Mixing
$[T,S]:=TS-ST=I$ cannot holds
$begingroup$
Let $mathcal{B}(mathcal{H})$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $mathcal{H}$.
Let $T,Sin mathcal{B}(mathcal{H})$. I want to prove that the equality
begin{equation}label{commz}
[T,S]:=TS-ST=I tag{1},
end{equation}
cannot hold.
To see this, assume that $(1)$ holds. We shall prove by induction that
begin{equation}label{tag2}
[T, S^n] = nS^{n - 1},;nin mathbb{N}^*.
end{equation}
By assumption, $[T,S]=S^{0}=I$. Suppose $[T,S^{n}]= nS^{n-1}$ for some $nin mathbb{N}^*$. Then
begin{align*}
[T,S^{n+1}]
& = TS^{n+1}-S^{n+1}T\
& =(TS^{n}-S^{n}T)S+S^{n}TS-S^{n+1}T \
& = [T,S^{n}]S+S^{n}[T,S] \
& = nS^{n-1}S+S^{n}=(n+1) S^{n}.
end{align*}
So,
begin{equation*}label{tag11}
TS^n - S^nT = n S^{n=1},
end{equation*}
holds for all $nin mathbb{N}^*$. Hence,
begin{align*}
n|S^{n-1}|
& = |TS^n - S^nT|\
&leq 2 |T|cdot|S^{n} |\
&leq 2 |T|cdot|S|cdot|S^{n-1} |.
end{align*}
If $|S^{n-1} |ne 0$, we get $n le 2 |T||S|$, for all $nin mathbb{N}^*$. This leads to a contradiction. So, $(1)$ cannot hold for $T,Sin mathcal{B}(mathcal{H})$.
Why $S^{n-1}ne 0$ for all $nin mathbb{N}^*$?
functional-analysis proof-verification operator-theory
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
$endgroup$
add a comment |
$begingroup$
Let $mathcal{B}(mathcal{H})$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $mathcal{H}$.
Let $T,Sin mathcal{B}(mathcal{H})$. I want to prove that the equality
begin{equation}label{commz}
[T,S]:=TS-ST=I tag{1},
end{equation}
cannot hold.
To see this, assume that $(1)$ holds. We shall prove by induction that
begin{equation}label{tag2}
[T, S^n] = nS^{n - 1},;nin mathbb{N}^*.
end{equation}
By assumption, $[T,S]=S^{0}=I$. Suppose $[T,S^{n}]= nS^{n-1}$ for some $nin mathbb{N}^*$. Then
begin{align*}
[T,S^{n+1}]
& = TS^{n+1}-S^{n+1}T\
& =(TS^{n}-S^{n}T)S+S^{n}TS-S^{n+1}T \
& = [T,S^{n}]S+S^{n}[T,S] \
& = nS^{n-1}S+S^{n}=(n+1) S^{n}.
end{align*}
So,
begin{equation*}label{tag11}
TS^n - S^nT = n S^{n=1},
end{equation*}
holds for all $nin mathbb{N}^*$. Hence,
begin{align*}
n|S^{n-1}|
& = |TS^n - S^nT|\
&leq 2 |T|cdot|S^{n} |\
&leq 2 |T|cdot|S|cdot|S^{n-1} |.
end{align*}
If $|S^{n-1} |ne 0$, we get $n le 2 |T||S|$, for all $nin mathbb{N}^*$. This leads to a contradiction. So, $(1)$ cannot hold for $T,Sin mathcal{B}(mathcal{H})$.
Why $S^{n-1}ne 0$ for all $nin mathbb{N}^*$?
functional-analysis proof-verification operator-theory
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
$endgroup$
3
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
2
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38
add a comment |
$begingroup$
Let $mathcal{B}(mathcal{H})$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $mathcal{H}$.
Let $T,Sin mathcal{B}(mathcal{H})$. I want to prove that the equality
begin{equation}label{commz}
[T,S]:=TS-ST=I tag{1},
end{equation}
cannot hold.
To see this, assume that $(1)$ holds. We shall prove by induction that
begin{equation}label{tag2}
[T, S^n] = nS^{n - 1},;nin mathbb{N}^*.
end{equation}
By assumption, $[T,S]=S^{0}=I$. Suppose $[T,S^{n}]= nS^{n-1}$ for some $nin mathbb{N}^*$. Then
begin{align*}
[T,S^{n+1}]
& = TS^{n+1}-S^{n+1}T\
& =(TS^{n}-S^{n}T)S+S^{n}TS-S^{n+1}T \
& = [T,S^{n}]S+S^{n}[T,S] \
& = nS^{n-1}S+S^{n}=(n+1) S^{n}.
end{align*}
So,
begin{equation*}label{tag11}
TS^n - S^nT = n S^{n=1},
end{equation*}
holds for all $nin mathbb{N}^*$. Hence,
begin{align*}
n|S^{n-1}|
& = |TS^n - S^nT|\
&leq 2 |T|cdot|S^{n} |\
&leq 2 |T|cdot|S|cdot|S^{n-1} |.
end{align*}
If $|S^{n-1} |ne 0$, we get $n le 2 |T||S|$, for all $nin mathbb{N}^*$. This leads to a contradiction. So, $(1)$ cannot hold for $T,Sin mathcal{B}(mathcal{H})$.
