Representation of negative Quantum entropy in terms of eigenvalues, i.e., $text{Tr}(Mlog M...
$begingroup$
Negative Quantum entropy or Negative Von Nuemann entropy is defined as $f(M)=text{Tr}(Mlog M -M)$.
Where $M$ is a positive definite matrix in $mathbb{S}_+^n$, $log$ is natural matrix logarithm for which $log(M)$ is defined as $log(M)=sum_{i=1}^{n}log(lambda_i)v_iv_i^T$ where $(lambda_i,v_i)$ are eigenpairs of $M$.
Show $f(M)=text{Tr}(Mlog M -M)=sum_{i=1}^{n}(lambda_ilog(lambda_i)-lambda_i)$.
linear-algebra matrices positive-definite symmetric-matrices
asked Jan 18 at 0:38
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Negative Quantum entropy or Negative Von Nuemann entropy is defined as $f(M)=text{Tr}(Mlog M -M)$.
Where $M$ is a positive definite matrix in $mathbb{S}_+^n$, $log$ is natural matrix logarithm for which $log(M)$ is defined as $log(M)=sum_{i=1}^{n}log(lambda_i)v_iv_i^T$ where $(lambda_i,v_i)$ are eigenpairs of $M$.
Show $f(M)=text{Tr}(Mlog M -M)=sum_{i=1}^{n}(lambda_ilog(lambda_i)-lambda_i)$.
linear-algebra matrices positive-definite symmetric-matrices
asked Jan 18 at 0:38
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Negative Quantum entropy or Negative Von Nuemann entropy is defined as $f(M)=text{Tr}(Mlog M -M)$.
Where $M$ is a positive definite matrix in $mathbb{S}_+^n$, $log$ is natural matrix logarithm for which $log(M)$ is defined as $log(M)=sum_{i=1}^{n}log(lambda_i)v_iv_i^T$ where $(lambda_i,v_i)$ are eigenpairs of $M$.
Show $f(M)=text{Tr}(Mlog M -M)=sum_{i=1}^{n}(lambda_ilog(lambda_i)-lambda_i)$.
linear-algebra matrices positive-definite symmetric-matrices
asked Jan 18 at 0:38
SaeedSaeed
1,036310
$endgroup$
Negative Quantum entropy or Negative Von Nuemann entropy is defined as $f(M)=text{Tr}(Mlog M -M)$.
Where $M$ is a positive definite matrix in $mathbb{S}_+^n$, $log$ is natural matrix logarithm for which $log(M)$ is defined as $log(M)=sum_{i=1}^{n}log(lambda_i)v_iv_i^T$ where $(lambda_i,v_i)$ are eigenpairs of $M$.
Show $f(M)=text{Tr}(Mlog M -M)=sum_{i=1}^{n}(lambda_ilog(lambda_i)-lambda_i)$.
linear-algebra matrices positive-definite symmetric-matrices
linear-algebra matrices positive-definite symmetric-matrices
asked Jan 18 at 0:38
SaeedSaeed
1,036310
asked Jan 18 at 0:38
SaeedSaeed
1,036310
edited Jan 19 at 4:05
Saeed
asked Jan 18 at 0:38
SaeedSaeed
1,036310
asked Jan 18 at 0:38
SaeedSaeed
1,036310
asked Jan 18 at 0:38
SaeedSaeed
1,036310
1,036310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $Minmathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $Lambda=text{diag}left{lambda_1,lambda_2,...,lambda_nright}$, with each $lambda_j>0$, such that
$$
M=ULambda U^{top}.
$$
Hence, using the definition of $log M$,
begin{align}
Mlog M-M&=left(ULambda U^{top}right)left(UlogLambda,U^{top}right)-ULambda U^{top}\
&=Uleft(LambdalogLambda-Lambdaright)U^{top}.
end{align}
Consequently,
begin{align}
f(M)&=text{tr}left(Mlog M-Mright)\
&=text{tr}left(Uleft(LambdalogLambda-Lambdaright)U^{top}right)\
&=text{tr}left(left(LambdalogLambda-Lambdaright)U^{top}Uright)\
&=text{tr}left(LambdalogLambda-Lambdaright)\
&=sum_{j=1}^nlambda_jleft(loglambda_j-1right).
end{align}
answered Jan 18 at 1:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
|
show 6 more comments
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Since $Minmathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $Lambda=text{diag}left{lambda_1,lambda_2,...,lambda_nright}$, with each $lambda_j>0$, such that
$$
M=ULambda U^{top}.
$$
Hence, using the definition of $log M$,
begin{align}
Mlog M-M&=left(ULambda U^{top}right)left(UlogLambda,U^{top}right)-ULambda U^{top}\
&=Uleft(LambdalogLambda-Lambdaright)U^{top}.
end{align}
Consequently,
begin{align}
f(M)&=text{tr}left(Mlog M-Mright)\
&=text{tr}left(Uleft(LambdalogLambda-Lambdaright)U^{top}right)\
&=text{tr}left(left(LambdalogLambda-Lambdaright)U^{top}Uright)\
&=text{tr}left(LambdalogLambda-Lambdaright)\
&=sum_{j=1}^nlambda_jleft(loglambda_j-1right).
