Orbits of the conjugation action of $GL_3 (mathbb{R})$ on the nonsingular symmetric $3times3$-matrices
$begingroup$
Let $S$ be the space of all symmetric $3 times 3$ matrices of full rank and with real entries. $GL_3 (mathbb{R})$ acts on this space by conjugation,
begin{align*}
g.A = (g^{-1})^T A g^{-1}, quad g in GL_3 (mathbb{R}), quad A in S.
end{align*}
I have read (in the book in the references, p. 165) that if $det A > 0$ (EDIT: And if $A$ is indefinite), then the orbit of $A$ contains the matrix
begin{align*}
sigma_0 = left( begin{matrix}
0 & 0 & 1\
0 & -1 & 0 \
1 & 0 & 0
end{matrix} right),
end{align*}
but for some reason I am stuck trying to prove this.
My attempt:
If $lambda_1, lambda_2, lambda_3$ are the eigenvalues of $A$, then $Q^T hspace{-0.1cm} A Q = text{diag}(lambda_1, lambda_2, lambda_3)$ for some $Q in O(3)$. Given that $det A > 0$, either all of the eigenvalues are positive, or exactly two of them are negative. So if
begin{align*}
B = left( begin{matrix}
text{sign } lambda_1 & 0 & 0\
0 & text{sign } lambda_2 & 0 \
0 & 0 & text{sign } lambda_3
end{matrix} right),
end{align*}
we can write $C^T Q^T cdot A cdot Q C = B$ where $C$ is the diagonal matrix whose entries are $1/|lambda_i|$ for $i = 1,2,3$. From this point, I hoped to be able to apply some permutation matrices to conjugate $B$ to an anti-diagonal matrix and then finally conjugate by some other matrix to ensure that the signs of this anti-diagonal matrix match those of $sigma_0$.
Any ideas? Thank you.
References: Ergodic Theory and Topological Dynamics of Group Actions on Homogeneous Spaces by M. Bekka and M. Mayer
matrices group-actions quadratic-forms symmetric-matrices
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
add a comment |
$begingroup$
Let $S$ be the space of all symmetric $3 times 3$ matrices of full rank and with real entries. $GL_3 (mathbb{R})$ acts on this space by conjugation,
begin{align*}
g.A = (g^{-1})^T A g^{-1}, quad g in GL_3 (mathbb{R}), quad A in S.
end{align*}
I have read (in the book in the references, p. 165) that if $det A > 0$ (EDIT: And if $A$ is indefinite), then the orbit of $A$ contains the matrix
begin{align*}
sigma_0 = left( begin{matrix}
0 & 0 & 1\
0 & -1 & 0 \
1 & 0 & 0
end{matrix} right),
end{align*}
but for some reason I am stuck trying to prove this.
My attempt:
If $lambda_1, lambda_2, lambda_3$ are the eigenvalues of $A$, then $Q^T hspace{-0.1cm} A Q = text{diag}(lambda_1, lambda_2, lambda_3)$ for some $Q in O(3)$. Given that $det A > 0$, either all of the eigenvalues are positive, or exactly two of them are negative. So if
begin{align*}
B = left( begin{matrix}
text{sign } lambda_1 & 0 & 0\
0 & text{sign } lambda_2 & 0 \
0 & 0 & text{sign } lambda_3
end{matrix} right),
end{align*}
we can write $C^T Q^T cdot A cdot Q C = B$ where $C$ is the diagonal matrix whose entries are $1/|lambda_i|$ for $i = 1,2,3$. From this point, I hoped to be able to apply some permutation matrices to conjugate $B$ to an anti-diagonal matrix and then finally conjugate by some other matrix to ensure that the signs of this anti-diagonal matrix match those of $sigma_0$.
Any ideas? Thank you.
References: Ergodic Theory and Topological Dynamics of Group Actions on Homogeneous Spaces by M. Bekka and M. Mayer
matrices group-actions quadratic-forms symmetric-matrices
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
1
$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59
add a comment |
$begingroup$
Let $S$ be the space of all symmetric $3 times 3$ matrices of full rank and with real entries. $GL_3 (mathbb{R})$ acts on this space by conjugation,
begin{align*}
g.A = (g^{-1})^T A g^{-1}, quad g in GL_3 (mathbb{R}), quad A in S.
end{align*}
I have read (in the book in the references, p. 165) that if $det A > 0$ (EDIT: And if $A$ is indefinite), then the orbit of $A$ contains the matrix
begin{align*}
sigma_0 = left( begin{matrix}
0 & 0 & 1\
0 & -1 & 0 \
1 & 0 & 0
end{matrix} right),
end{align*}
but for some reason I am stuck trying to prove this.
My attempt:
If $lambda_1, lambda_2, lambda_3$ are the eigenvalues of $A$, then $Q^T hspace{-0.1cm} A Q = text{diag}(lambda_1, lambda_2, lambda_3)$ for some $Q in O(3)$. Given that $det A > 0$, either all of the eigenvalues are positive, or exactly two of them are negative. So if
begin{align*}
B = left( begin{matrix}
text{sign } lambda_1 & 0 & 0\
0 & text{sign } lambda_2 & 0 \
0 & 0 & text{sign } lambda_3
end{matrix} right),
end{align*}
we can write $C^T Q^T cdot A cdot Q C = B$ where $C$ is the diagonal matrix whose entries are $1/|lambda_i|$ for $i = 1,2,3$. From this point, I hoped to be able to apply some permutation matrices to conjugate $B$ to an anti-diagonal matrix and then finally conjugate by some other matrix to ensure that the signs of this anti-diagonal matrix match those of $sigma_0$.
Any ideas? Thank you.
