Uniqueness of representations of matrix product states












2












$begingroup$


Let ${A_k}_{k=1dots N}$ and ${B_k}_{k=1dots N}$ be two sets of $dtimes d$ matrices over the complex numbers such that for any length $L$ and any sets of indices ${j_1,j_2,dots j_L=1dots N}$ the following trace equality holds:



$text{Tr}(A_{j_1}A_{j_2}dots A_{j_L})=text{Tr}(B_{j_1}B_{j_2}dots B_{j_L})$



This equality is obviously satisfied if there exists an invertible matrix $S$ such that $B_k=SA_kS^{-1}$ for all $k=1dots N$.



Under what conditions can we guarantee that there is such an $S$ intertwiner?






linear-algebra matrices lie-algebras mathematical-physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:31










  • $begingroup$
    Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
    $endgroup$
    – Balázs Pozsgay
    Dec 18 '18 at 14:34










  • $begingroup$
    This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:36










  • $begingroup$
    @Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 14:41










  • $begingroup$
    in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:45


















2












$begingroup$


Let ${A_k}_{k=1dots N}$ and ${B_k}_{k=1dots N}$ be two sets of $dtimes d$ matrices over the complex numbers such that for any length $L$ and any sets of indices ${j_1,j_2,dots j_L=1dots N}$ the following trace equality holds:



$text{Tr}(A_{j_1}A_{j_2}dots A_{j_L})=text{Tr}(B_{j_1}B_{j_2}dots B_{j_L})$



This equality is obviously satisfied if there exists an invertible matrix $S$ such that $B_k=SA_kS^{-1}$ for all $k=1dots N$.



Under what conditions can we guarantee that there is such an $S$ intertwiner?






linear-algebra matrices lie-algebras mathematical-physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:31










  • $begingroup$
    Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
    $endgroup$
    – Balázs Pozsgay
    Dec 18 '18 at 14:34










  • $begingroup$
    This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:36










  • $begingroup$
    @Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 14:41










  • $begingroup$
    in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:45
















2












2








2





$begingroup$


Let ${A_k}_{k=1dots N}$ and ${B_k}_{k=1dots N}$ be two sets of $dtimes d$ matrices over the complex numbers such that for any length $L$ and any sets of indices ${j_1,j_2,dots j_L=1dots N}$ the following trace equality holds:



$text{Tr}(A_{j_1}A_{j_2}dots A_{j_L})=text{Tr}(B_{j_1}B_{j_2}dots B_{j_L})$



This equality is obviously satisfied if there exists an invertible matrix $S$ such that $B_k=SA_kS^{-1}$ for all $k=1dots N$.



Under what conditions can we guarantee that there is such an $S$ intertwiner?






linear-algebra matrices lie-algebras mathematical-physics







share|cite|improve this question











$endgroup$




Let ${A_k}_{k=1dots N}$ and ${B_k}_{k=1dots N}$ be two sets of $dtimes d$ matrices over the complex numbers such that for any length $L$ and any sets of indices ${j_1,j_2,dots j_L=1dots N}$ the following trace equality holds:



$text{Tr}(A_{j_1}A_{j_2}dots A_{j_L})=text{Tr}(B_{j_1}B_{j_2}dots B_{j_L})$



This equality is obviously satisfied if there exists an invertible matrix $S$ such that $B_k=SA_kS^{-1}$ for all $k=1dots N$.



Under what conditions can we guarantee that there is such an $S$ intertwiner?






linear-algebra matrices lie-algebras mathematical-physics




linear-algebra matrices lie-algebras mathematical-physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:31









Omnomnomnom

127k790178




127k790178










asked Dec 18 '18 at 14:22









Balázs PozsgayBalázs Pozsgay

19818




19818












  • $begingroup$
    do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:31










  • $begingroup$
    Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
    $endgroup$
    – Balázs Pozsgay
    Dec 18 '18 at 14:34










  • $begingroup$
    This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:36










  • $begingroup$
    @Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 14:41










  • $begingroup$
    in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:45




















  • $begingroup$
    do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:31










  • $begingroup$
    Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
    $endgroup$
    – Balázs Pozsgay
    Dec 18 '18 at 14:34










