Mixing
$X = C[0,1]$, $Y = C[a,b]$, $T: X to Y$ isomorphism, can it be injective?
$X = C[0,1]$, $Y = C[a,b]$, $T: X to Y$ isomorphism, can it be injective?
This isomorphism is just a change of variables of the form:
$$ f(t) to f(at+b)$$
It is surjective and it satisfies:
$$ Vert Tx Vert = Vert x Vert $$
I need to say if it can also be injective.
I do not think it can, so I want to prove it by contradiction:
If it was then T would be an homomorphism, I want to say something regarding the two spaces but I am struggling to find the contradiction.
functional-analysis
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
add a comment |
$X = C[0,1]$, $Y = C[a,b]$, $T: X to Y$ isomorphism, can it be injective?
This isomorphism is just a change of variables of the form:
$$ f(t) to f(at+b)$$
It is surjective and it satisfies:
$$ Vert Tx Vert = Vert x Vert $$
I need to say if it can also be injective.
I do not think it can, so I want to prove it by contradiction:
If it was then T would be an homomorphism, I want to say something regarding the two spaces but I am struggling to find the contradiction.
functional-analysis
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
1
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39
add a comment |
$X = C[0,1]$, $Y = C[a,b]$, $T: X to Y$ isomorphism, can it be injective?
This isomorphism is just a change of variables of the form:
$$ f(t) to f(at+b)$$
It is surjective and it satisfies:
$$ Vert Tx Vert = Vert x Vert $$
I need to say if it can also be injective.
I do not think it can, so I want to prove it by contradiction:
If it was then T would be an homomorphism, I want to say something regarding the two spaces but I am struggling to find the contradiction.
functional-analysis
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
$X = C[0,1]$, $Y = C[a,b]$, $T: X to Y$ isomorphism, can it be injective?
This isomorphism is just a change of variables of the form:
$$ f(t) to f(at+b)$$
It is surjective and it satisfies:
$$ Vert Tx Vert = Vert x Vert $$
I need to say if it can also be injective.
I do not think it can, so I want to prove it by contradiction:
If it was then T would be an homomorphism, I want to say something regarding the two spaces but I am struggling to find the contradiction.
functional-analysis
functional-analysis
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
edited Nov 20 '18 at 20:42
Batominovski
33.8k33292
33.8k33292
asked Nov 20 '18 at 20:34
qcc101
458113
asked Nov 20 '18 at 20:34
qcc101
458113
asked Nov 20 '18 at 20:34
qcc101
458113
458113
1
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39
add a comment |
1
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39
1
1
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39
add a comment |
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1
It is a linear map with trivial kernel...indeed $F[f](x)=f(g(x))$ is always injective when $g$ is.
– Ian
Nov 20 '18 at 20:35
I do not think I am following you.
– qcc101
Nov 20 '18 at 20:38
A linear mapping between vector spaces is injective if and only if its kernel contains only the zero vector of its domain.
– Ian
Nov 20 '18 at 20:39