# Orbits of the conjugation action of $GL_3 (\mathbb{R})$ on the nonsingular symmetric $3\times3$-matrices

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

• 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. – Will Jagy Jan 10 at 19:06
• Right, we want $A$ to be indefinite. Thank you! – Teddan the Terran Jan 11 at 8:59

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.
• 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. – Teddan the Terran Jan 11 at 12:01