  • $begingroup$
    This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:36










  • $begingroup$
    @Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 14:41










  • $begingroup$
    in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
    $endgroup$
    – Enkidu
    Dec 18 '18 at 14:45


















$begingroup$
do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
$endgroup$
– Enkidu
Dec 18 '18 at 14:31




$begingroup$
do you know anything about the ring you are working over? is it a field? may we assume it is algebraically closed?
$endgroup$
– Enkidu
Dec 18 '18 at 14:31












$begingroup$
Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
$endgroup$
– Balázs Pozsgay
Dec 18 '18 at 14:34




$begingroup$
Sorry, I was only thinking about the complex numbers. Perhaps I should correct it.
$endgroup$
– Balázs Pozsgay
Dec 18 '18 at 14:34












$begingroup$
This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
$endgroup$
– Enkidu
Dec 18 '18 at 14:36




$begingroup$
This $S$ should then exist by simultaneaous jordanizing, if and only if those morphisms commute, however, your two things have nothing to do with each other, since the trace does not see conjugacion with invertible morphisms, so your conjugation with $S$ can help you to get the above identity, but it is quite the overkill
$endgroup$
– Enkidu
Dec 18 '18 at 14:36












$begingroup$
@Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
$endgroup$
– Omnomnomnom
Dec 18 '18 at 14:41




$begingroup$
@Enkidu my understanding of your comment is "for a fixed $k$, there exists a matrix $S$ such that $B_k = SA_k S^{-1}$ if and only if $A_kB_k = B_k A_k$". I don't really see how this addresses the question being asked. If this isn't what you mean, then I'm not sure what you're trying to say.
$endgroup$
– Omnomnomnom
Dec 18 '18 at 14:41












$begingroup$
in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
$endgroup$
– Enkidu
Dec 18 '18 at 14:45






$begingroup$
in my opinion this question is just about the existence of such an $S$, hence it has nothing to do with the trace above "Under what conditions can we guarantee that there is such an S intertwiner?", and hence if you know that you need to triangularize all of the morphisms simultaneously (the $S$ should work for all $k$). So you get a quite clear characterization of those ${A_k},{B_k}$ such tht such an $S$ exists. (There being more than two might be essential, since if you have only have $B_1=S A_1 S^{-1}$ it only fixes the Jordan type)
$endgroup$
– Enkidu
Dec 18 '18 at 14:45












1 Answer
1






active

oldest

votes


















1












$begingroup$

With the help of some tips from @Omnomnomnom I found the answer. It is written in
"Shirshov's theorem and representations of semigroups" by A. Freedman, R.N. Gupta, R.M. Guralnick. Their Corollary 2.7 on page 163 deals with this problem.



The answer is: If the matrix algebras generated by the two sets are semisimple, then there is a desired common similarity transformation. The semisimple property was the one I was looking for.



Perhaps a simpler explanation is given in "On the Unitary Similarity of Matrix Families" by Yu. A. Al'pinKh. D. Ikramov. Theorem 1 is what I needed. Here the condition is the complete reducibility, which is equivalent to the semisimple property.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:42












  • $begingroup$
    Thanks! What is the difference?
    $endgroup$
    – Balázs Pozsgay
    Dec 19 '18 at 17:45










  • $begingroup$
    mathoverflow, though it gets less traffic, is better suited to research-level questions
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:57











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

With the help of some tips from @Omnomnomnom I found the answer. It is written in
"Shirshov's theorem and representations of semigroups" by A. Freedman, R.N. Gupta, R.M. Guralnick. Their Corollary 2.7 on page 163 deals with this problem.



The answer is: If the matrix algebras generated by the two sets are semisimple, then there is a desired common similarity transformation. The semisimple property was the one I was looking for.



