An explicit realization of the similarity of the transpose of a matrix in function field.
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
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
add a comment |
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
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
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
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
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
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
edited Dec 12 '16 at 8:23
user26857
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
edited Dec 12 '16 at 8:23
user26857
39.2k123882
edited Dec 12 '16 at 8:23
user26857
39.2k123882
edited Dec 12 '16 at 8:23
user26857
39.2k123882
39.2k123882
asked Dec 12 '16 at 1:23
Mathemagician
557411
asked Dec 12 '16 at 1:23
Mathemagician
557411
asked Dec 12 '16 at 1:23
Mathemagician
557411
557411
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12
add a comment |
1 Answer
1
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oldest
votes
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix},
$$
and
$$
A^T P= begin{pmatrix} frac{a(a-d)}b + c & a \ a & b end{pmatrix}.
$$
Thus, $PA=A^T P$, and $P$ is nonsingular.
answered Nov 24 at 2:40
i707107
11.9k21447
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
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix},
$$
and
$$
A^T P= begin{pmatrix} frac{a(a-d)}b + c & a \ a & b end{pmatrix}.
$$
Thus, $PA=A^T P$, and $P$ is nonsingular.
answered Nov 24 at 2:40
i707107
11.9k21447
add a comment |
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix},
$$
and
$$
A^T P= begin{pmatrix} frac{a(a-d)}b + c & a \ a & b end{pmatrix}.
$$
Thus, $PA=A^T P$, and $P$ is nonsingular.
answered Nov 24 at 2:40
i707107
11.9k21447
add a comment |
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix},
$$
and
$$
A^T P= begin{pmatrix} frac{a(a-d)}b + c & a \ a & b end{pmatrix}.
$$
Thus, $PA=A^T P$, and $P$ is nonsingular.
answered Nov 24 at 2:40
i707107
11.9k21447
We may take
$$
P=begin{pmatrix} frac{a-d}b & 1 \ 1 & 0 end{pmatrix}.
$$
Then
$$
PA=begin{pmatrix} frac{a(a-d)}b + c & a \ a & bend{pmatrix},
$$
and
$$
A^T P= begin{pmatrix} frac{a(a-d)}b + c & a \ a & b end{pmatrix}.
$$
Thus, $PA=A^T P$, and $P$ is nonsingular.
answered Nov 24 at 2:40
i707107
11.9k21447
answered Nov 24 at 2:40
i707107
11.9k21447
answered Nov 24 at 2:40
i707107
11.9k21447
answered Nov 24 at 2:40
i707107
11.9k21447
11.9k21447
add a comment |
add a comment |
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If $P$ works, then any scalar times $P$ will also work. So can't you just multiply $P$ with the product of the denominators?
– Jyrki Lahtonen
Dec 12 '16 at 7:47
Solve the linear equation $M^tP-PM = 0$.
– Marc Bogaerts
Dec 15 '16 at 17:12