end{align}
answered Jan 18 at 1:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
|
show 6 more comments
$begingroup$
Since $Minmathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $Lambda=text{diag}left{lambda_1,lambda_2,...,lambda_nright}$, with each $lambda_j>0$, such that
$$
M=ULambda U^{top}.
$$
Hence, using the definition of $log M$,
begin{align}
Mlog M-M&=left(ULambda U^{top}right)left(UlogLambda,U^{top}right)-ULambda U^{top}\
&=Uleft(LambdalogLambda-Lambdaright)U^{top}.
end{align}
Consequently,
begin{align}
f(M)&=text{tr}left(Mlog M-Mright)\
&=text{tr}left(Uleft(LambdalogLambda-Lambdaright)U^{top}right)\
&=text{tr}left(left(LambdalogLambda-Lambdaright)U^{top}Uright)\
&=text{tr}left(LambdalogLambda-Lambdaright)\
&=sum_{j=1}^nlambda_jleft(loglambda_j-1right).
end{align}
answered Jan 18 at 1:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
|
show 6 more comments
$begingroup$
Since $Minmathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $Lambda=text{diag}left{lambda_1,lambda_2,...,lambda_nright}$, with each $lambda_j>0$, such that
$$
M=ULambda U^{top}.
$$
Hence, using the definition of $log M$,
begin{align}
Mlog M-M&=left(ULambda U^{top}right)left(UlogLambda,U^{top}right)-ULambda U^{top}\
&=Uleft(LambdalogLambda-Lambdaright)U^{top}.
end{align}
Consequently,
begin{align}
f(M)&=text{tr}left(Mlog M-Mright)\
&=text{tr}left(Uleft(LambdalogLambda-Lambdaright)U^{top}right)\
&=text{tr}left(left(LambdalogLambda-Lambdaright)U^{top}Uright)\
&=text{tr}left(LambdalogLambda-Lambdaright)\
&=sum_{j=1}^nlambda_jleft(loglambda_j-1right).
end{align}
answered Jan 18 at 1:48
hypernovahypernova
4,834414
$endgroup$
Since $Minmathbb{S}_+^n$, there must exist an orthogonal matrix $U$ and a diagonal matrix $Lambda=text{diag}left{lambda_1,lambda_2,...,lambda_nright}$, with each $lambda_j>0$, such that
$$
M=ULambda U^{top}.
$$
Hence, using the definition of $log M$,
begin{align}
Mlog M-M&=left(ULambda U^{top}right)left(UlogLambda,U^{top}right)-ULambda U^{top}\
&=Uleft(LambdalogLambda-Lambdaright)U^{top}.
end{align}
Consequently,
begin{align}
f(M)&=text{tr}left(Mlog M-Mright)\
&=text{tr}left(Uleft(LambdalogLambda-Lambdaright)U^{top}right)\
&=text{tr}left(left(LambdalogLambda-Lambdaright)U^{top}Uright)\
&=text{tr}left(LambdalogLambda-Lambdaright)\
&=sum_{j=1}^nlambda_jleft(loglambda_j-1right).
end{align}
answered Jan 18 at 1:48
hypernovahypernova
4,834414
edited Jan 18 at 17:53
answered Jan 18 at 1:48
hypernovahypernova
4,834414
answered Jan 18 at 1:48
hypernovahypernova
4,834414
answered Jan 18 at 1:48
hypernovahypernova
4,834414
4,834414
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
|
show 6 more comments
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
$begingroup$
Excuse me but I think you miss understood this question. You have copied the irrelevant answer which is relevant to my other question. Please delete this answer so that others try to solve it.
$endgroup$
– Saeed
Jan 18 at 16:34
1
1
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
@Saeed: Apologies for my misunderstanding. I corrected my answer. Perhaps you want to have a look at it. Besides, I would really appreciate it if you could explain your desired identity with this special case $M=I_n$ (I am afraid an additional "$-lambda_i$" term is needed).
$endgroup$
– hypernova
Jan 18 at 17:55
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
I think you are right but I am following this paper jmlr.org/papers/volume6/tsuda05a/tsuda05a.pdf. On page 999, in the second line they have this equality. Am I missing something there?
$endgroup$
– Saeed
Jan 18 at 19:11
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
@Saeed: I suspect the paper has some problem here, but not that serious. First, the choice of the quantum entropy in this paper seems unconventional. You may compare it with this link. Secondly, it states an additional normalization condition $text{tr}M=1$ at the 4th line from the last on page 998, with which we would have $f(M)=sum_{j=1}^nlambda_jloglambda_j-1$. The constant $1$ could then be ignored since it appears to set some base level (like the potential energy).
$endgroup$
– hypernova
Jan 19 at 0:00
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
$begingroup$
The link is different. I think my question should be negative Von Neumann entropy. I revised them all.
$endgroup$
– Saeed
Jan 19 at 4:07
|
show 6 more comments
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