Representation of negative Quantum entropy in terms of eigenvalues, i.e., $text{Tr}(Mlog M...












6












$begingroup$


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







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 19 at 4:05







      Saeed

















      asked Jan 18 at 0:38









      SaeedSaeed

      1,036310




      1,036310






















          1 Answer
          1






          active

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          3












          $begingroup$

          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}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          3












          $begingroup$

          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}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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}






          share|cite|improve this answer











          $endgroup$



          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}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 17:53

























          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


















          • $begingroup$
            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


















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