Mixing
About a problem of linear transformation $ ST-TS=I $
$begingroup$
The problem is, $S$ & $T$ are two linear operator i.e. belongs to $$L(V)$$ such that $ST-TS=I_n$ then prove that $L(V)$ is infinite dimensional vector space.
Now, I have showed that for all $nin mathbb{N}$ $S^{n+1}T-TS^{n+1}=(n+1)S^n$ by induction procedure. And I have showed that $forall ninmathbb{N}$ $S^nneq 0$, I mean zero linear operator.
How then I prove that $L(V)$ is infinite dimensional. Obviously I have to find a linearly independent set of $n$ many linear operators of for each $ninmathbb{N}$. But how construct it. May be $S^n$ are linearly independent to each other; not proved yet.
Any help will be appreciated.
linear-algebra linear-transformations
edited Jan 21 at 6:27
user549397
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
$endgroup$
add a comment |
$begingroup$
The problem is, $S$ & $T$ are two linear operator i.e. belongs to $$L(V)$$ such that $ST-TS=I_n$ then prove that $L(V)$ is infinite dimensional vector space.
Now, I have showed that for all $nin mathbb{N}$ $S^{n+1}T-TS^{n+1}=(n+1)S^n$ by induction procedure. And I have showed that $forall ninmathbb{N}$ $S^nneq 0$, I mean zero linear operator.
How then I prove that $L(V)$ is infinite dimensional. Obviously I have to find a linearly independent set of $n$ many linear operators of for each $ninmathbb{N}$. But how construct it. May be $S^n$ are linearly independent to each other; not proved yet.
Any help will be appreciated.
linear-algebra linear-transformations
edited Jan 21 at 6:27
user549397
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
$endgroup$
add a comment |
$begingroup$
The problem is, $S$ & $T$ are two linear operator i.e. belongs to $$L(V)$$ such that $ST-TS=I_n$ then prove that $L(V)$ is infinite dimensional vector space.
Now, I have showed that for all $nin mathbb{N}$ $S^{n+1}T-TS^{n+1}=(n+1)S^n$ by induction procedure. And I have showed that $forall ninmathbb{N}$ $S^nneq 0$, I mean zero linear operator.
How then I prove that $L(V)$ is infinite dimensional. Obviously I have to find a linearly independent set of $n$ many linear operators of for each $ninmathbb{N}$. But how construct it. May be $S^n$ are linearly independent to each other; not proved yet.
Any help will be appreciated.
linear-algebra linear-transformations
edited Jan 21 at 6:27
user549397
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
$endgroup$
The problem is, $S$ & $T$ are two linear operator i.e. belongs to $$L(V)$$ such that $ST-TS=I_n$ then prove that $L(V)$ is infinite dimensional vector space.
Now, I have showed that for all $nin mathbb{N}$ $S^{n+1}T-TS^{n+1}=(n+1)S^n$ by induction procedure. And I have showed that $forall ninmathbb{N}$ $S^nneq 0$, I mean zero linear operator.
How then I prove that $L(V)$ is infinite dimensional. Obviously I have to find a linearly independent set of $n$ many linear operators of for each $ninmathbb{N}$. But how construct it. May be $S^n$ are linearly independent to each other; not proved yet.
Any help will be appreciated.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 21 at 6:27
user549397
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
edited Jan 21 at 6:27
user549397
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
edited Jan 21 at 6:27
user549397
1,5061418
edited Jan 21 at 6:27
user549397
1,5061418
edited Jan 21 at 6:27
user549397
1,5061418
1,5061418
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
asked Jan 21 at 3:06
Karambir.kdKarambir.kd
885
885
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.
In cases of positive characteristic, there exists a counterexample: over $mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where
$$
S = begin{pmatrix} 0&1 \0 & 0end{pmatrix}, quad T = begin{pmatrix} 0 & 0 \ 1& 0end{pmatrix}.
$$
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
$endgroup$
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
add a comment |
$begingroup$
By induction, we find that
$$ S^nT - TS^n = begin{cases} nS^{n-1}, & n geq 1 \ 0, & n = 0 end{cases} $$
We claim the following:
Claim. $p(S) neq 0$ for any polynomial $p in F[x]$.
In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.
Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal ${p(x) in F[x] : p(S) = 0 }$.) If we write $m(x) = sum_{k} a_k x^k$, then
$$ 0 = m(S)T - Tm(S) = sum_{kgeq 0} a_k (S^k T - TS^k) = sum_{kgeq 1} k a_k S^{k-1} = m'(S). $$
This contradicts the minimality of $m$.
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
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%2f3081444%2fabout-a-problem-of-linear-transformation-st-ts-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.
In cases of positive characteristic, there exists a counterexample: over $mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where
$$
S = begin{pmatrix} 0&1 \0 & 0end{pmatrix}, quad T = begin{pmatrix} 0 & 0 \ 1& 0end{pmatrix}.
$$
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
$endgroup$
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
add a comment |
$begingroup$
If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.
