Prove $Tx=x$, for $xin H$, if and only if $(Tx,x)=|x|^2$ and $ker(I-T)=ker(I-T^*)$
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
Let $H$ be a complex Hilbert space and $T:Hrightarrow H$ an operation such that $|T|leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=|x|^2$
$ker(I-T)=ker(I-T^*)$.
My attempt
1. $(Tx,x)=(x,x)=|x|^2$ if $Tx=x$. Conversely, WTS $|Tx-x|=0$ $forall xin H$.
We get $|Tx-x|=(Tx-x,Tx-x)=|Tx|^2-(x,Tx)$. But $|T|leq 1$ has not been used.
Please how do I proceed?
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
linear-algebra functional-analysis operator-theory hilbert-spaces adjoint-operators
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
edited Jan 6 at 16:36
Muhammad Mubarak
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 6 at 15:17
Muhammad MubarakMuhammad Mubarak
9810
9810
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063982%2fprove-tx-x-for-x-in-h-if-and-only-if-tx-x-x-2-and-keri-t-ker%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
$endgroup$
add a comment |
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
$endgroup$
add a comment |
$begingroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
$endgroup$
For the first point, first off since $left langle Tx,xrightrangle=|x|^2$ you also have $left langle x,Txrightrangle = overline{|x|^2}=|x|^2$. Thus
$$0 leqleftlangle Tx-x,Tx-xrightrangle =|Tx|^2+|x|^2-leftlangle Tx, xrightrangle-leftlangle x,Txrightrangle= |Tx|^2-|x|^2 $$
Since $|T|leq 1$,
$$0 leq |Tx|^2-|x|^2 leq |x|^2-|x|^2=0 $$
and hence $Tx=x$.
For the second point, using the first point
begin{align*}xin ker(I-T)iff Tx=x iff left langle Tx,xrightrangle =|x|^2end{align*}
Now observe that if $|T|leq 1$ then also $|T^*|leq 1$. Indeed,
begin{align*}|T^*|&=sup_{xneq 0}frac{|T^*x|}{|x|}=sup_{xneq 0}frac{leftlangle T^*x,T^*xrightrangle^{1/2}}{|x|}=sup_{xneq 0}frac{leftlangle TT^*x,xrightrangle^{1/2}}{|x|}leq sup_{xneq 0}frac{|TT^*x|^{1/2}}{|x|^{1/2}}leq\
&leqsup_{xneq 0}frac{|T^*x|^{1/2}}{|x|^{1/2}}=|T^*|^{1/2}end{align*}
i.e. $|T^*|leq |T^*|^{1/2}$ and thus $|T^*|leq 1$ (remark: this argument can be easily extended to prove that in general $|T|=|T^*|$ for any bounded linear operator $T$).
Therefore the first point may be applied to $T^*$, too.
begin{align*}
xin ker (I-T^*)iff T^*x=xiff left langle T^*x,xrightrangle = |x|^2iff left langle x,Txrightrangle =|x|^2
end{align*}
But $left langle Tx,xrightrangle =|x|^2 $ implies $left langle x, Txrightrangle = left langle Tx,xrightrangle =|x|^2$, as shown above, and viceversa, thus the equivalence is proved.
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
edited Jan 6 at 16:18
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
answered Jan 6 at 15:49
Lorenzo QuarisaLorenzo Quarisa
3,765623
3,765623
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063982%2fprove-tx-x-for-x-in-h-if-and-only-if-tx-x-x-2-and-keri-t-ker%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