Why $S^{n-1}ne 0$ for all $nin mathbb{N}^*$?
functional-analysis proof-verification operator-theory
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
$endgroup$
Let $mathcal{B}(mathcal{H})$ the algebra of all bounded linear operators on an infinite dimensional complex Hilbert space $mathcal{H}$.
Let $T,Sin mathcal{B}(mathcal{H})$. I want to prove that the equality
begin{equation}label{commz}
[T,S]:=TS-ST=I tag{1},
end{equation}
cannot hold.
To see this, assume that $(1)$ holds. We shall prove by induction that
begin{equation}label{tag2}
[T, S^n] = nS^{n - 1},;nin mathbb{N}^*.
end{equation}
By assumption, $[T,S]=S^{0}=I$. Suppose $[T,S^{n}]= nS^{n-1}$ for some $nin mathbb{N}^*$. Then
begin{align*}
[T,S^{n+1}]
& = TS^{n+1}-S^{n+1}T\
& =(TS^{n}-S^{n}T)S+S^{n}TS-S^{n+1}T \
& = [T,S^{n}]S+S^{n}[T,S] \
& = nS^{n-1}S+S^{n}=(n+1) S^{n}.
end{align*}
So,
begin{equation*}label{tag11}
TS^n - S^nT = n S^{n=1},
end{equation*}
holds for all $nin mathbb{N}^*$. Hence,
begin{align*}
n|S^{n-1}|
& = |TS^n - S^nT|\
&leq 2 |T|cdot|S^{n} |\
&leq 2 |T|cdot|S|cdot|S^{n-1} |.
end{align*}
If $|S^{n-1} |ne 0$, we get $n le 2 |T||S|$, for all $nin mathbb{N}^*$. This leads to a contradiction. So, $(1)$ cannot hold for $T,Sin mathcal{B}(mathcal{H})$.
Why $S^{n-1}ne 0$ for all $nin mathbb{N}^*$?
functional-analysis proof-verification operator-theory
functional-analysis proof-verification operator-theory
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
edited Jan 20 at 18:37
Schüler
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
asked Jan 20 at 18:20
SchülerSchüler
1,5391421
1,5391421
3
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
2
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38
add a comment |
3
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
2
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38
3
3
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
2
2
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed, the conclusion holds only if $S^{n-1}ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.
answered Jan 20 at 19:07
BerciBerci
61.3k23674
$endgroup$
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
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%2f3080944%2ft-s-ts-st-i-cannot-holds%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
$begingroup$
Indeed, the conclusion holds only if $S^{n-1}ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.
answered Jan 20 at 19:07
BerciBerci
61.3k23674
$endgroup$
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
add a comment |
$begingroup$
Indeed, the conclusion holds only if $S^{n-1}ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.
answered Jan 20 at 19:07
BerciBerci
61.3k23674
$endgroup$
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
add a comment |
$begingroup$
Indeed, the conclusion holds only if $S^{n-1}ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.
answered Jan 20 at 19:07
BerciBerci
61.3k23674
$endgroup$
Indeed, the conclusion holds only if $S^{n-1}ne 0$.
But, assume $S^k=0$ for some $k$, then by your formula $[T, S^k] =kS^{k-1}$, we get $S^{k-1}=0$, which in turn implies $S^{k-2}=0$, and so on, until $S=0$ which is impossible, provided that $[T, S] =I$.
answered Jan 20 at 19:07
BerciBerci
61.3k23674
answered Jan 20 at 19:07
BerciBerci
61.3k23674
answered Jan 20 at 19:07
BerciBerci
61.3k23674
answered Jan 20 at 19:07
BerciBerci
61.3k23674
61.3k23674
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
add a comment |
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
You have show that $S^k$ is non zero for all positive integer k?
$endgroup$
– Schüler
Jan 20 at 19:17
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
Yes, based on your identity.
$endgroup$
– Berci
Jan 20 at 19:19
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
$begingroup$
According to my question, We can prove also that the equality $$[T, S] = lambda I, ; 0 ne lambda in mathbb{C},$$ cannot hold, since the above equality may be written as $[T,lambda^{-1}S]=I$ and this is impossible. Do you agree with me? Thanks.
$endgroup$
– Schüler
Jan 20 at 19:20
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%2f3080944%2ft-s-ts-st-i-cannot-holds%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
3
$begingroup$
If $S^n=0$ for some $n$, take $n$ to be the smallest such; what can you say about $nS^{n-1}$, and so about $S^{n-1}$ ?
$endgroup$
– Max
Jan 20 at 18:28
$begingroup$
Why not just take $S=I$? Then it's always 0. Not sure if I'm missing something.
$endgroup$
– Matt Samuel
Jan 20 at 18:32
$begingroup$
@MattSamuel I assume that $(1)$ holds for some $T,S$ not every.
$endgroup$
– Schüler
Jan 20 at 18:34
2
$begingroup$
I was so confused reading this because "$[T,S] = I$ cannot hold for every $T, S$" is logically equivalent to "there exists some $T, S$ such that $[T,S] ne I$" What should be said instead is "for every $T, S, [T, S] ne I$" @Matt Samuel
$endgroup$
– Trevor Gunn
Jan 20 at 18:34
$begingroup$
@TrevorGunn Thank you for the remark. Please see my edit.
$endgroup$
– Schüler
Jan 20 at 18:38