Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, if [A,[A,B]]=0$?
Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?
I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:
$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
$$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$
But now I do not see how to use $[A,[A,B]] = 0$.
Does anyone has hints for this or a hint how to advance?
linear-algebra matrices
edited Nov 28 at 19:29
Bernard
117k637111
asked Nov 28 at 18:22
MPB94
23816
add a comment |
Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?
I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:
$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
$$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$
But now I do not see how to use $[A,[A,B]] = 0$.
Does anyone has hints for this or a hint how to advance?
linear-algebra matrices
edited Nov 28 at 19:29
Bernard
117k637111
asked Nov 28 at 18:22
MPB94
23816
add a comment |
Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?
I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:
$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
$$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$
But now I do not see how to use $[A,[A,B]] = 0$.
Does anyone has hints for this or a hint how to advance?
linear-algebra matrices
edited Nov 28 at 19:29
Bernard
117k637111
asked Nov 28 at 18:22
MPB94
23816
Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?
I tried to rewrite this and reduce it to $operatorname{Tr}([A,B]^{m-1} AB) = operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:
$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$
$$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$
But now I do not see how to use $[A,[A,B]] = 0$.
Does anyone has hints for this or a hint how to advance?
linear-algebra matrices
linear-algebra matrices
edited Nov 28 at 19:29
Bernard
117k637111
asked Nov 28 at 18:22
MPB94
23816
edited Nov 28 at 19:29
Bernard
117k637111
asked Nov 28 at 18:22
MPB94
23816
edited Nov 28 at 19:29
Bernard
117k637111
edited Nov 28 at 19:29
Bernard
117k637111
edited Nov 28 at 19:29
Bernard
117k637111
117k637111
asked Nov 28 at 18:22
MPB94
23816
asked Nov 28 at 18:22
MPB94
23816
asked Nov 28 at 18:22
MPB94
23816
23816
add a comment |
add a comment |
1 Answer
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From
$$[A,[A,B]]=0$$
we have that
$$A[A,B]=[A,B]A$$
And
$$A[A,B]^n=[A,B]^nA$$
So using your work:
$$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
which is what you wanted to prove.
answered Nov 28 at 19:12
Botond
5,4332732
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
From
$$[A,[A,B]]=0$$
we have that
$$A[A,B]=[A,B]A$$
And
$$A[A,B]^n=[A,B]^nA$$
So using your work:
$$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
which is what you wanted to prove.
answered Nov 28 at 19:12
Botond
5,4332732
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
add a comment |
From
$$[A,[A,B]]=0$$
we have that
$$A[A,B]=[A,B]A$$
And
$$A[A,B]^n=[A,B]^nA$$
So using your work:
$$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
which is what you wanted to prove.
answered Nov 28 at 19:12
Botond
5,4332732
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
add a comment |
From
$$[A,[A,B]]=0$$
we have that
$$A[A,B]=[A,B]A$$
And
$$A[A,B]^n=[A,B]^nA$$
So using your work:
$$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
which is what you wanted to prove.
answered Nov 28 at 19:12
Botond
5,4332732
From
$$[A,[A,B]]=0$$
we have that
$$A[A,B]=[A,B]A$$
And
$$A[A,B]^n=[A,B]^nA$$
So using your work:
$$text{Tr}([A,B]^{m-1}AB)=text{Tr}(B[A,B]^{m-1}A)=text{Tr}(BA[A,B]^{m-1})$$
which is what you wanted to prove.
answered Nov 28 at 19:12
Botond
5,4332732
answered Nov 28 at 19:12
Botond
5,4332732
answered Nov 28 at 19:12
Botond
5,4332732
answered Nov 28 at 19:12
Botond
5,4332732
5,4332732
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
add a comment |
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
Thanks, I looked for such an identity and somehow did not see it. :)
– MPB94
Nov 28 at 19:37
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
@MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :)
– Botond
Nov 28 at 19:43
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
$A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha.
– MPB94
Nov 28 at 19:47
add a comment |
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