Perhaps a simpler explanation is given in "On the Unitary Similarity of Matrix Families" by Yu. A. Al'pinKh. D. Ikramov. Theorem 1 is what I needed. Here the condition is the complete reducibility, which is equivalent to the semisimple property.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:42












  • $begingroup$
    Thanks! What is the difference?
    $endgroup$
    – Balázs Pozsgay
    Dec 19 '18 at 17:45










  • $begingroup$
    mathoverflow, though it gets less traffic, is better suited to research-level questions
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:57
















1












$begingroup$

With the help of some tips from @Omnomnomnom I found the answer. It is written in
"Shirshov's theorem and representations of semigroups" by A. Freedman, R.N. Gupta, R.M. Guralnick. Their Corollary 2.7 on page 163 deals with this problem.



The answer is: If the matrix algebras generated by the two sets are semisimple, then there is a desired common similarity transformation. The semisimple property was the one I was looking for.



Perhaps a simpler explanation is given in "On the Unitary Similarity of Matrix Families" by Yu. A. Al'pinKh. D. Ikramov. Theorem 1 is what I needed. Here the condition is the complete reducibility, which is equivalent to the semisimple property.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:42












  • $begingroup$
    Thanks! What is the difference?
    $endgroup$
    – Balázs Pozsgay
    Dec 19 '18 at 17:45










  • $begingroup$
    mathoverflow, though it gets less traffic, is better suited to research-level questions
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:57














1












1








1





$begingroup$

With the help of some tips from @Omnomnomnom I found the answer. It is written in
"Shirshov's theorem and representations of semigroups" by A. Freedman, R.N. Gupta, R.M. Guralnick. Their Corollary 2.7 on page 163 deals with this problem.



The answer is: If the matrix algebras generated by the two sets are semisimple, then there is a desired common similarity transformation. The semisimple property was the one I was looking for.



Perhaps a simpler explanation is given in "On the Unitary Similarity of Matrix Families" by Yu. A. Al'pinKh. D. Ikramov. Theorem 1 is what I needed. Here the condition is the complete reducibility, which is equivalent to the semisimple property.






share|cite|improve this answer











$endgroup$



With the help of some tips from @Omnomnomnom I found the answer. It is written in
"Shirshov's theorem and representations of semigroups" by A. Freedman, R.N. Gupta, R.M. Guralnick. Their Corollary 2.7 on page 163 deals with this problem.



The answer is: If the matrix algebras generated by the two sets are semisimple, then there is a desired common similarity transformation. The semisimple property was the one I was looking for.



Perhaps a simpler explanation is given in "On the Unitary Similarity of Matrix Families" by Yu. A. Al'pinKh. D. Ikramov. Theorem 1 is what I needed. Here the condition is the complete reducibility, which is equivalent to the semisimple property.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 21:05

























answered Dec 19 '18 at 17:31









Balázs PozsgayBalázs Pozsgay

19818




19818












  • $begingroup$
    I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:42












  • $begingroup$
    Thanks! What is the difference?
    $endgroup$
    – Balázs Pozsgay
    Dec 19 '18 at 17:45










  • $begingroup$
    mathoverflow, though it gets less traffic, is better suited to research-level questions
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:57


















  • $begingroup$
    I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:42












  • $begingroup$
    Thanks! What is the difference?
    $endgroup$
    – Balázs Pozsgay
    Dec 19 '18 at 17:45










  • $begingroup$
    mathoverflow, though it gets less traffic, is better suited to research-level questions
    $endgroup$
    – Omnomnomnom
    Dec 19 '18 at 17:57
















$begingroup$
I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
$endgroup$
– Omnomnomnom
Dec 19 '18 at 17:42






$begingroup$
I'm glad that you found your answer. In the future, you may have more success posting a question like this on math overflow
$endgroup$
– Omnomnomnom
Dec 19 '18 at 17:42














$begingroup$
Thanks! What is the difference?
$endgroup$
– Balázs Pozsgay
Dec 19 '18 at 17:45




$begingroup$
Thanks! What is the difference?
$endgroup$
– Balázs Pozsgay
Dec 19 '18 at 17:45












$begingroup$
mathoverflow, though it gets less traffic, is better suited to research-level questions
$endgroup$
– Omnomnomnom
Dec 19 '18 at 17:57




$begingroup$
mathoverflow, though it gets less traffic, is better suited to research-level questions
$endgroup$
– Omnomnomnom
Dec 19 '18 at 17:57


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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