In Hilbert space $ dim(operatorname{coker}(T))=dim(ker(T^{*}))$
$begingroup$
$H$ is a Hilbert space, and $Tin B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)
$operatorname{ind}(T)=dim(ker(T))-dim(operatorname{coker}(T))$
and another's (when work with Hilbert space)
$operatorname{ind}(T)=dim(ker(T))-dim(ker(T^{*}))$ , $T^{*}$ is the adjoint operator
so how can i show that
$dim(operatorname{coker}(T))=dim(ker(T^{*}))$
$operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
thanks
operator-theory
asked Nov 30 '18 at 22:03
user89940user89940
797
$endgroup$
add a comment |
$begingroup$
$H$ is a Hilbert space, and $Tin B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)
$operatorname{ind}(T)=dim(ker(T))-dim(operatorname{coker}(T))$
and another's (when work with Hilbert space)
$operatorname{ind}(T)=dim(ker(T))-dim(ker(T^{*}))$ , $T^{*}$ is the adjoint operator
so how can i show that
$dim(operatorname{coker}(T))=dim(ker(T^{*}))$
$operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
thanks
operator-theory
asked Nov 30 '18 at 22:03
user89940user89940
797
$endgroup$
$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00
add a comment |
$begingroup$
$H$ is a Hilbert space, and $Tin B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)
$operatorname{ind}(T)=dim(ker(T))-dim(operatorname{coker}(T))$
and another's (when work with Hilbert space)
$operatorname{ind}(T)=dim(ker(T))-dim(ker(T^{*}))$ , $T^{*}$ is the adjoint operator
so how can i show that
$dim(operatorname{coker}(T))=dim(ker(T^{*}))$
$operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
thanks
operator-theory
asked Nov 30 '18 at 22:03
user89940user89940
797
$endgroup$
$H$ is a Hilbert space, and $Tin B(H)$ continuous and Fredholm operator.
Same books in definition of Fredholm use (when work in Banach space)
$operatorname{ind}(T)=dim(ker(T))-dim(operatorname{coker}(T))$
and another's (when work with Hilbert space)
$operatorname{ind}(T)=dim(ker(T))-dim(ker(T^{*}))$ , $T^{*}$ is the adjoint operator
so how can i show that
$dim(operatorname{coker}(T))=dim(ker(T^{*}))$
$operatorname{coker}(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
thanks
operator-theory
operator-theory
asked Nov 30 '18 at 22:03
user89940user89940
797
asked Nov 30 '18 at 22:03
user89940user89940
797
edited Dec 1 '18 at 0:12
user89940
asked Nov 30 '18 at 22:03
user89940user89940
797
asked Nov 30 '18 at 22:03
user89940user89940
797
asked Nov 30 '18 at 22:03
user89940user89940
797
797
$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00
add a comment |
$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00
$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For any closed subspace $K$ of a Hilbert space $H$, we have
$$H/Kcong K^perp.$$
Proof.
$newcommand{co}{operatorname{coker}}$Let begin{align*}Phi: & K^perpto H/K\
& hmapsto [h]end{align*}
(1) Of course, $Phi$ is linear bounded.
(2) $Phi$ is injective. Suppose $hin K^perp$ such that $Phi(h)=0$, then $hin K^perpcap K$, so $h=0$.
(3) $Phi$ is surjective. For any $hin H$, there is some $h'in K^perp$ such that $h-h'in K$, thus $Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/THcong (TH)^perp$.
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
$endgroup$
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
|
show 4 more comments
Your Answer
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any closed subspace $K$ of a Hilbert space $H$, we have
$$H/Kcong K^perp.$$
Proof.
$newcommand{co}{operatorname{coker}}$Let begin{align*}Phi: & K^perpto H/K\
& hmapsto [h]end{align*}
(1) Of course, $Phi$ is linear bounded.
(2) $Phi$ is injective. Suppose $hin K^perp$ such that $Phi(h)=0$, then $hin K^perpcap K$, so $h=0$.
(3) $Phi$ is surjective. For any $hin H$, there is some $h'in K^perp$ such that $h-h'in K$, thus $Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/THcong (TH)^perp$.
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
$endgroup$
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
|
show 4 more comments
$begingroup$
For any closed subspace $K$ of a Hilbert space $H$, we have
$$H/Kcong K^perp.$$
Proof.
$newcommand{co}{operatorname{coker}}$Let begin{align*}Phi: & K^perpto H/K\
& hmapsto [h]end{align*}
(1) Of course, $Phi$ is linear bounded.
(2) $Phi$ is injective. Suppose $hin K^perp$ such that $Phi(h)=0$, then $hin K^perpcap K$, so $h=0$.
(3) $Phi$ is surjective. For any $hin H$, there is some $h'in K^perp$ such that $h-h'in K$, thus $Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/THcong (TH)^perp$.
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
$endgroup$
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
|
show 4 more comments
$begingroup$
For any closed subspace $K$ of a Hilbert space $H$, we have
$$H/Kcong K^perp.$$
Proof.
$newcommand{co}{operatorname{coker}}$Let begin{align*}Phi: & K^perpto H/K\
& hmapsto [h]end{align*}
(1) Of course, $Phi$ is linear bounded.
(2) $Phi$ is injective. Suppose $hin K^perp$ such that $Phi(h)=0$, then $hin K^perpcap K$, so $h=0$.
(3) $Phi$ is surjective. For any $hin H$, there is some $h'in K^perp$ such that $h-h'in K$, thus $Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/THcong (TH)^perp$.
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
$endgroup$
For any closed subspace $K$ of a Hilbert space $H$, we have
$$H/Kcong K^perp.$$
Proof.
$newcommand{co}{operatorname{coker}}$Let begin{align*}Phi: & K^perpto H/K\
& hmapsto [h]end{align*}
(1) Of course, $Phi$ is linear bounded.
(2) $Phi$ is injective. Suppose $hin K^perp$ such that $Phi(h)=0$, then $hin K^perpcap K$, so $h=0$.
(3) $Phi$ is surjective. For any $hin H$, there is some $h'in K^perp$ such that $h-h'in K$, thus $Phi(h')=[h']=[h]$.
Noting that $TH$ is closed subspace in the question, $H/THcong (TH)^perp$.
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
edited Dec 1 '18 at 4:22
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
answered Dec 1 '18 at 2:31
C.DingC.Ding
1,3911321
1,3911321
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
|
show 4 more comments
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
1
1
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed
$endgroup$
– user89940
Dec 1 '18 at 3:10
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
For any subset $S$ of $H$, $H/overline{span{S}}=S^perp$. The very usage of $T(H)$ is closed is $T(H)=overline{span{T(H)}}$
$endgroup$
– C.Ding
Dec 1 '18 at 4:13
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
A better answer has been updated.
$endgroup$
– C.Ding
Dec 1 '18 at 4:23
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
$begingroup$
the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/…
$endgroup$
– user89940
Dec 1 '18 at 16:15
1
1
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
$begingroup$
Yes. If you define coker of an operator $u:Xto Y$ between Banach spaces as $ker u^#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version.
$endgroup$
– C.Ding
Dec 5 '18 at 7:53
|
show 4 more comments
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$begingroup$
you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem?
$endgroup$
– Picaud Vincent
Nov 30 '18 at 23:16
$begingroup$
@PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal
$endgroup$
– user89940
Nov 30 '18 at 23:18
$begingroup$
$coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{perp}$. so in Hilbert is true $H/T(H)approx (T(H))^{perp}$ ?
$endgroup$
– user89940
Dec 1 '18 at 0:00