Normal Operator || T^2|| = ||T||^2
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
add a comment |
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
add a comment |
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
linear-algebra
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
798
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
add a comment |
Your Answer
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%2f3197793%2fnormal-operator-t2-t2%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
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
add a comment |
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
add a comment |
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 46 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
add a comment |
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
30 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
29 mins ago
1
1
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
29 mins ago
1
1
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
26 mins ago
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%2f3197793%2fnormal-operator-t2-t2%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
