Orthogonal transformation with additional constraints
$begingroup$
Let $A$ be an orthogonal matrix, i.e. $AA^{T}=mathbb{I}$. It is given that $A$ satisfies an additional constraint, $AMA^{T}=PMP^{T}$, where $P$ is some permutation matrix and $M_{ij}=sgn(i-j)$. Can $A$ be still an arbitrary orthogonal matrix or are there any new constraints on it? In two dimension there are no new restrictions.
linear-algebra matrices linear-transformations orthogonal-matrices
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
$endgroup$
add a comment |
$begingroup$
Let $A$ be an orthogonal matrix, i.e. $AA^{T}=mathbb{I}$. It is given that $A$ satisfies an additional constraint, $AMA^{T}=PMP^{T}$, where $P$ is some permutation matrix and $M_{ij}=sgn(i-j)$. Can $A$ be still an arbitrary orthogonal matrix or are there any new constraints on it? In two dimension there are no new restrictions.
linear-algebra matrices linear-transformations orthogonal-matrices
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
$endgroup$
add a comment |
$begingroup$
Let $A$ be an orthogonal matrix, i.e. $AA^{T}=mathbb{I}$. It is given that $A$ satisfies an additional constraint, $AMA^{T}=PMP^{T}$, where $P$ is some permutation matrix and $M_{ij}=sgn(i-j)$. Can $A$ be still an arbitrary orthogonal matrix or are there any new constraints on it? In two dimension there are no new restrictions.
linear-algebra matrices linear-transformations orthogonal-matrices
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
$endgroup$
Let $A$ be an orthogonal matrix, i.e. $AA^{T}=mathbb{I}$. It is given that $A$ satisfies an additional constraint, $AMA^{T}=PMP^{T}$, where $P$ is some permutation matrix and $M_{ij}=sgn(i-j)$. Can $A$ be still an arbitrary orthogonal matrix or are there any new constraints on it? In two dimension there are no new restrictions.
linear-algebra matrices linear-transformations orthogonal-matrices
linear-algebra matrices linear-transformations orthogonal-matrices
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
edited Dec 11 '18 at 6:49
SanJoy
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
asked Dec 10 '18 at 14:23
SanJoySanJoy
215
215
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
No. The matrix $PMP^T$ has integer entries. That's not usually the case with $AMA^T$. For instance, with
$$
A=begin{bmatrix}
tfrac1{sqrt{14}}&tfrac3{sqrt{14}}&tfrac2{sqrt{14}}\
tfrac5{sqrt{42}}&tfrac1{sqrt{42}}&-tfrac4{sqrt{42}}\
tfrac1{sqrt3}&-tfrac1{sqrt3}&tfrac1{sqrt3}
end{bmatrix},
$$
The 1,2 entry of $AMA^T$ is $sqrt3$.
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
add a comment |
$begingroup$
Your claim is wrong even when $n=2$. E.g. when $P=I_2$, we need $AM=MA$. Since $M$ in this case is precisely the rotation matrix $R$ for angle $pi/2$, it only commutes with scalar multiples of rotation matrices. Therefore $A$ must have determinant $1$. Similarly, if $P$ is a transposition matrix, $A$ must have determinant $-1$.
In general, since $M$ is a skew symmetric matrix with distinct eigenvalues (namely, $tanleft(frac{kpi}{2n}right)i$ for $k=1-n,,3-n,ldots,,n-3,,n-1$), it is orthogonally similar to a direct sum of different nonzero scalar multiples of $R$ (and also the scalar $0$ if $n$ is odd). It follows that $P^TA$, via the same similarity transform, must be equal to a direct sum of $2times2$ rotation matrices (and also a scalar $pm1$ when $n$ is odd).
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
$endgroup$
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. The matrix $PMP^T$ has integer entries. That's not usually the case with $AMA^T$. For instance, with
$$
A=begin{bmatrix}
tfrac1{sqrt{14}}&tfrac3{sqrt{14}}&tfrac2{sqrt{14}}\
tfrac5{sqrt{42}}&tfrac1{sqrt{42}}&-tfrac4{sqrt{42}}\
tfrac1{sqrt3}&-tfrac1{sqrt3}&tfrac1{sqrt3}
end{bmatrix},
$$
The 1,2 entry of $AMA^T$ is $sqrt3$.
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
add a comment |
$begingroup$
No. The matrix $PMP^T$ has integer entries. That's not usually the case with $AMA^T$. For instance, with
$$
A=begin{bmatrix}
tfrac1{sqrt{14}}&tfrac3{sqrt{14}}&tfrac2{sqrt{14}}\
tfrac5{sqrt{42}}&tfrac1{sqrt{42}}&-tfrac4{sqrt{42}}\
tfrac1{sqrt3}&-tfrac1{sqrt3}&tfrac1{sqrt3}
end{bmatrix},
$$
The 1,2 entry of $AMA^T$ is $sqrt3$.
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
add a comment |
$begingroup$
No. The matrix $PMP^T$ has integer entries. That's not usually the case with $AMA^T$. For instance, with
$$
A=begin{bmatrix}
tfrac1{sqrt{14}}&tfrac3{sqrt{14}}&tfrac2{sqrt{14}}\
tfrac5{sqrt{42}}&tfrac1{sqrt{42}}&-tfrac4{sqrt{42}}\
tfrac1{sqrt3}&-tfrac1{sqrt3}&tfrac1{sqrt3}
end{bmatrix},
$$
The 1,2 entry of $AMA^T$ is $sqrt3$.
