A Symmetric bilinear variation for self congruence?
0
$begingroup$
--edited--
May I know the group or condition that satisfies $$S^T A S = A$$ possibly when $A_{ntimes n}$ is symmetric?
Meanwhile, below is an example for $S_{3times 3}$.
MATLAB code:
A = [2.9603 1.3855 1.8212 %symmetric
1.3855 4.7366 1.2824
1.8212 1.2824 2.653];
S = [-0.29412 -0.35451 0.71379
0.15026 1.0698 0.18701
1.15 -0.0036714 0.17226];
S'*A*S % returns symmetric matrix A as 'ans' below
% it's off a bit as I intentionally rounded the elements in the
% matrices above for the sake of communicating this.
% ans =
% 2.9604 1.3856 1.8212
% 1.3856 4.7367 1.2824
% 1.8212 1.2824 2.653
Thanks in advance.
bilinear-form
asked Nov 28 '18 at 5:57
KayKay
144
$endgroup$
|
show 1 more comment
0
$begingroup$
--edited--
May I know the group or condition that satisfies $$S^T A S = A$$ possibly when $A_{ntimes n}$ is symmetric?
Meanwhile, below is an example for $S_{3times 3}$.
MATLAB code:
A = [2.9603 1.3855 1.8212 %symmetric
1.3855 4.7366 1.2824
1.8212 1.2824 2.653];
S = [-0.29412 -0.35451 0.71379
0.15026 1.0698 0.18701
1.15 -0.0036714 0.17226];
S'*A*S % returns symmetric matrix A as 'ans' below
% it's off a bit as I intentionally rounded the elements in the
% matrices above for the sake of communicating this.
% ans =
% 2.9604 1.3856 1.8212
% 1.3856 4.7367 1.2824
% 1.8212 1.2824 2.653
Thanks in advance.
bilinear-form
asked Nov 28 '18 at 5:57
KayKay
144
$endgroup$
$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56
|
show 1 more comment
0
0
0
$begingroup$
--edited--
May I know the group or condition that satisfies $$S^T A S = A$$ possibly when $A_{ntimes n}$ is symmetric?
Meanwhile, below is an example for $S_{3times 3}$.
MATLAB code:
A = [2.9603 1.3855 1.8212 %symmetric
1.3855 4.7366 1.2824
1.8212 1.2824 2.653];
S = [-0.29412 -0.35451 0.71379
0.15026 1.0698 0.18701
1.15 -0.0036714 0.17226];
S'*A*S % returns symmetric matrix A as 'ans' below
% it's off a bit as I intentionally rounded the elements in the
% matrices above for the sake of communicating this.
% ans =
% 2.9604 1.3856 1.8212
% 1.3856 4.7367 1.2824
% 1.8212 1.2824 2.653
Thanks in advance.
bilinear-form
asked Nov 28 '18 at 5:57
KayKay
144
$endgroup$
--edited--
May I know the group or condition that satisfies $$S^T A S = A$$ possibly when $A_{ntimes n}$ is symmetric?
Meanwhile, below is an example for $S_{3times 3}$.
MATLAB code:
A = [2.9603 1.3855 1.8212 %symmetric
1.3855 4.7366 1.2824
1.8212 1.2824 2.653];
S = [-0.29412 -0.35451 0.71379
0.15026 1.0698 0.18701
1.15 -0.0036714 0.17226];
S'*A*S % returns symmetric matrix A as 'ans' below
% it's off a bit as I intentionally rounded the elements in the
% matrices above for the sake of communicating this.
% ans =
% 2.9604 1.3856 1.8212
% 1.3856 4.7367 1.2824
% 1.8212 1.2824 2.653
Thanks in advance.
bilinear-form
bilinear-form
asked Nov 28 '18 at 5:57
KayKay
144
asked Nov 28 '18 at 5:57
KayKay
144
edited Nov 29 '18 at 23:30
Kay
asked Nov 28 '18 at 5:57
KayKay
144
asked Nov 28 '18 at 5:57
KayKay
144
asked Nov 28 '18 at 5:57
KayKay
144
144
$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56
|
show 1 more comment
$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56
$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56
|
show 1 more comment
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$begingroup$
What do you mean by "group or condition"? Isn't $S^T A S = A$ precisely the condition on $S$? Note that if $A$ is too general, namely degenerate, then the set of $S$ satisfying the above condition won't form a group as they won't always be invertible (e.g. $A = 0$ puts no constraint on S). If $A$ is invertible and of signature (p,q), then by use of the spectral theorem one can show that the group of those $S$ satisfying $S^TAS = A$ is conjugated to the "orthogonal group in signature (p,q)", $O(p,q)$. But perhaps are you imposing other constraints on $S$, such as being symplectic?
$endgroup$
– Jordan Payette
Nov 29 '18 at 14:03
$begingroup$
@JordanPayette, thanks for the insight on spectral theorem. The conditions are that $A$ is Symmetric, positive definite, and invertible. No, it is not symplectic since simplectic matrices are of the order $2n times 2n$. That makes sense now. I do appreciate.
$endgroup$
– Kay
Nov 29 '18 at 17:21
$begingroup$
Of course $S$ can't be symplectic if your matrices are of odd size, but you mentioned symplectic matrices and you chose the symplectic-geometry tag, so that needed precisions. If $A_{n times n}$ is positive definite, then it only has positive eigenvalues ; by the spectral theorem it's of the form $A = M^TM$ where $M = D^{1/2}Q$ where $D$ is diagonal and formed by the eigenvalues of $A$ and $Q$ is orthogonal. Note that $M$ is invertible, since $A$ is positive definite. Then ${S : S^T A S = A } = { S : MSM^{-1} in O(n)} = M^{-1} O(n) M$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 17:34
$begingroup$
@JordanPayette. Thanks for the reply. However, I honestly couldn't understand the last part: ${ S : MSM^{-1} in O(n)} = M^{-1} O(n) M$ Can you KINDLY explain with words? Thanks.
$endgroup$
– Kay
Nov 29 '18 at 20:42
$begingroup$
Write $I_n$ for the identity matrix of size n and $O(n) = { Q in M(n, mathbb{R}) : Q^TQ = I_{n}}$ for the orthogonal group in dim n. Put differently $Q$ is orthogonal iff $Q^T I_n Q = I_n$. Now by the spectral theorem, for any symmetric positive definite matrix $A$ there are $Q$ orthogonal and $D$ diagonal s.t. $$A = Q^TDQ = Q^T D^{1/2} D^{1/2} Q = (D^{1/2}Q)^T(D^{1/2}Q) = M^TM.$$ Now $S$ satisfies $S^TAS = A$ iff $S^TM^TMS = M^TM$ iff $(M^T)^{-1}S^TM^TMSM^{-1} = I$ iff $MSM^{-1} in O(n)$ iff $S = M^{-1}PM$ for some $P$ orthogonal iff $S in M^{-1}O(n)M := {M^{-1}RM : R in O(n)}$.
$endgroup$
– Jordan Payette
Nov 29 '18 at 22:56