Equivalence of injectivity of a lineal operator $T$ between two Banach spaces and and existence of another...
Given $X$ and $Y$ two Banach spaces and $T in L(X,Y)$, where $L(X,Y)$ is the set of all the linear operators of $X$ in $Y$, show that the following sentences are equivalent:
a) There exists $S in L(X,Y)$ such that $S(T(x))=x , , forall x in X$.
b) $T$ is injective and $T(X)$ is a complemented space of $Y$.
functional-analysis linear-transformations banach-spaces
asked yesterday
Maggie94
854
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Given $X$ and $Y$ two Banach spaces and $T in L(X,Y)$, where $L(X,Y)$ is the set of all the linear operators of $X$ in $Y$, show that the following sentences are equivalent:
a) There exists $S in L(X,Y)$ such that $S(T(x))=x , , forall x in X$.
b) $T$ is injective and $T(X)$ is a complemented space of $Y$.
functional-analysis linear-transformations banach-spaces
asked yesterday
Maggie94
854
1
What where your attempts in proving this?
– James
yesterday
add a comment |
Given $X$ and $Y$ two Banach spaces and $T in L(X,Y)$, where $L(X,Y)$ is the set of all the linear operators of $X$ in $Y$, show that the following sentences are equivalent:
a) There exists $S in L(X,Y)$ such that $S(T(x))=x , , forall x in X$.
b) $T$ is injective and $T(X)$ is a complemented space of $Y$.
functional-analysis linear-transformations banach-spaces
asked yesterday
Maggie94
854
Given $X$ and $Y$ two Banach spaces and $T in L(X,Y)$, where $L(X,Y)$ is the set of all the linear operators of $X$ in $Y$, show that the following sentences are equivalent:
a) There exists $S in L(X,Y)$ such that $S(T(x))=x , , forall x in X$.
b) $T$ is injective and $T(X)$ is a complemented space of $Y$.
functional-analysis linear-transformations banach-spaces
functional-analysis linear-transformations banach-spaces
asked yesterday
Maggie94
854
asked yesterday
Maggie94
854
asked yesterday
Maggie94
854
asked yesterday
Maggie94
854
asked yesterday
Maggie94
854
854
1
What where your attempts in proving this?
– James
yesterday
add a comment |
1
What where your attempts in proving this?
– James
yesterday
1
1
What where your attempts in proving this?
– James
yesterday
What where your attempts in proving this?
– James
yesterday
add a comment |
1 Answer
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Your $S$ should be a linear operator on $Y$, otherwise the composition makes no sense. The direction a) to b) is simple, just work with the definition of an injective map -- you don't need anything else. If you can't manage, I will add a proof here but at least try to solve it yourself.
For b) implies a), let ${x_imid iin I}$ be a basis for $X$. By infectivity of $T$, ${T(x_i)mid iin I}$ forms a basis of the image of $T$ -- convince yourself that this is true and let me know if you can't. Now define $$S:Yto X, smapstobegin{cases}T^{-1}(s),& if;sinmathrm{img}(T),\ 0,&else.end{cases}$$ You can then check by writing $s=sum alpha_ix_i$ that $S$ is as you want it to. Can you fill the missing details yourself?
answered yesterday
James
703115
add a comment |
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1 Answer
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Your $S$ should be a linear operator on $Y$, otherwise the composition makes no sense. The direction a) to b) is simple, just work with the definition of an injective map -- you don't need anything else. If you can't manage, I will add a proof here but at least try to solve it yourself.
For b) implies a), let ${x_imid iin I}$ be a basis for $X$. By infectivity of $T$, ${T(x_i)mid iin I}$ forms a basis of the image of $T$ -- convince yourself that this is true and let me know if you can't. Now define $$S:Yto X, smapstobegin{cases}T^{-1}(s),& if;sinmathrm{img}(T),\ 0,&else.end{cases}$$ You can then check by writing $s=sum alpha_ix_i$ that $S$ is as you want it to. Can you fill the missing details yourself?
answered yesterday
James
703115
add a comment |
Your $S$ should be a linear operator on $Y$, otherwise the composition makes no sense. The direction a) to b) is simple, just work with the definition of an injective map -- you don't need anything else. If you can't manage, I will add a proof here but at least try to solve it yourself.
For b) implies a), let ${x_imid iin I}$ be a basis for $X$. By infectivity of $T$, ${T(x_i)mid iin I}$ forms a basis of the image of $T$ -- convince yourself that this is true and let me know if you can't. Now define $$S:Yto X, smapstobegin{cases}T^{-1}(s),& if;sinmathrm{img}(T),\ 0,&else.end{cases}$$ You can then check by writing $s=sum alpha_ix_i$ that $S$ is as you want it to. Can you fill the missing details yourself?
answered yesterday
James
703115
add a comment |
Your $S$ should be a linear operator on $Y$, otherwise the composition makes no sense. The direction a) to b) is simple, just work with the definition of an injective map -- you don't need anything else. If you can't manage, I will add a proof here but at least try to solve it yourself.
For b) implies a), let ${x_imid iin I}$ be a basis for $X$. By infectivity of $T$, ${T(x_i)mid iin I}$ forms a basis of the image of $T$ -- convince yourself that this is true and let me know if you can't. Now define $$S:Yto X, smapstobegin{cases}T^{-1}(s),& if;sinmathrm{img}(T),\ 0,&else.end{cases}$$ You can then check by writing $s=sum alpha_ix_i$ that $S$ is as you want it to. Can you fill the missing details yourself?
answered yesterday
James
703115
Your $S$ should be a linear operator on $Y$, otherwise the composition makes no sense. The direction a) to b) is simple, just work with the definition of an injective map -- you don't need anything else. If you can't manage, I will add a proof here but at least try to solve it yourself.
For b) implies a), let ${x_imid iin I}$ be a basis for $X$. By infectivity of $T$, ${T(x_i)mid iin I}$ forms a basis of the image of $T$ -- convince yourself that this is true and let me know if you can't. Now define $$S:Yto X, smapstobegin{cases}T^{-1}(s),& if;sinmathrm{img}(T),\ 0,&else.end{cases}$$ You can then check by writing $s=sum alpha_ix_i$ that $S$ is as you want it to. Can you fill the missing details yourself?
answered yesterday
James
703115
answered yesterday
James
703115
answered yesterday
James
703115
answered yesterday
James
703115
703115
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1
What where your attempts in proving this?
– James
yesterday