Mixing
How to prove that a vector space homomorphism $phi$ is injective if and only if $ker (phi)={0}$?
$begingroup$
A vector space homomorphism $$phi: Vrightarrow W$$ is a map between two vector spaces which satisfies the rule
$$phi(lambda v + mu v')=lambdaphi(v)+muphi(v').$$
Injectivity means
$$forall {x,x'in V}phi(x)=phi(x')Rightarrow x=x'.$$
I have already proved the statement for group homomorphisms.
And $phi$ is a group homomorphism under addition, i.e.,
$$phi(v+w)=phi(v)+phi(w).$$
How can I make the transition to vector spaces; I hope it is clear what I am asking for.
I am confused because there are two operations under $phi$: the scalar multiplication and the addition.
linear-algebra
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
$endgroup$
add a comment |
$begingroup$
A vector space homomorphism $$phi: Vrightarrow W$$ is a map between two vector spaces which satisfies the rule
$$phi(lambda v + mu v')=lambdaphi(v)+muphi(v').$$
Injectivity means
$$forall {x,x'in V}phi(x)=phi(x')Rightarrow x=x'.$$
I have already proved the statement for group homomorphisms.
And $phi$ is a group homomorphism under addition, i.e.,
$$phi(v+w)=phi(v)+phi(w).$$
How can I make the transition to vector spaces; I hope it is clear what I am asking for.
I am confused because there are two operations under $phi$: the scalar multiplication and the addition.
linear-algebra
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
$endgroup$
1
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45
add a comment |
$begingroup$
A vector space homomorphism $$phi: Vrightarrow W$$ is a map between two vector spaces which satisfies the rule
$$phi(lambda v + mu v')=lambdaphi(v)+muphi(v').$$
Injectivity means
$$forall {x,x'in V}phi(x)=phi(x')Rightarrow x=x'.$$
I have already proved the statement for group homomorphisms.
And $phi$ is a group homomorphism under addition, i.e.,
$$phi(v+w)=phi(v)+phi(w).$$
How can I make the transition to vector spaces; I hope it is clear what I am asking for.
I am confused because there are two operations under $phi$: the scalar multiplication and the addition.
linear-algebra
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
$endgroup$
A vector space homomorphism $$phi: Vrightarrow W$$ is a map between two vector spaces which satisfies the rule
$$phi(lambda v + mu v')=lambdaphi(v)+muphi(v').$$
Injectivity means
$$forall {x,x'in V}phi(x)=phi(x')Rightarrow x=x'.$$
I have already proved the statement for group homomorphisms.
And $phi$ is a group homomorphism under addition, i.e.,
$$phi(v+w)=phi(v)+phi(w).$$
How can I make the transition to vector spaces; I hope it is clear what I am asking for.
I am confused because there are two operations under $phi$: the scalar multiplication and the addition.
linear-algebra
linear-algebra
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
edited Jan 30 at 19:15
J. W. Tanner
4,4691320
4,4691320
asked Jan 30 at 18:56
New2MathNew2Math
16715
asked Jan 30 at 18:56
New2MathNew2Math
16715
asked Jan 30 at 18:56
New2MathNew2Math
16715
16715
1
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45
add a comment |
1
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45
1
1
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The proof is the same. A vector space is a group wrt the internal operation (i.e. $(V,+)$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
answered Jan 30 at 19:04
HarnakHarnak
1,369512
$endgroup$
add a comment |
$begingroup$
The proof should look very similar to the one for group homomorphisms.
If $phi$ is injective, then $phi(x)=0=phi(0)$ implies that $x=0$, so the kernel is ${0}$. Conversely, suppose the kernel is ${0}$ and that $phi(x)=phi(y)$.
Then $phi(x)-phi(y)=0$. Using homomorphism property, this means $phi(x-y)=0$ so that $x-y$ is in the kernel, which only consists of $0$. Thus $x-y=0$, or $x=y$, proving injectivity.
answered Jan 30 at 19:03
pwerthpwerth
3,310417
$endgroup$
add a comment |
$begingroup$
As you observed, if you take in your definition of $phi$, $lambda=mu=1$, it follows that, in particular:
$$phi(x+y)=phi(x)+phi(y), forall x,yin V.$$
Therefore $phi:Vto W$ is, in particular, a morphism between the groups $V$ and $W$. Applying your result for groups, it follows that $ker(phi)$ (seen as the kernel of the group morphism is $0$).
The external operation does not intervene in this statement.
answered Jan 30 at 21:41
IuliaIulia
1,025415
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
The proof is the same. A vector space is a group wrt the internal operation (i.e. $(V,+)$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
answered Jan 30 at 19:04
HarnakHarnak
1,369512
$endgroup$
add a comment |
$begingroup$
The proof is the same. A vector space is a group wrt the internal operation (i.e. $(V,+)$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
answered Jan 30 at 19:04
HarnakHarnak
1,369512
$endgroup$
add a comment |
$begingroup$
The proof is the same. A vector space is a group wrt the internal operation (i.e. $(V,+)$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
answered Jan 30 at 19:04
HarnakHarnak
1,369512
$endgroup$
The proof is the same. A vector space is a group wrt the internal operation (i.e. $(V,+)$ is a group) and a vector space homomorphism is, in particular, a group homomorphism.
answered Jan 30 at 19:04
HarnakHarnak
1,369512
edited Jan 31 at 9:36
answered Jan 30 at 19:04
HarnakHarnak
1,369512
answered Jan 30 at 19:04
HarnakHarnak
1,369512
answered Jan 30 at 19:04
HarnakHarnak
1,369512
1,369512
add a comment |
add a comment |
$begingroup$
The proof should look very similar to the one for group homomorphisms.
