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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

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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
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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.

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  • $\begingroup$ 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

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