Isomorphisms in the proof of the Fredholm alternative/Theorem of Riesz-Schauder (for compact operators)
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In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following:
Let $X$ be a Banach space, $T:X to X$ be a compact operator and $A:=T-I$. We proved that $mbox{ker}(A)< infty$ and that there is a closed subspace $Vsubset X$ such that $X=Voplus mbox{ker}(A)$. Why does it follow that
$X/V cong mbox{ker}(A)?$
We also showed that $mbox{Im}(A)$ is closed and that $V cong mbox{Im}(A)$ and from this it should follow that
$X/V cong X/mbox{Im}(A)$.
Why is this step true?
Thanks in advance!
functional-analysis proof-explanation compact-operators
asked Dec 29 '18 at 22:17
Max93Max93
30529
$endgroup$
add a comment |
$begingroup$
In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following:
Let $X$ be a Banach space, $T:X to X$ be a compact operator and $A:=T-I$. We proved that $mbox{ker}(A)< infty$ and that there is a closed subspace $Vsubset X$ such that $X=Voplus mbox{ker}(A)$. Why does it follow that
$X/V cong mbox{ker}(A)?$
We also showed that $mbox{Im}(A)$ is closed and that $V cong mbox{Im}(A)$ and from this it should follow that
$X/V cong X/mbox{Im}(A)$.
Why is this step true?
Thanks in advance!
functional-analysis proof-explanation compact-operators
asked Dec 29 '18 at 22:17
Max93Max93
30529
$endgroup$
add a comment |
$begingroup$
In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following:
Let $X$ be a Banach space, $T:X to X$ be a compact operator and $A:=T-I$. We proved that $mbox{ker}(A)< infty$ and that there is a closed subspace $Vsubset X$ such that $X=Voplus mbox{ker}(A)$. Why does it follow that
$X/V cong mbox{ker}(A)?$
We also showed that $mbox{Im}(A)$ is closed and that $V cong mbox{Im}(A)$ and from this it should follow that
$X/V cong X/mbox{Im}(A)$.
Why is this step true?
Thanks in advance!
functional-analysis proof-explanation compact-operators
asked Dec 29 '18 at 22:17
Max93Max93
30529
$endgroup$
In a proof of the Fredholm alternative/Theorem of Riesz-Schauder, I came across the following:
Let $X$ be a Banach space, $T:X to X$ be a compact operator and $A:=T-I$. We proved that $mbox{ker}(A)< infty$ and that there is a closed subspace $Vsubset X$ such that $X=Voplus mbox{ker}(A)$. Why does it follow that
$X/V cong mbox{ker}(A)?$
We also showed that $mbox{Im}(A)$ is closed and that $V cong mbox{Im}(A)$ and from this it should follow that
$X/V cong X/mbox{Im}(A)$.
Why is this step true?
Thanks in advance!
functional-analysis proof-explanation compact-operators
functional-analysis proof-explanation compact-operators
asked Dec 29 '18 at 22:17
Max93Max93
30529
asked Dec 29 '18 at 22:17
Max93Max93
30529
edited Dec 29 '18 at 22:38
Max93
asked Dec 29 '18 at 22:17
Max93Max93
30529
asked Dec 29 '18 at 22:17
Max93Max93
30529
asked Dec 29 '18 at 22:17
Max93Max93
30529
30529
add a comment |
add a comment |
1 Answer
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Consider the linear operators
$$i: ker A to X: x mapsto x$$ and
$$pi: X to X/V: x mapsto [x].$$
$i$ is the inclusion map and $pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y in ker pi$ iff $y in V$ and equivalent $y notin ker A$ as by assumption $X = ker A oplus V$.
The composition $Phi := pi circ i$ is the isomorphism you are looking for: Let $x in ker Phi$. Then $pi (ix) = [0]$ and therefore $ix in V$. Since $X = ker A oplus V$ this means $x = 0$ and hence $Phi$ is injective.
