 0 is in W. In particular, (-1)a = -a is in W. Finally, if a and b are in W, then a + b = 1a + b is in W.Thus W is a subspace of V.
Conversely, if W is a subspace of V, a and b are in W, and c is a scalar, certainly ca + b is in W.
What I don't get is:
How (-1)p+p= 0 is concluded to be in W?
How proving a + b = 1a + b is in W helps?
Please help in solving my above queries.
Thanks.
linear-algebra proof-explanation
linear-algebra proof-explanation
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
asked Dec 31 '18 at 12:06
dheeraj suthardheeraj suthar
72
72
add a comment |
add a comment |
2 Answers
2
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oldest
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- Take $c = -1$, $a = b = p$.
- Recall the definition of a subspace: it's a subset containing $0$ such that, for all $a, b$ contained in it, $a + b$ is contained in it, and for all $a$ contained in it and all scalars $lambda$, $lambda a$ is contained in it.
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
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add a comment |
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- In the property of the theorem, take $a=b=p$ and $c=-1$.
- Stability for addition is one of the properties which define a subspace, along with stability for scalar multiplication.
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- Take $c = -1$, $a = b = p$.
- Recall the definition of a subspace: it's a subset containing $0$ such that, for all $a, b$ contained in it, $a + b$ is contained in it, and for all $a$ contained in it and all scalars $lambda$, $lambda a$ is contained in it.
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
$endgroup$
add a comment |
$begingroup$
- Take $c = -1$, $a = b = p$.
- Recall the definition of a subspace: it's a subset containing $0$ such that, for all $a, b$ contained in it, $a + b$ is contained in it, and for all $a$ contained in it and all scalars $lambda$, $lambda a$ is contained in it.
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
$endgroup$
add a comment |
$begingroup$
- Take $c = -1$, $a = b = p$.
- Recall the definition of a subspace: it's a subset containing $0$ such that, for all $a, b$ contained in it, $a + b$ is contained in it, and for all $a$ contained in it and all scalars $lambda$, $lambda a$ is contained in it.
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
$endgroup$
- Take $c = -1$, $a = b = p$.
- Recall the definition of a subspace: it's a subset containing $0$ such that, for all $a, b$ contained in it, $a + b$ is contained in it, and for all $a$ contained in it and all scalars $lambda$, $lambda a$ is contained in it.
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
answered Dec 31 '18 at 12:12
user3482749user3482749
4,3541119
4,3541119
add a comment |
add a comment |
$begingroup$
- In the property of the theorem, take $a=b=p$ and $c=-1$.
- Stability for addition is one of the properties which define a subspace, along with stability for scalar multiplication.
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
- In the property of the theorem, take $a=b=p$ and $c=-1$.
- Stability for addition is one of the properties which define a subspace, along with stability for scalar multiplication.
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
$endgroup$
add a comment |
$begingroup$
- In the property of the theorem, take $a=b=p$ and $c=-1$.
- Stability for addition is one of the properties which define a subspace, along with stability for scalar multiplication.
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
$endgroup$
- In the property of the theorem, take $a=b=p$ and $c=-1$.
- Stability for addition is one of the properties which define a subspace, along with stability for scalar multiplication.
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
answered Dec 31 '18 at 12:11
BernardBernard
125k743119
125k743119
add a comment |
add a comment |
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