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.

I assume that you mean $\Bbb R^3$ with $[x,y] = x \times y$ – Ben Grossmann Sep 21 '15 at 18:25 — Ben Grossmann, Sep 21 '15 at 18:25

Ben Grossmann · Accepted Answer · 2015-09-21T20:35:06.463

3

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 $0$.

edited Sep 21 '15 at 20:35

answered Sep 21 '15 at 18:23

Ben Grossmann

225,327

The Lie algebra of the cross product has brackets $[x,y]=z,[z,x]=y,[y,z]=x$, so $ad(x)=\begin{pmatrix} 0 & 0 & 0\cr 0 & 0 & -1\cr 0 & 1 & 0\end{pmatrix}$. But we have $ad(x)^2\neq 0$. So I am not sure. – Dietrich Burde Sep 21 '15 at 19:02
@DietrichBurde well... that wasn't so clever of me then. Fixed it. – Ben Grossmann Sep 21 '15 at 20:32
@DietrichBurde I had stupidly forgotten that we brackets fail to be associative. – Ben Grossmann Sep 21 '15 at 20:36

Dietrich Burde · Answer 2 · 2015-09-21T19:23:58.357

1

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.

edited Sep 21 '15 at 19:23

answered Sep 21 '15 at 19:09

Dietrich Burde

130,978

In $R^3, ad_x $ is not diagonalizable

2 Answers2