References: Ergodic Theory and Topological Dynamics of Group Actions on Homogeneous Spaces by M. Bekka and M. Mayer
matrices group-actions quadratic-forms symmetric-matrices
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
Let $S$ be the space of all symmetric $3 times 3$ matrices of full rank and with real entries. $GL_3 (mathbb{R})$ acts on this space by conjugation,
begin{align*}
g.A = (g^{-1})^T A g^{-1}, quad g in GL_3 (mathbb{R}), quad A in S.
end{align*}
I have read (in the book in the references, p. 165) that if $det A > 0$ (EDIT: And if $A$ is indefinite), then the orbit of $A$ contains the matrix
begin{align*}
sigma_0 = left( begin{matrix}
0 & 0 & 1\
0 & -1 & 0 \
1 & 0 & 0
end{matrix} right),
end{align*}
but for some reason I am stuck trying to prove this.
My attempt:
If $lambda_1, lambda_2, lambda_3$ are the eigenvalues of $A$, then $Q^T hspace{-0.1cm} A Q = text{diag}(lambda_1, lambda_2, lambda_3)$ for some $Q in O(3)$. Given that $det A > 0$, either all of the eigenvalues are positive, or exactly two of them are negative. So if
begin{align*}
B = left( begin{matrix}
text{sign } lambda_1 & 0 & 0\
0 & text{sign } lambda_2 & 0 \
0 & 0 & text{sign } lambda_3
end{matrix} right),
end{align*}
we can write $C^T Q^T cdot A cdot Q C = B$ where $C$ is the diagonal matrix whose entries are $1/|lambda_i|$ for $i = 1,2,3$. From this point, I hoped to be able to apply some permutation matrices to conjugate $B$ to an anti-diagonal matrix and then finally conjugate by some other matrix to ensure that the signs of this anti-diagonal matrix match those of $sigma_0$.
Any ideas? Thank you.
References: Ergodic Theory and Topological Dynamics of Group Actions on Homogeneous Spaces by M. Bekka and M. Mayer
matrices group-actions quadratic-forms symmetric-matrices
matrices group-actions quadratic-forms symmetric-matrices
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
edited Jan 11 at 11:59
Teddan the Terran
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
asked Jan 10 at 16:59
Teddan the TerranTeddan the Terran
1,206210
1,206210
1
$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59
add a comment |
1
$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59
1
1
$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That certainly is false. The action in question is a matrix congruence. By Sylvester's law of inertia, the inertia of a matrix is an invariant under congruence. This means every matrix in the orbit containing $sigma_0$ must have the same inertia as $sigma_0$, i.e. it must have one positive eigenvalues and two negative eigenvalues. Clearly, this is not always the case if you only require that $det(A)>0$, because $A$ may have three positive eigenvalues.
answered Jan 11 at 11:45
user1551user1551
73.4k566128
$endgroup$
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
add a comment |
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$begingroup$
That certainly is false. The action in question is a matrix congruence. By Sylvester's law of inertia, the inertia of a matrix is an invariant under congruence. This means every matrix in the orbit containing $sigma_0$ must have the same inertia as $sigma_0$, i.e. it must have one positive eigenvalues and two negative eigenvalues. Clearly, this is not always the case if you only require that $det(A)>0$, because $A$ may have three positive eigenvalues.
answered Jan 11 at 11:45
user1551user1551
73.4k566128
$endgroup$
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
add a comment |
$begingroup$
That certainly is false. The action in question is a matrix congruence. By Sylvester's law of inertia, the inertia of a matrix is an invariant under congruence. This means every matrix in the orbit containing $sigma_0$ must have the same inertia as $sigma_0$, i.e. it must have one positive eigenvalues and two negative eigenvalues. Clearly, this is not always the case if you only require that $det(A)>0$, because $A$ may have three positive eigenvalues.
answered Jan 11 at 11:45
user1551user1551
73.4k566128
$endgroup$
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
add a comment |
$begingroup$
That certainly is false. The action in question is a matrix congruence. By Sylvester's law of inertia, the inertia of a matrix is an invariant under congruence. This means every matrix in the orbit containing $sigma_0$ must have the same inertia as $sigma_0$, i.e. it must have one positive eigenvalues and two negative eigenvalues. Clearly, this is not always the case if you only require that $det(A)>0$, because $A$ may have three positive eigenvalues.
answered Jan 11 at 11:45
user1551user1551
73.4k566128
$endgroup$
That certainly is false. The action in question is a matrix congruence. By Sylvester's law of inertia, the inertia of a matrix is an invariant under congruence. This means every matrix in the orbit containing $sigma_0$ must have the same inertia as $sigma_0$, i.e. it must have one positive eigenvalues and two negative eigenvalues. Clearly, this is not always the case if you only require that $det(A)>0$, because $A$ may have three positive eigenvalues.
answered Jan 11 at 11:45
user1551user1551
73.4k566128
answered Jan 11 at 11:45
user1551user1551
73.4k566128
answered Jan 11 at 11:45
user1551user1551
73.4k566128
answered Jan 11 at 11:45
user1551user1551
73.4k566128
73.4k566128
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
add a comment |
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
$begingroup$
Thank you, you are absolutely right. As Will Jagy also pointed out, the claim in the book also relies on $A$ being indefinite. I just edited the question to emphasize this.
$endgroup$
– Teddan the Terran
Jan 11 at 12:01
add a comment |
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$begingroup$
You are missing a hypothesis. If there is a nonzero real vector $v$ such that $v^T Av = 0,$ add in $det A > 0,$ then you get the result.
$endgroup$
– Will Jagy
Jan 10 at 19:06
$begingroup$
Right, we want $A$ to be indefinite. Thank you!
$endgroup$
– Teddan the Terran
Jan 11 at 8:59