Is the requirement that $u+v$ be in $V$ if $u$ and $v$ are in $V$ a valid axiom for the definition of a...

Multi tool use
$begingroup$
For example, at: https://en.wikipedia.org/wiki/Vector_space
There are 8 axioms that a qualify a set to be a vector space. My professor also gave us 8.
However, a textbook I'm reading states 10 axioms. It has all 8 of the above mentioned axioms, as well as the following two:
If $u$ and $v$ are in $V$, then $u+v$ is also in $V$
If $u$ is in $V$, then $cu$ is in V (where $c$ is an arbritrary scalar)
Why is it that these two axioms aren't explicitly stated in some places? As far as I can't tell it's not that they're included within the 8 existing axioms. I know that these are the requirements for something to qualify as a subspace of a vector space...does that have anything to do with it?
I'd just like to know why these two axioms aren't explicitly stated as they seem very important and don't seem to be indirectly covered by any of the other axioms. Any help is appreciated!
linear-algebra vector-spaces vectors axioms
$endgroup$
add a comment |
$begingroup$
For example, at: https://en.wikipedia.org/wiki/Vector_space
There are 8 axioms that a qualify a set to be a vector space. My professor also gave us 8.
However, a textbook I'm reading states 10 axioms. It has all 8 of the above mentioned axioms, as well as the following two:
If $u$ and $v$ are in $V$, then $u+v$ is also in $V$
If $u$ is in $V$, then $cu$ is in V (where $c$ is an arbritrary scalar)
Why is it that these two axioms aren't explicitly stated in some places? As far as I can't tell it's not that they're included within the 8 existing axioms. I know that these are the requirements for something to qualify as a subspace of a vector space...does that have anything to do with it?
I'd just like to know why these two axioms aren't explicitly stated as they seem very important and don't seem to be indirectly covered by any of the other axioms. Any help is appreciated!
linear-algebra vector-spaces vectors axioms
$endgroup$
add a comment |
$begingroup$
For example, at: https://en.wikipedia.org/wiki/Vector_space
There are 8 axioms that a qualify a set to be a vector space. My professor also gave us 8.
However, a textbook I'm reading states 10 axioms. It has all 8 of the above mentioned axioms, as well as the following two:
If $u$ and $v$ are in $V$, then $u+v$ is also in $V$
If $u$ is in $V$, then $cu$ is in V (where $c$ is an arbritrary scalar)
Why is it that these two axioms aren't explicitly stated in some places? As far as I can't tell it's not that they're included within the 8 existing axioms. I know that these are the requirements for something to qualify as a subspace of a vector space...does that have anything to do with it?
I'd just like to know why these two axioms aren't explicitly stated as they seem very important and don't seem to be indirectly covered by any of the other axioms. Any help is appreciated!
linear-algebra vector-spaces vectors axioms
$endgroup$
For example, at: https://en.wikipedia.org/wiki/Vector_space
There are 8 axioms that a qualify a set to be a vector space. My professor also gave us 8.
However, a textbook I'm reading states 10 axioms. It has all 8 of the above mentioned axioms, as well as the following two:
If $u$ and $v$ are in $V$, then $u+v$ is also in $V$
If $u$ is in $V$, then $cu$ is in V (where $c$ is an arbritrary scalar)
Why is it that these two axioms aren't explicitly stated in some places? As far as I can't tell it's not that they're included within the 8 existing axioms. I know that these are the requirements for something to qualify as a subspace of a vector space...does that have anything to do with it?
I'd just like to know why these two axioms aren't explicitly stated as they seem very important and don't seem to be indirectly covered by any of the other axioms. Any help is appreciated!
linear-algebra vector-spaces vectors axioms
linear-algebra vector-spaces vectors axioms
edited Jan 17 at 18:16