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
No. The matrix $PMP^T$ has integer entries. That's not usually the case with $AMA^T$. For instance, with
$$
A=begin{bmatrix}
tfrac1{sqrt{14}}&tfrac3{sqrt{14}}&tfrac2{sqrt{14}}\
tfrac5{sqrt{42}}&tfrac1{sqrt{42}}&-tfrac4{sqrt{42}}\
tfrac1{sqrt3}&-tfrac1{sqrt3}&tfrac1{sqrt3}
end{bmatrix},
$$
The 1,2 entry of $AMA^T$ is $sqrt3$.
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
answered Dec 10 '18 at 15:32
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
add a comment |
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
$begingroup$
I would like to what precisely this constraints do for an arbitrary n. Or put it in another way what all matrices will respect this constraint.
$endgroup$
– SanJoy
Dec 10 '18 at 19:59
add a comment |
$begingroup$
Your claim is wrong even when $n=2$. E.g. when $P=I_2$, we need $AM=MA$. Since $M$ in this case is precisely the rotation matrix $R$ for angle $pi/2$, it only commutes with scalar multiples of rotation matrices. Therefore $A$ must have determinant $1$. Similarly, if $P$ is a transposition matrix, $A$ must have determinant $-1$.
In general, since $M$ is a skew symmetric matrix with distinct eigenvalues (namely, $tanleft(frac{kpi}{2n}right)i$ for $k=1-n,,3-n,ldots,,n-3,,n-1$), it is orthogonally similar to a direct sum of different nonzero scalar multiples of $R$ (and also the scalar $0$ if $n$ is odd). It follows that $P^TA$, via the same similarity transform, must be equal to a direct sum of $2times2$ rotation matrices (and also a scalar $pm1$ when $n$ is odd).
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
$endgroup$
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
add a comment |
$begingroup$
Your claim is wrong even when $n=2$. E.g. when $P=I_2$, we need $AM=MA$. Since $M$ in this case is precisely the rotation matrix $R$ for angle $pi/2$, it only commutes with scalar multiples of rotation matrices. Therefore $A$ must have determinant $1$. Similarly, if $P$ is a transposition matrix, $A$ must have determinant $-1$.
In general, since $M$ is a skew symmetric matrix with distinct eigenvalues (namely, $tanleft(frac{kpi}{2n}right)i$ for $k=1-n,,3-n,ldots,,n-3,,n-1$), it is orthogonally similar to a direct sum of different nonzero scalar multiples of $R$ (and also the scalar $0$ if $n$ is odd). It follows that $P^TA$, via the same similarity transform, must be equal to a direct sum of $2times2$ rotation matrices (and also a scalar $pm1$ when $n$ is odd).
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
$endgroup$
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
add a comment |
$begingroup$
Your claim is wrong even when $n=2$. E.g. when $P=I_2$, we need $AM=MA$. Since $M$ in this case is precisely the rotation matrix $R$ for angle $pi/2$, it only commutes with scalar multiples of rotation matrices. Therefore $A$ must have determinant $1$. Similarly, if $P$ is a transposition matrix, $A$ must have determinant $-1$.
In general, since $M$ is a skew symmetric matrix with distinct eigenvalues (namely, $tanleft(frac{kpi}{2n}right)i$ for $k=1-n,,3-n,ldots,,n-3,,n-1$), it is orthogonally similar to a direct sum of different nonzero scalar multiples of $R$ (and also the scalar $0$ if $n$ is odd). It follows that $P^TA$, via the same similarity transform, must be equal to a direct sum of $2times2$ rotation matrices (and also a scalar $pm1$ when $n$ is odd).
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
$endgroup$
Your claim is wrong even when $n=2$. E.g. when $P=I_2$, we need $AM=MA$. Since $M$ in this case is precisely the rotation matrix $R$ for angle $pi/2$, it only commutes with scalar multiples of rotation matrices. Therefore $A$ must have determinant $1$. Similarly, if $P$ is a transposition matrix, $A$ must have determinant $-1$.
In general, since $M$ is a skew symmetric matrix with distinct eigenvalues (namely, $tanleft(frac{kpi}{2n}right)i$ for $k=1-n,,3-n,ldots,,n-3,,n-1$), it is orthogonally similar to a direct sum of different nonzero scalar multiples of $R$ (and also the scalar $0$ if $n$ is odd). It follows that $P^TA$, via the same similarity transform, must be equal to a direct sum of $2times2$ rotation matrices (and also a scalar $pm1$ when $n$ is odd).
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
answered Dec 10 '18 at 17:43
user1551user1551
72.5k566127
72.5k566127
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
add a comment |
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
$begingroup$
I mean to say that, when n=2, A can be either a rotation when Det(A)=1, or a reflection matrix when Det(A)=-1. In the first case, P is identity and second case P is the nontrivial permutation matrix. So any orthogonal matrix belongs to either of the classes.
$endgroup$
– SanJoy
Dec 10 '18 at 19:00
add a comment |
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