An explicit realization of the similarity of the transpose of a matrix in function field.
$begingroup$
Let $K=mathbb{F}(a,b,c,d)$ be the field of rational functions in four variables over a field $mathbb{F}$. The matrix
$$ A=left( {begin{array}{cc}
a & b \ c & d end{array} } right)$$ over $K$ is conjugate to its transpose. Hence there exists an invertible matrix $P$ over $K$ such that $A^t=PAP^{-1}$. What is an explicit formula for $P$, in terms of $a,b,c,d$ ? Can we choose $P$ to have polynomial entries in $a,b,c,d$ ?
linear-algebra abstract-algebra matrices
edited Dec 12 '16 at 8:23
user26857
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
$endgroup$
add a comment |
$begingroup$
Let $K=mathbb{F}(a,b,c,d)$ be the field of rational functions in four variables over a field $mathbb{F}$. The matrix
$$ A=left( {begin{array}{cc}
a & b \ c & d end{array} } right)$$ over $K$ is conjugate to its transpose. Hence there exists an invertible matrix $P$ over $K$ such that $A^t=PAP^{-1}$. What is an explicit formula for $P$, in terms of $a,b,c,d$ ? Can we choose $P$ to have polynomial entries in $a,b,c,d$ ?
linear-algebra abstract-algebra matrices
edited Dec 12 '16 at 8:23
user26857
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
$endgroup$
$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
$begingroup$
Let $K=mathbb{F}(a,b,c,d)$ be the field of rational functions in four variables over a field $mathbb{F}$. The matrix
$$ A=left( {begin{array}{cc}
a & b \ c & d end{array} } right)$$ over $K$ is conjugate to its transpose. Hence there exists an invertible matrix $P$ over $K$ such that $A^t=PAP^{-1}$. What is an explicit formula for $P$, in terms of $a,b,c,d$ ? Can we choose $P$ to have polynomial entries in $a,b,c,d$ ?
linear-algebra abstract-algebra matrices
edited Dec 12 '16 at 8:23
user26857
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
$endgroup$
Let $K=mathbb{F}(a,b,c,d)$ be the field of rational functions in four variables over a field $mathbb{F}$. The matrix
$$ A=left( {begin{array}{cc}
a & b \ c & d end{array} } right)$$ over $K$ is conjugate to its transpose. Hence there exists an invertible matrix $P$ over $K$ such that $A^t=PAP^{-1}$. What is an explicit formula for $P$, in terms of $a,b,c,d$ ? Can we choose $P$ to have polynomial entries in $a,b,c,d$ ?
linear-algebra abstract-algebra matrices
linear-algebra abstract-algebra matrices
edited Dec 12 '16 at 8:23
user26857
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
edited Dec 12 '16 at 8:23
user26857
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
edited Dec 12 '16 at 8:23
user26857
39.3k124183
edited Dec 12 '16 at 8:23
user26857
39.3k124183
edited Dec 12 '16 at 8:23
user26857
39.3k124183
39.3k124183
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
asked Dec 12 '16 at 1:23
MathemagicianMathemagician
577411
577411
$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12
$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An example of $P$
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix}=A^TP,
$$
and $P$ is nonsingular.
Polynomial Entries
As suggested by @Jyrki Lahtonen's comment, it is possible to choose $P$ to have polynomial entries by multiplying $b$. If $$
P=begin{pmatrix} a-d & b\ b & 0 end{pmatrix}.
$$
Then we have
$$
PA=begin{pmatrix} a(a-d) + bc & ab\ ab & b^2end{pmatrix}=A^TP.
$$
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An example of $P$
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix}=A^TP,
$$
and $P$ is nonsingular.
Polynomial Entries
As suggested by @Jyrki Lahtonen's comment, it is possible to choose $P$ to have polynomial entries by multiplying $b$. If $$
P=begin{pmatrix} a-d & b\ b & 0 end{pmatrix}.
$$
Then we have
$$
PA=begin{pmatrix} a(a-d) + bc & ab\ ab & b^2end{pmatrix}=A^TP.
$$
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
$endgroup$
add a comment |
$begingroup$
An example of $P$
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix}=A^TP,
$$
and $P$ is nonsingular.
Polynomial Entries
As suggested by @Jyrki Lahtonen's comment, it is possible to choose $P$ to have polynomial entries by multiplying $b$. If $$
P=begin{pmatrix} a-d & b\ b & 0 end{pmatrix}.
$$
Then we have
$$
PA=begin{pmatrix} a(a-d) + bc & ab\ ab & b^2end{pmatrix}=A^TP.
$$
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
$endgroup$
add a comment |
$begingroup$
An example of $P$
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix}=A^TP,
$$
and $P$ is nonsingular.
Polynomial Entries
As suggested by @Jyrki Lahtonen's comment, it is possible to choose $P$ to have polynomial entries by multiplying $b$. If $$
P=begin{pmatrix} a-d & b\ b & 0 end{pmatrix}.
$$
Then we have
$$
PA=begin{pmatrix} a(a-d) + bc & ab\ ab & b^2end{pmatrix}=A^TP.
$$
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
$endgroup$
An example of $P$
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix}=A^TP,
$$
and $P$ is nonsingular.
Polynomial Entries
As suggested by @Jyrki Lahtonen's comment, it is possible to choose $P$ to have polynomial entries by multiplying $b$. If $$
P=begin{pmatrix} a-d & b\ b & 0 end{pmatrix}.
$$
Then we have
$$
PA=begin{pmatrix} a(a-d) + bc & ab\ ab & b^2end{pmatrix}=A^TP.
$$
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
edited Jan 19 at 4:48
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
answered Nov 24 '18 at 2:40
i707107i707107
12.3k21547
12.3k21547
add a comment |
add a comment |
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$begingroup$
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
$endgroup$
– Jyrki Lahtonen
Dec 12 '16 at 7:47
$begingroup$
Solve the linear equation $M^tP-PM = 0$.
$endgroup$
– Marc Bogaerts
Dec 15 '16 at 17:12