Find an element of $SO(1,2)$ such that is not diagonalisable in $\mathbb{C}$

Question

I am starting to work with Lie Groups and I was given the following exercise:

Find an example of $A$ be a $3 \times 3$ matrix with real coefficients such that preserves a inner product with signature $(1,2)$ such that $A$ is not diagonalisable.

My thoughts:

Characterising the whole $SO(1,2)$ by hand seems to be an complicated approach, as it is a nonlinear system with 9 variables. I suppose it could be done but does not seem to be very enlightening.

As any matrix in $\mathbb{C}$ has a Jordan form, I thought it was a good place to start. So I tried to compute what Jordan forms could be in $SO(1,2)$. Turns out that none of then do (beside the diagonal ones). The thing is that $SO(1,2)$ is not closed by conjugations of invertible matrices.

I know that would be easy to disprove in the case we were talking about $SO(n)$, but the signature changes everything.

Answer 1

A good approach is using a theorem due to Witt:

Lemma: Let $k$ be a field and $\operatorname{char}k\ne 2$. Let $\phi$ be a non-degenerate symmetric bilinear map $\phi:k^n\times k^n\to k$, let $W$ be a subspace of $k^n$ and let $A:W\to k^n$ be an injective map such that $\phi(Aw,Av)=\phi(w,v)$ for all $w,v\in W$. Then, there is a (necessarily injective) map $\widetilde A:k^n\to k^n$ such that $\phi\left(\widetilde Ax,\widetilde Ay\right)=\phi(x,y)$ for all $x,y\in k^n$ and $\left.\widetilde A\right\rvert_W=A$.

Now, we can choose a system of coordinates on $\Bbb R^3$ such that $$\phi(x,y)=x^t\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}y$$ Let's call $W=\operatorname{span}(e_1,e_2)$, so that $\phi^W:=\left.\phi\right\rvert_{W\times W}$ is the symmetric bilinear function $$a^t\begin{pmatrix}1&0\\0&0\end{pmatrix}b$$

Now, you can find rather easily an isometry $A:W\to W$ of $(W, \phi^W)$ which is not diagonalisable. Any extension of $A$ to $\Bbb R^3$ is a map the restriction of which to the invariant subspace $W$ is not diagonalisable. Thus, it cannot be diagonalised.

Added: If you're lost mid-way, the point of the reasoning above is giving an abstract reason for the existence of a real matrix $A=\begin{pmatrix}1&0&a\\ 1&1&b\\ 0&0&c\end{pmatrix}$ such that $A^t\begin{pmatrix}1&0&0\\0&0&1\\ 0&1&0\end{pmatrix}A=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}$. Of course, such a matrix can be found explicitly.