For a given $A \in \mathrm{O}_n(\mathbb{R})$, consider the map \begin{align*} \phi_A: \mathfrak{o}_n(\mathbb{R}) & \to \mathrm{Mat}_{n \times n}(\mathbb{R}) \\ x &\mapsto Ax-xA^T. \end{align*} Recall that $\mathfrak{o}_n(\mathbb{R})$, the Lie algebra of $\mathrm{O}_n(\mathbb{R})$, is isomorphic as a vector space to the space of $n \times n$ antisymmetric matrices . I've found myself stuck trying to answer the following question: for which $A$ is the kernel of $\phi_A$ trivial?
My observations so far:
- The image of $\phi_A$ is contained in the subspace of symmetric matrices.
- The kernel of $\phi_A$ is isomorphic to the $-1$ eigenspace of $(D \iota)_A$, the derivative of the map \begin{align*} \iota: \mathrm{O}_n(\mathbb{R}) & \to \mathrm{O}_n(\mathbb{R}) \\ B & \mapsto B^{-1} \end{align*} evaluated at $A$.
- There exist $A$ for which the kernel of $\phi_A$ is trivial (e.g., $n=2$ and $A$ is rotation by $\pi/2$), and there exist $A$ for which the kernel of $\phi_A$ is nontrivial (e.g., $n$ is arbitrary and $A$ is the identity).
My investigations so far lead me to believe that the fixed subspace of $A$ is playing a major role here, but making this intuition rigorous is proving difficult for me. S.O.S.! Any and all insights are welcome.
Hint. By a change of orthonormal basis, we may assume that $A=I_p\oplus-I_q\oplus R_{\theta_1}\oplus\cdots\oplus R_{\theta_m}$, where each each $R_{\theta_k}$ denotes a $2\times2$ rotation matrix for an angle $\theta_k\in(0,\pi)$. It is not hard to see that $Ax=xA^T$ has a nonzero skew symmetric solution if and only if $p\ge2,\ q\ge2$ or some two $\theta_k$s are equal to each other. In other words, $\phi_A$ has a non-trivial kernel if and only if $A$ has a repeated eigenvalue over $\mathbb C$.