In cases of positive characteristic, there exists a counterexample: over $mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where
$$
S = begin{pmatrix} 0&1 \0 & 0end{pmatrix}, quad T = begin{pmatrix} 0 & 0 \ 1& 0end{pmatrix}.
$$
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
$endgroup$
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
add a comment |
$begingroup$
If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.
In cases of positive characteristic, there exists a counterexample: over $mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where
$$
S = begin{pmatrix} 0&1 \0 & 0end{pmatrix}, quad T = begin{pmatrix} 0 & 0 \ 1& 0end{pmatrix}.
$$
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
$endgroup$
If $V$ is a finite-dimensional vector space, you can show that such a thing cannot happen by taking traces on both sides.
In cases of positive characteristic, there exists a counterexample: over $mathbb{F}_{2}$, we have $ST-TS = I_{2}$ where
$$
S = begin{pmatrix} 0&1 \0 & 0end{pmatrix}, quad T = begin{pmatrix} 0 & 0 \ 1& 0end{pmatrix}.
$$
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
edited Jan 21 at 8:35
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
answered Jan 21 at 3:11
Seewoo LeeSeewoo Lee
7,009927
7,009927
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
add a comment |
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
1
1
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
by taking what? Can't understand!
$endgroup$
– Karambir.kd
Jan 21 at 3:14
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
@Karambir.kd I edited! Now you can see the answer.
$endgroup$
– Seewoo Lee
Jan 21 at 4:34
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
I didn't tell about the field. If the characteristic of field is zero then done. But how about characteristic is non zero.
$endgroup$
– Karambir.kd
Jan 21 at 6:57
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
$begingroup$
@Karambir.kd I added a counterexample in that case.
$endgroup$
– Seewoo Lee
Jan 21 at 8:35
add a comment |
$begingroup$
By induction, we find that
$$ S^nT - TS^n = begin{cases} nS^{n-1}, & n geq 1 \ 0, & n = 0 end{cases} $$
We claim the following:
Claim. $p(S) neq 0$ for any polynomial $p in F[x]$.
In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.
Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal ${p(x) in F[x] : p(S) = 0 }$.) If we write $m(x) = sum_{k} a_k x^k$, then
$$ 0 = m(S)T - Tm(S) = sum_{kgeq 0} a_k (S^k T - TS^k) = sum_{kgeq 1} k a_k S^{k-1} = m'(S). $$
This contradicts the minimality of $m$.
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
add a comment |
$begingroup$
By induction, we find that
$$ S^nT - TS^n = begin{cases} nS^{n-1}, & n geq 1 \ 0, & n = 0 end{cases} $$
We claim the following:
Claim. $p(S) neq 0$ for any polynomial $p in F[x]$.
In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.
Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal ${p(x) in F[x] : p(S) = 0 }$.) If we write $m(x) = sum_{k} a_k x^k$, then
$$ 0 = m(S)T - Tm(S) = sum_{kgeq 0} a_k (S^k T - TS^k) = sum_{kgeq 1} k a_k S^{k-1} = m'(S). $$
This contradicts the minimality of $m$.
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
add a comment |
$begingroup$
By induction, we find that
$$ S^nT - TS^n = begin{cases} nS^{n-1}, & n geq 1 \ 0, & n = 0 end{cases} $$
We claim the following:
Claim. $p(S) neq 0$ for any polynomial $p in F[x]$.
In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.
Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal ${p(x) in F[x] : p(S) = 0 }$.) If we write $m(x) = sum_{k} a_k x^k$, then
$$ 0 = m(S)T - Tm(S) = sum_{kgeq 0} a_k (S^k T - TS^k) = sum_{kgeq 1} k a_k S^{k-1} = m'(S). $$
This contradicts the minimality of $m$.
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
By induction, we find that
$$ S^nT - TS^n = begin{cases} nS^{n-1}, & n geq 1 \ 0, & n = 0 end{cases} $$
We claim the following:
Claim. $p(S) neq 0$ for any polynomial $p in F[x]$.
In view of the Cayley-Hamilton theorem, this is enough to argue that $V$ is not finite-dimensional.
Proof of Claim. Assume otherwise and let $m(x)$ be the minimal polynomial of $S$. (This is simply the generator of the principal ideal ${p(x) in F[x] : p(S) = 0 }$.) If we write $m(x) = sum_{k} a_k x^k$, then
$$ 0 = m(S)T - Tm(S) = sum_{kgeq 0} a_k (S^k T - TS^k) = sum_{kgeq 1} k a_k S^{k-1} = m'(S). $$
This contradicts the minimality of $m$.
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
edited Jan 27 at 11:40
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
answered Jan 21 at 7:33
Sangchul LeeSangchul Lee
95.6k12171279
95.6k12171279
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%2f3081444%2fabout-a-problem-of-linear-transformation-st-ts-i%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
