Traceless matrices and commutators The 2019 Stack Overflow Developer Survey Results Are InHow can I prove that if $ trace(C)=0 $ then $C=AB-BA$?Existence of $p times p $ matrices $A$ and $B$ over the field $mathbb F_p$, $p$ prime, such that $AB-BA=I$.Subspace of the real matrices generated by the matrices of the form $AB-BA$Characterizing the image of the commutatoris it possible to find such nonsingular matrices?Can we write a matrix with zero trace as a commutator?Universal parametrization for orthogonal matricesHall's identity and beyond?Exponential restricts to special linear matricesProving two matrices are cogredient over $mathbbQ$Meaning of $[A,B]$ when $A$ and $B$ are self-adjointSkew-Hermitian matricesCharacterizing the image of the commutatorCommutators of Almost Diagonal MatricesQuestion related to the trace and the commutator of matricesLie algebra of a matrix group
How to create dashed lines/arrows in Illustrator
What is the best strategy for white in this position?
Dual Citizen. Exited the US on Italian passport recently
Spanish for "widget"
What is the use of option -o in the useradd command?
What does "sndry explns" mean in one of the Hitchhiker's guide books?
Does a dangling wire really electrocute me if I'm standing in water?
On the insanity of kings as an argument against monarchy
Is domain driven design an anti-SQL pattern?
aging parents with no investments
Idiomatic way to prevent slicing?
What are my rights when I have a Sparpreis ticket but can't board an overcrowded train?
What do the Banks children have against barley water?
Is flight data recorder erased after every flight?
Where to refill my bottle in India?
Are USB sockets on wall outlets live all the time, even when the switch is off?
How are circuits which use complex ICs normally simulated?
What does "rabbited" mean/imply in this sentence?
How to reverse every other sublist of a list?
Is this food a bread or a loaf?
Why do UK politicians seemingly ignore opinion polls on Brexit?
How long do I have to send payment?
What is the meaning of Triage in Cybersec world?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
Traceless matrices and commutators
The 2019 Stack Overflow Developer Survey Results Are InHow can I prove that if $ trace(C)=0 $ then $C=AB-BA$?Existence of $p times p $ matrices $A$ and $B$ over the field $mathbb F_p$, $p$ prime, such that $AB-BA=I$.Subspace of the real matrices generated by the matrices of the form $AB-BA$Characterizing the image of the commutatoris it possible to find such nonsingular matrices?Can we write a matrix with zero trace as a commutator?Universal parametrization for orthogonal matricesHall's identity and beyond?Exponential restricts to special linear matricesProving two matrices are cogredient over $mathbbQ$Meaning of $[A,B]$ when $A$ and $B$ are self-adjointSkew-Hermitian matricesCharacterizing the image of the commutatorCommutators of Almost Diagonal MatricesQuestion related to the trace and the commutator of matricesLie algebra of a matrix group
$begingroup$
Any traceless $ntimes n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?
When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:
- multiples of the identity may have trace $=0$.
- a matrix may have a spectrum equal to the whole field.
Are all traceless matrices commutators? If not, for which $ninmathbbNsetminuslbrace 0 rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.
matrices finite-fields
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
$endgroup$
add a comment |
$begingroup$
Any traceless $ntimes n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?
When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:
- multiples of the identity may have trace $=0$.
- a matrix may have a spectrum equal to the whole field.
Are all traceless matrices commutators? If not, for which $ninmathbbNsetminuslbrace 0 rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.
matrices finite-fields
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
$endgroup$
add a comment |
$begingroup$
Any traceless $ntimes n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?
When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:
- multiples of the identity may have trace $=0$.
- a matrix may have a spectrum equal to the whole field.
Are all traceless matrices commutators? If not, for which $ninmathbbNsetminuslbrace 0 rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.
matrices finite-fields
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
$endgroup$
Any traceless $ntimes n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?
When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:
- multiples of the identity may have trace $=0$.
- a matrix may have a spectrum equal to the whole field.
