Normal Operator || T^2|| = ||T||^2 Announcing the arrival of Valued Associate #679: Cesar...
lm and glm function in R
Will I be more secure with my own router behind my ISP's router?
What's the difference between using dependency injection with a container and using a service locator?
tabularx column has extra padding at right?
Why these surprising proportionalities of integrals involving odd zeta values?
Why doesn't the university give past final exams' answers?
Has a Nobel Peace laureate ever been accused of war crimes?
What could prevent concentrated local exploration?
How to ask rejected full-time candidates to apply to teach individual courses?
How do I deal with an erroneously large refund?
Who can become a wight?
If gravity precedes the formation of a solar system, where did the mass come from that caused the gravity?
Converting a text document with special format to Pandas DataFrame
Suing a Police Officer Instead of the Police Department
Can I ask an author to send me his ebook?
Does traveling In The United States require a passport or can I use my green card if not a US citizen?
Does Prince Arnaud cause someone holding the Princess to lose?
“Since the train was delayed for more than an hour, passengers were given a full refund.” – Why is there no article before “passengers”?
Is "ein Herz wie das meine" an antiquated or colloquial use of the possesive pronoun?
Why aren't road bike wheels tiny?
Pointing to problems without suggesting solutions
"Destructive force" carried by a B-52?
Is Bran literally the world's memory?
false 'Security alert' from Google - every login generates mails from 'no-reply@accounts.google.com'
Normal Operator || T^2|| = ||T||^2
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)If $S$ and $T$ are commuting, normal operators, then $ST$ is normal$T^2=I$ implies that $T$ is a normal operatorProving an operator is Self-adjoint using the Spectral TheoremEigenvalues of adjoint operator [General Case]diagonalizability implies existence of an inner product wrt an operator is normalNormal operator over real inner product spaceNormal matrix over real inner product space with real eigenvalues is Hermitianpolar form of unitary operatorWe have a linear operator T. Show $T^2=Id$ implies $T=T^*$Some property of Normal Operator
$begingroup$
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 23 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
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 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 23 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
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 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 23 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
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 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 23 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 23 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 23 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 23 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 23 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
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 mins ago
add a comment |
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 mins ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
7 mins ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
7 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
6 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 mins ago
$begingroup$
@eric $|left<x,Axright>|le|x||Ax|le|A||x|^2$.
$endgroup$
– Lord Shark the Unknown
4 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
