Mixing
Is there a function that defines the trace of a matrix A divided by its determinant?
$begingroup$
Given an $n times n$ invertible matrix $A$, does there exist a function $f(n,A)$ in any field of mathematics such that:
$$
f(n,A) = dfrac{mbox{tr}(A)}{det(A)}
$$
If so, has its behavior been studied in any depth? Also is there a connection between eigenvalues defining the trace and determinant of A and additive number theory?
matrices
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
$endgroup$
add a comment |
$begingroup$
Given an $n times n$ invertible matrix $A$, does there exist a function $f(n,A)$ in any field of mathematics such that:
$$
f(n,A) = dfrac{mbox{tr}(A)}{det(A)}
$$
If so, has its behavior been studied in any depth? Also is there a connection between eigenvalues defining the trace and determinant of A and additive number theory?
matrices
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
$endgroup$
1
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
5
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
1
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
4
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53
add a comment |
$begingroup$
Given an $n times n$ invertible matrix $A$, does there exist a function $f(n,A)$ in any field of mathematics such that:
$$
f(n,A) = dfrac{mbox{tr}(A)}{det(A)}
$$
If so, has its behavior been studied in any depth? Also is there a connection between eigenvalues defining the trace and determinant of A and additive number theory?
matrices
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
$endgroup$
Given an $n times n$ invertible matrix $A$, does there exist a function $f(n,A)$ in any field of mathematics such that:
$$
f(n,A) = dfrac{mbox{tr}(A)}{det(A)}
$$
If so, has its behavior been studied in any depth? Also is there a connection between eigenvalues defining the trace and determinant of A and additive number theory?
matrices
matrices
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
edited Jan 13 at 19:08
Mostafa Ayaz
15.6k3939
15.6k3939
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
asked Jul 2 '15 at 21:41
Fermat271Fermat271
284
284
1
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
5
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
1
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
4
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53
add a comment |
1
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
5
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
1
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
4
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53
1
1
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
5
5
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
1
1
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
4
4
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
when $A$ is a 2 by 2 matrix then we can define a 4 by 4 matrix $A'$ which is an operator on the 4 dimensional vector space of 2 by 2 matrices. The operator $A'$ is defined by the equation
$$A'(M)=(A')^{-1}MA$$
where $M$ is an arbitrary 2 by 2 matrix. Then it is the case that
$$Tr(A')=Tr(A)/ det(A)$$
I just figured this out this morning while doing part (a) of #6 on the August 2012 University of Maryland algebra qualifying exam. If you are interested in how this fact is applied to character theory / representation theory of finite groups you should read the full problem, it's a really neat question!
I don't know of any relation to additive number theory.
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
$endgroup$
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
add a comment |
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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
$begingroup$
when $A$ is a 2 by 2 matrix then we can define a 4 by 4 matrix $A'$ which is an operator on the 4 dimensional vector space of 2 by 2 matrices. The operator $A'$ is defined by the equation
$$A'(M)=(A')^{-1}MA$$
where $M$ is an arbitrary 2 by 2 matrix. Then it is the case that
$$Tr(A')=Tr(A)/ det(A)$$
I just figured this out this morning while doing part (a) of #6 on the August 2012 University of Maryland algebra qualifying exam. If you are interested in how this fact is applied to character theory / representation theory of finite groups you should read the full problem, it's a really neat question!
I don't know of any relation to additive number theory.
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
$endgroup$
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
add a comment |
$begingroup$
when $A$ is a 2 by 2 matrix then we can define a 4 by 4 matrix $A'$ which is an operator on the 4 dimensional vector space of 2 by 2 matrices. The operator $A'$ is defined by the equation
$$A'(M)=(A')^{-1}MA$$
where $M$ is an arbitrary 2 by 2 matrix. Then it is the case that
$$Tr(A')=Tr(A)/ det(A)$$
I just figured this out this morning while doing part (a) of #6 on the August 2012 University of Maryland algebra qualifying exam. If you are interested in how this fact is applied to character theory / representation theory of finite groups you should read the full problem, it's a really neat question!
I don't know of any relation to additive number theory.
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
$endgroup$
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
add a comment |
$begingroup$
when $A$ is a 2 by 2 matrix then we can define a 4 by 4 matrix $A'$ which is an operator on the 4 dimensional vector space of 2 by 2 matrices. The operator $A'$ is defined by the equation
$$A'(M)=(A')^{-1}MA$$
where $M$ is an arbitrary 2 by 2 matrix. Then it is the case that
$$Tr(A')=Tr(A)/ det(A)$$
I just figured this out this morning while doing part (a) of #6 on the August 2012 University of Maryland algebra qualifying exam. If you are interested in how this fact is applied to character theory / representation theory of finite groups you should read the full problem, it's a really neat question!
I don't know of any relation to additive number theory.
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
$endgroup$
when $A$ is a 2 by 2 matrix then we can define a 4 by 4 matrix $A'$ which is an operator on the 4 dimensional vector space of 2 by 2 matrices. The operator $A'$ is defined by the equation
$$A'(M)=(A')^{-1}MA$$
where $M$ is an arbitrary 2 by 2 matrix. Then it is the case that
$$Tr(A')=Tr(A)/ det(A)$$
I just figured this out this morning while doing part (a) of #6 on the August 2012 University of Maryland algebra qualifying exam. If you are interested in how this fact is applied to character theory / representation theory of finite groups you should read the full problem, it's a really neat question!
I don't know of any relation to additive number theory.
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
edited Jan 13 at 19:13
Taroccoesbrocco
5,42271839
5,42271839
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
answered Jan 13 at 18:48
Ian TeixeiraIan Teixeira
111
111
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
add a comment |
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
$begingroup$
Welcome to stackexchange. Interesting observation. Please format with mathjax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:54
add a comment |
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1
$begingroup$
What is interesting about it?
$endgroup$
– user230734
Jul 2 '15 at 21:43
5
$begingroup$
You just defined this function, so it certainly exists... Another way to describe it is that it is the sum of the eigenvalues divided by the product of the eigenvalues. What do you want to know about its behavior?
$endgroup$
– Lukas Geyer
Jul 2 '15 at 21:45
1
$begingroup$
@LukasGeyer: why would it be so? Eigenvalues of a matrix needn't be the diagonal entries unless it's triangular.
$endgroup$
– user230734
Jul 2 '15 at 21:46
4
$begingroup$
The trace is always the sum of the eigenvalues (listed by algebraic multiplicity) and the determinant is always the product of the eigenvalues (listed by algebraic multiplicity). See e.g. en.wikipedia.org/wiki/Characteristic_polynomial
$endgroup$
– Robert Israel
Jul 2 '15 at 21:53