Proving a subset of Vector space as a Subspacenecessary condition for subspace of a vector spaceVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceSubset of a vector spaceProof for the necessity of conditions for a subspaceLinear Algebra: which of the definition of subspace of a vector space is more correct?Proof that something is a subspace given it's a subset of a vector space.Set $S= x,y,z in mathbbZ$ is a subset of vector space $mathbbR^3$, how do I show that it is not a subspace of $mathbbR^3$.zero vector subspaceProof of $mathcalL(V,W)$ is a vector space
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Proving a subset of Vector space as a Subspace
necessary condition for subspace of a vector spaceVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceSubset of a vector spaceProof for the necessity of conditions for a subspaceLinear Algebra: which of the definition of subspace of a vector space is more correct?Proof that something is a subspace given it's a subset of a vector space.Set $S=(x,y,z)$ is a subset of vector space $mathbbR^3$, how do I show that it is not a subspace of $mathbbR^3$.zero vector subspaceProof of $mathcalL(V,W)$ is a vector space
$begingroup$
I know that if the conditions of being a vector space is fulfilled by a subset of a vector space then it will be a subspace. And these conditions reduce to the three conditions that is:
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
But how come these condition fulfill that additive inverse of a vector in the subset also exists in the subset.
I searched for this answer where it told that
$ -1.x=-x $
and hence additive inverse exists in the subset, but how can we assume that the field always contains $ -1 $ ?
linear-algebra abstract-algebra vector-spaces
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I know that if the conditions of being a vector space is fulfilled by a subset of a vector space then it will be a subspace. And these conditions reduce to the three conditions that is:
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
But how come these condition fulfill that additive inverse of a vector in the subset also exists in the subset.
I searched for this answer where it told that
$ -1.x=-x $
and hence additive inverse exists in the subset, but how can we assume that the field always contains $ -1 $ ?
linear-algebra abstract-algebra vector-spaces
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32
add a comment |
$begingroup$
I know that if the conditions of being a vector space is fulfilled by a subset of a vector space then it will be a subspace. And these conditions reduce to the three conditions that is:
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
But how come these condition fulfill that additive inverse of a vector in the subset also exists in the subset.
I searched for this answer where it told that
$ -1.x=-x $
and hence additive inverse exists in the subset, but how can we assume that the field always contains $ -1 $ ?
linear-algebra abstract-algebra vector-spaces
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I know that if the conditions of being a vector space is fulfilled by a subset of a vector space then it will be a subspace. And these conditions reduce to the three conditions that is:
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
But how come these condition fulfill that additive inverse of a vector in the subset also exists in the subset.
I searched for this answer where it told that
$ -1.x=-x $
and hence additive inverse exists in the subset, but how can we assume that the field always contains $ -1 $ ?
linear-algebra abstract-algebra vector-spaces
linear-algebra abstract-algebra vector-spaces
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 9:29
ShivamShivam
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 9:29
ShivamShivam
62
asked Mar 13 at 9:29
ShivamShivam
62
62
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shivam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32
add a comment |
1
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32
1
1
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
add a comment |
$begingroup$
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
add a comment |
$begingroup$
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
But, for any field $K$, every element of $K$ has an additive inverse. That's part of the definition of field.
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
answered Mar 13 at 9:32
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
add a comment |
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
but is it necessary that both $ 1 $ and $ -1 $ exist in the field
$endgroup$
– Shivam
Mar 13 at 9:50
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
$begingroup$
By the definition of field, $1in K$ and, again by the definition of field, every element of $K$ has an additive inverse. In particular, $-1in K$.
$endgroup$
– José Carlos Santos
Mar 13 at 9:54
add a comment |
Shivam is a new contributor. Be nice, and check out our Code of Conduct.
Shivam is a new contributor. Be nice, and check out our Code of Conduct.
Shivam is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
If the field contains $a$ it must also contain $-a$.
$endgroup$
– An.Ditlev
Mar 13 at 9:32