José Carlos Santos
160k22127232
160k22127232
asked Jan 17 at 18:05
James RonaldJames Ronald
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3 Answers
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$begingroup$
Note that Wikipedia also lists those two axioms of closure under addition and scalar multiplication. They just don't list them among the axioms. See the below quote from the article you linked:
A vector space over a field $F$ is a set $V$ together with two operations that satisfy the eight axioms listed below.
- The first operation, called vector addition or simply addition $+ : V × V → V$, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set $V$).
- The second operation, called scalar multiplication $· : F × V → V$, takes any scalar $a$ and any vector $v$ and gives another vector $av$. (Similarly, the vector $av$ is an element of the set $V$ ).
$endgroup$
add a comment |
$begingroup$
The assertion $(forall u,vin V):u+vin V$ is implicit in the assertion that addition is an operation, since an operation in $V$ is a map from $Vtimes V$ into $V$.
$endgroup$
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
add a comment |
$begingroup$
In the wikipedia article, addition and multiplication are defined as maps
begin{align}
+&:Vtimes Vto V, ;(v,w)mapsto v+w\
cdot&:Ftimes Vto V, ;(lambda,v)mapsto lambdacdot v
end{align}
which, by definition imply closure (since the codomain of the maps is $V$).
A question that comes to my mind regarding the 10 axiom treatment: How are addition and scalar multiplication defined that closure is not immediate?
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that Wikipedia also lists those two axioms of closure under addition and scalar multiplication. They just don't list them among the axioms. See the below quote from the article you linked:
A vector space over a field $F$ is a set $V$ together with two operations that satisfy the eight axioms listed below.
- The first operation, called vector addition or simply addition $+ : V × V → V$, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set $V$).
- The second operation, called scalar multiplication $· : F × V → V$, takes any scalar $a$ and any vector $v$ and gives another vector $av$. (Similarly, the vector $av$ is an element of the set $V$ ).
$endgroup$
add a comment |
$begingroup$
Note that Wikipedia also lists those two axioms of closure under addition and scalar multiplication. They just don't list them among the axioms. See the below quote from the article you linked:
A vector space over a field $F$ is a set $V$ together with two operations that satisfy the eight axioms listed below.
- The first operation, called vector addition or simply addition $+ : V × V → V$, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set $V$).
- The second operation, called scalar multiplication $· : F × V → V$, takes any scalar $a$ and any vector $v$ and gives another vector $av$. (Similarly, the vector $av$ is an element of the set $V$ ).
$endgroup$
add a comment |
$begingroup$
Note that Wikipedia also lists those two axioms of closure under addition and scalar multiplication. They just don't list them among the axioms. See the below quote from the article you linked:
A vector space over a field $F$ is a set $V$ together with two operations that satisfy the eight axioms listed below.
- The first operation, called vector addition or simply addition $+ : V × V → V$, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set $V$).
- The second operation, called scalar multiplication $· : F × V → V$, takes any scalar $a$ and any vector $v$ and gives another vector $av$. (Similarly, the vector $av$ is an element of the set $V$ ).
$endgroup$
Note that Wikipedia also lists those two axioms of closure under addition and scalar multiplication. They just don't list them among the axioms. See the below quote from the article you linked:
A vector space over a field $F$ is a set $V$ together with two operations that satisfy the eight axioms listed below.
- The first operation, called vector addition or simply addition $+ : V × V → V$, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. (Note that the resultant vector is also an element of the set $V$).
- The second operation, called scalar multiplication $· : F × V → V$, takes any scalar $a$ and any vector $v$ and gives another vector $av$. (Similarly, the vector $av$ is an element of the set $V$ ).
answered Jan 17 at 18:11


ArthurArthur
115k7116198
115k7116198
add a comment |
add a comment |
$begingroup$
The assertion $(forall u,vin V):u+vin V$ is implicit in the assertion that addition is an operation, since an operation in $V$ is a map from $Vtimes V$ into $V$.
$endgroup$
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
add a comment |
$begingroup$
The assertion $(forall u,vin V):u+vin V$ is implicit in the assertion that addition is an operation, since an operation in $V$ is a map from $Vtimes V$ into $V$.
$endgroup$
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
add a comment |
$begingroup$
The assertion $(forall u,vin V):u+vin V$ is implicit in the assertion that addition is an operation, since an operation in $V$ is a map from $Vtimes V$ into $V$.
$endgroup$
The assertion $(forall u,vin V):u+vin V$ is implicit in the assertion that addition is an operation, since an operation in $V$ is a map from $Vtimes V$ into $V$.
answered Jan 17 at 18:11


José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
add a comment |
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Ahh I see, thank you! I'm guessing that, in the same way, the fact that the vector space must be closed under scalar multiplication is implicit in the fact that scalar multiplication is an operation?
$endgroup$
– James Ronald
Jan 17 at 21:11
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
$begingroup$
Yes, it is the same thing.
$endgroup$
– José Carlos Santos
Jan 17 at 21:15
add a comment |
$begingroup$
In the wikipedia article, addition and multiplication are defined as maps
begin{align}
+&:Vtimes Vto V, ;(v,w)mapsto v+w\
cdot&:Ftimes Vto V, ;(lambda,v)mapsto lambdacdot v
end{align}
which, by definition imply closure (since the codomain of the maps is $V$).
A question that comes to my mind regarding the 10 axiom treatment: How are addition and scalar multiplication defined that closure is not immediate?
$endgroup$
add a comment |
$begingroup$
In the wikipedia article, addition and multiplication are defined as maps
begin{align}
+&:Vtimes Vto V, ;(v,w)mapsto v+w\
cdot&:Ftimes Vto V, ;(lambda,v)mapsto lambdacdot v
end{align}
which, by definition imply closure (since the codomain of the maps is $V$).
A question that comes to my mind regarding the 10 axiom treatment: How are addition and scalar multiplication defined that closure is not immediate?
$endgroup$
add a comment |
$begingroup$
In the wikipedia article, addition and multiplication are defined as maps
begin{align}
+&:Vtimes Vto V, ;(v,w)mapsto v+w\
cdot&:Ftimes Vto V, ;(lambda,v)mapsto lambdacdot v
end{align}
which, by definition imply closure (since the codomain of the maps is $V$).
A question that comes to my mind regarding the 10 axiom treatment: How are addition and scalar multiplication defined that closure is not immediate?
$endgroup$
In the wikipedia article, addition and multiplication are defined as maps
begin{align}
+&:Vtimes Vto V, ;(v,w)mapsto v+w\
cdot&:Ftimes Vto V, ;(lambda,v)mapsto lambdacdot v
end{align}
which, by definition imply closure (since the codomain of the maps is $V$).
A question that comes to my mind regarding the 10 axiom treatment: How are addition and scalar multiplication defined that closure is not immediate?
answered Jan 17 at 18:11
David HillDavid Hill
9,0761619
9,0761619
add a comment |
add a comment |
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