In $R^3, ad_x $ is not diagonalizable

Question

I'm doing HW in Lie algebra, there is one question ask me to prove that in $\mathbb R_\text{^}^3$ there is no non-zero element $x$ such that $ad_x$ is diagonalizable. I try to present it in some basis and prove eigenvalues are not in $\mathbb R$ but failed. I couldn't come up with any other thoughts. Can somebody give me some hints?

Thank you.

Answer 1

Hint: One helpful observation is that for all $y$, $(\operatorname{ad}_x^3 + \operatorname{ad}_x)y = 0$. If $\operatorname{ad}_x$ is diagonalizable over the reals, its only eigenvalue can be $0$.

Answer 2

The Lie algebra $L$ of the cross product is isomorphic to $\mathfrak{so}_3(\mathbb{R})$, and $ad(L)$ consists of $3\times 3$ real skew-symmetric matrices. Since the eigenvalues of a real nonzero skew-symmetric matrix are imaginary it is not possible to diagonalize them over $\mathbb{R}$ - except for the zero matrix.