Uniqueness of representations of matrix product states
$begingroup$
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
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
|
show 8 more comments
$begingroup$
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
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
$begingroup$
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
|
show 8 more comments
$begingroup$
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
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
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
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
edited Dec 18 '18 at 15:31
Omnomnomnom
128k791184
128k791184
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
asked Dec 18 '18 at 14:22
Balázs PozsgayBalázs Pozsgay
19818
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
|
show 8 more comments
$begingroup$
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
|
show 8 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
$endgroup$
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.
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
edited Dec 19 '18 at 21:05
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
answered Dec 19 '18 at 17:31
Balázs PozsgayBalázs Pozsgay
19818
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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