If $phi$ is injective, then $phi(x)=0=phi(0)$ implies that $x=0$, so the kernel is ${0}$. Conversely, suppose the kernel is ${0}$ and that $phi(x)=phi(y)$.
Then $phi(x)-phi(y)=0$. Using homomorphism property, this means $phi(x-y)=0$ so that $x-y$ is in the kernel, which only consists of $0$. Thus $x-y=0$, or $x=y$, proving injectivity.
answered Jan 30 at 19:03
pwerthpwerth
3,310417
$endgroup$
add a comment |
$begingroup$
The proof should look very similar to the one for group homomorphisms.
If $phi$ is injective, then $phi(x)=0=phi(0)$ implies that $x=0$, so the kernel is ${0}$. Conversely, suppose the kernel is ${0}$ and that $phi(x)=phi(y)$.
Then $phi(x)-phi(y)=0$. Using homomorphism property, this means $phi(x-y)=0$ so that $x-y$ is in the kernel, which only consists of $0$. Thus $x-y=0$, or $x=y$, proving injectivity.
answered Jan 30 at 19:03
pwerthpwerth
3,310417
$endgroup$
add a comment |
$begingroup$
The proof should look very similar to the one for group homomorphisms.
If $phi$ is injective, then $phi(x)=0=phi(0)$ implies that $x=0$, so the kernel is ${0}$. Conversely, suppose the kernel is ${0}$ and that $phi(x)=phi(y)$.
Then $phi(x)-phi(y)=0$. Using homomorphism property, this means $phi(x-y)=0$ so that $x-y$ is in the kernel, which only consists of $0$. Thus $x-y=0$, or $x=y$, proving injectivity.
answered Jan 30 at 19:03
pwerthpwerth
3,310417
$endgroup$
The proof should look very similar to the one for group homomorphisms.
If $phi$ is injective, then $phi(x)=0=phi(0)$ implies that $x=0$, so the kernel is ${0}$. Conversely, suppose the kernel is ${0}$ and that $phi(x)=phi(y)$.
Then $phi(x)-phi(y)=0$. Using homomorphism property, this means $phi(x-y)=0$ so that $x-y$ is in the kernel, which only consists of $0$. Thus $x-y=0$, or $x=y$, proving injectivity.
answered Jan 30 at 19:03
pwerthpwerth
3,310417
answered Jan 30 at 19:03
pwerthpwerth
3,310417
answered Jan 30 at 19:03
pwerthpwerth
3,310417
answered Jan 30 at 19:03
pwerthpwerth
3,310417
3,310417
add a comment |
add a comment |
$begingroup$
As you observed, if you take in your definition of $phi$, $lambda=mu=1$, it follows that, in particular:
$$phi(x+y)=phi(x)+phi(y), forall x,yin V.$$
Therefore $phi:Vto W$ is, in particular, a morphism between the groups $V$ and $W$. Applying your result for groups, it follows that $ker(phi)$ (seen as the kernel of the group morphism is $0$).
The external operation does not intervene in this statement.
answered Jan 30 at 21:41
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
As you observed, if you take in your definition of $phi$, $lambda=mu=1$, it follows that, in particular:
$$phi(x+y)=phi(x)+phi(y), forall x,yin V.$$
Therefore $phi:Vto W$ is, in particular, a morphism between the groups $V$ and $W$. Applying your result for groups, it follows that $ker(phi)$ (seen as the kernel of the group morphism is $0$).
The external operation does not intervene in this statement.
answered Jan 30 at 21:41
IuliaIulia
1,025415
$endgroup$
add a comment |
$begingroup$
As you observed, if you take in your definition of $phi$, $lambda=mu=1$, it follows that, in particular:
$$phi(x+y)=phi(x)+phi(y), forall x,yin V.$$
Therefore $phi:Vto W$ is, in particular, a morphism between the groups $V$ and $W$. Applying your result for groups, it follows that $ker(phi)$ (seen as the kernel of the group morphism is $0$).
The external operation does not intervene in this statement.
answered Jan 30 at 21:41
IuliaIulia
1,025415
$endgroup$
As you observed, if you take in your definition of $phi$, $lambda=mu=1$, it follows that, in particular:
$$phi(x+y)=phi(x)+phi(y), forall x,yin V.$$
Therefore $phi:Vto W$ is, in particular, a morphism between the groups $V$ and $W$. Applying your result for groups, it follows that $ker(phi)$ (seen as the kernel of the group morphism is $0$).
The external operation does not intervene in this statement.
answered Jan 30 at 21:41
IuliaIulia
1,025415
answered Jan 30 at 21:41
IuliaIulia
1,025415
answered Jan 30 at 21:41
IuliaIulia
1,025415
answered Jan 30 at 21:41
IuliaIulia
1,025415
1,025415
add a comment |
add a comment |
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1
$begingroup$
The definition of the kernel is the same in both cases, though.
$endgroup$
– Randall
Jan 30 at 18:58
$begingroup$
A vector space is an abelian group (plus an additional structure)., so already proved.
$endgroup$
– Bernard
Jan 30 at 19:45