In Hilbert space $ dim(operatorname{coker}(T))=dim(ker(T^{*}))$












0












$begingroup$


$H$ 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


$H$ 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


$H$ 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







share|cite|improve this question











$endgroup$




$H$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 0:12







user89940

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















1












$begingroup$


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, $H/THcong (TH)^perp$.






share|cite|improve this answer











$endgroup$









  • 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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$


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, $H/THcong (TH)^perp$.






share|cite|improve this answer











$endgroup$









  • 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
















1












$begingroup$


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, $H/THcong (TH)^perp$.






share|cite|improve this answer











$endgroup$









  • 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














1












1








1





$begingroup$


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, $H/THcong (TH)^perp$.






share|cite|improve this answer











$endgroup$




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, $H/THcong (TH)^perp$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 4:22

























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














  • 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


















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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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