It is also surjective. Let $[x] in X/V$. If $[x] = [0]$ then $Phi(0) = [x]$. If $[x] neq [0]$ then $x notin V$ and therefore $x in ker A$ and $Phi(x) = [x]$.
answered Dec 30 '18 at 13:36
eddieeddie
350110
$endgroup$
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
add a comment |
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1 Answer
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$begingroup$
Consider the linear operators
$$i: ker A to X: x mapsto x$$ and
$$pi: X to X/V: x mapsto [x].$$
$i$ is the inclusion map and $pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y in ker pi$ iff $y in V$ and equivalent $y notin ker A$ as by assumption $X = ker A oplus V$.
The composition $Phi := pi circ i$ is the isomorphism you are looking for: Let $x in ker Phi$. Then $pi (ix) = [0]$ and therefore $ix in V$. Since $X = ker A oplus V$ this means $x = 0$ and hence $Phi$ is injective.
It is also surjective. Let $[x] in X/V$. If $[x] = [0]$ then $Phi(0) = [x]$. If $[x] neq [0]$ then $x notin V$ and therefore $x in ker A$ and $Phi(x) = [x]$.
answered Dec 30 '18 at 13:36
eddieeddie
350110
$endgroup$
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
add a comment |
$begingroup$
Consider the linear operators
$$i: ker A to X: x mapsto x$$ and
$$pi: X to X/V: x mapsto [x].$$
$i$ is the inclusion map and $pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y in ker pi$ iff $y in V$ and equivalent $y notin ker A$ as by assumption $X = ker A oplus V$.
The composition $Phi := pi circ i$ is the isomorphism you are looking for: Let $x in ker Phi$. Then $pi (ix) = [0]$ and therefore $ix in V$. Since $X = ker A oplus V$ this means $x = 0$ and hence $Phi$ is injective.
It is also surjective. Let $[x] in X/V$. If $[x] = [0]$ then $Phi(0) = [x]$. If $[x] neq [0]$ then $x notin V$ and therefore $x in ker A$ and $Phi(x) = [x]$.
answered Dec 30 '18 at 13:36
eddieeddie
350110
$endgroup$
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
add a comment |
$begingroup$
Consider the linear operators
$$i: ker A to X: x mapsto x$$ and
$$pi: X to X/V: x mapsto [x].$$
$i$ is the inclusion map and $pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y in ker pi$ iff $y in V$ and equivalent $y notin ker A$ as by assumption $X = ker A oplus V$.
The composition $Phi := pi circ i$ is the isomorphism you are looking for: Let $x in ker Phi$. Then $pi (ix) = [0]$ and therefore $ix in V$. Since $X = ker A oplus V$ this means $x = 0$ and hence $Phi$ is injective.
It is also surjective. Let $[x] in X/V$. If $[x] = [0]$ then $Phi(0) = [x]$. If $[x] neq [0]$ then $x notin V$ and therefore $x in ker A$ and $Phi(x) = [x]$.
answered Dec 30 '18 at 13:36
eddieeddie
350110
$endgroup$
Consider the linear operators
$$i: ker A to X: x mapsto x$$ and
$$pi: X to X/V: x mapsto [x].$$
$i$ is the inclusion map and $pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y in ker pi$ iff $y in V$ and equivalent $y notin ker A$ as by assumption $X = ker A oplus V$.
The composition $Phi := pi circ i$ is the isomorphism you are looking for: Let $x in ker Phi$. Then $pi (ix) = [0]$ and therefore $ix in V$. Since $X = ker A oplus V$ this means $x = 0$ and hence $Phi$ is injective.
It is also surjective. Let $[x] in X/V$. If $[x] = [0]$ then $Phi(0) = [x]$. If $[x] neq [0]$ then $x notin V$ and therefore $x in ker A$ and $Phi(x) = [x]$.
answered Dec 30 '18 at 13:36
eddieeddie
350110
edited Dec 30 '18 at 16:49
answered Dec 30 '18 at 13:36
eddieeddie
350110
answered Dec 30 '18 at 13:36
eddieeddie
350110
answered Dec 30 '18 at 13:36
eddieeddie
350110
350110
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
add a comment |
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
$begingroup$
Thank you! This proofs $mbox{ker}(A) cong X/V$; what about the other isomorphism? Why do we have $X/mbox{Im}(A) cong X/V$?
$endgroup$
– Max93
Dec 30 '18 at 16:40
add a comment |
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