Are all traceless matrices commutators? If not, for which $ninmathbbNsetminuslbrace 0 rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.
matrices finite-fields
matrices finite-fields
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
edited Mar 27 '12 at 22:00
Olivier Bégassat
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
asked Mar 27 '12 at 21:34
Olivier BégassatOlivier Bégassat
13.4k12373
13.4k12373
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.
In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]cong k.$$
Now, the trace is a non-zero linear map $mathrmtr:M_n(k)to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.
N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.
Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
add a comment |
$begingroup$
Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^p^n - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
$endgroup$
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
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%2f125219%2ftraceless-matrices-and-commutators%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 $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.
In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]cong k.$$
Now, the trace is a non-zero linear map $mathrmtr:M_n(k)to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.
N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.
Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
add a comment |
$begingroup$
If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.
In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]cong k.$$
Now, the trace is a non-zero linear map $mathrmtr:M_n(k)to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.
N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.
Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
add a comment |
$begingroup$
If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.
In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]cong k.$$
Now, the trace is a non-zero linear map $mathrmtr:M_n(k)to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.
N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.
Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
If $k$ is any field, the $k$-algebra $M_n(k)$ is Morita equivalent (as a $k$-algebra) to $k$. It follows that $M_n(k)$ and $k$ have isomorphic Hochschild homologies. In particular, they have isomorphic $0$th Hochschild homology.
In general, if $A$ is a $k$-algebra, the zeroth homology is $HH_0(A)=A/[A,A]$, the quotient of $A$ by the subspace generated by commutators. Since $k$ is a commutative $k$-algebra, it is obvious that $HH_0(k)=k$. The first paragraph, then, tells us that $$M_n(k)/[M_n(k),M_n(k)]cong k.$$
Now, the trace is a non-zero linear map $mathrmtr:M_n(k)to k$ which vanishes on $[M_n(k),M_n(k)]$. It follows by the above isomorphism that $[M_n(k),M_n(k)]$ is precisely the kernel of the trace.
N.B.: The details underlying my first paragraph above are explain in Jean-Louis Loday's book on cyclic homology or in Chuck Weibel's one on homological algebra, among other places.
Later. Olivier wanted matrices to be actual commutators, not sums thereof. It is a result of Albert and Muckenhoupt that this is always possible over a field. See http://projecteuclid.org/euclid.mmj/1028990168
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
edited Mar 27 '12 at 23:55
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
answered Mar 27 '12 at 21:51
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
112k7159291
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
add a comment |
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
1
1
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
I see, but wouldn't this show that traceless matrices are sums of commutators? I want to know wether any traceless matrix is a commutator. I'll edit my question to make it clearer.
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 21:57
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
Yes. That's what I wrote :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 27 '12 at 22:01
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
$begingroup$
But are traceless matrices always a particular commutator?
$endgroup$
– Olivier Bégassat
Mar 27 '12 at 22:08
1
1
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
Thank you for your last edit. Unfortunately I don't have access to the whole article but only to the first page. I think it might be improper for me to ask you to send me the file via email, but if it isn't and you'd do it I could give you my email address. Thanks in any case!
$endgroup$
– Olivier Bégassat
Apr 29 '12 at 10:57
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
$begingroup$
This is a very nice application of abstract machinery! This question popped up in a different form, see here: math.stackexchange.com/questions/2677496/…
$endgroup$
– Mathematician 42
Mar 5 '18 at 9:47
add a comment |
$begingroup$
Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^p^n - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
$endgroup$
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
add a comment |
$begingroup$
Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^p^n - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
$endgroup$
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
add a comment |
$begingroup$
Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^p^n - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
$endgroup$
Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^p^n - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
answered Apr 13 '18 at 18:30
doge5everdoge5ever
364
364
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
add a comment |
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
Why is that matrix not a commutator?
$endgroup$
– darij grinberg
Apr 6 at 17:26
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
$begingroup$
I might have been answering a different question with my posted answer but yeah--this should be a commutator.
$endgroup$
– doge5ever
Apr 7 at 1:01
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%2f125219%2ftraceless-matrices-and-commutators%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
