Question on indefinite special orthogonal group I'm stuck on this question:
Construct an element $x \in SO(1,2)$ which is not diagonalizable.
Any help would be appreciated

You may construct $M\in M_3({\Bbb R})$ so that it maps (independent) vectors $u,v,w$ to e.g. $u,v+u,w+v$. It is then clearly in a Jordan form, so not diagonalizable. So suppose we have such a matrix. How to find $u,v,w$?
Writing $(x,y)=x^t g y$ with $g={\rm diag} (1,-1,-1)$ for the $SO(1,2)$ scalar product we need $M$ to preserve the scalar product. 
First, $(u,v)=(Mu,Mv)=(u,v+u)$ implies $(u,u)=0$ so the vector $u$ must be a null vector (in the light cone). It is unique up to normalization and rotations in the $y-z$ plane. We may take $u=[1 \ 1 \ 0]^t$. 
Second, $(v,v)=(v+u,v+u)  \Rightarrow (u,v)=0$ so $v$ must be ($g$)-orthogonal to $u$. Incidently this shows that the Jordan situation may not occur in $SO(1,1)$ because only $u$ is orthogonal to $u$ when $u$ is null in $SO(1,1)$. Anyway, we are in $SO(1,2)$ and
we may take $v=[0 \ 0 \ 1]^t$. This is unique up to adding multiples of $u$.
Finally $(w,v)=(w+v,v+u)$ and $(w,w)=(w+v,w+v)$ implies $2(w,v)+(v,v)=0$ and $(w,u)+(v,v)=0$ which leads to $w=[0 \ -1 \ -1/2]^t$. This is unique up to adding multiples of $u$ and $v$. Putting this together we want the matrix $M$ to verify:
$$ M 
\left( \begin{matrix} 1 & 0 & 0 \\ 1 & 0 & -1\\ 0 & 1 & -1/2 \end{matrix} \right) 
=\left( \begin{matrix} 1 & 1 & 0 \\ 1 & 1 & -1\\ 0 & 1 & 1/2 \end{matrix} \right) 
$$
and solving for $M$ yields:
$$ M =
\left( \begin{matrix} 1.5 & -0.5 & 1 \\ 0.5 & 0.5 & 1\\ 1 & -1 & 1 \end{matrix} \right) $$
You may verify that $M$ is the identity matrix + a nilpotent and that $M^t g M = g$ as wished.
Given the various choices above I think that the nil-manifold, i.e. the manifold of $M$'s for which $M$ is not diagonalizable has dimension $1+1+2=4$ in $M_3({\Bbb R})$. But if you only need one example, there is no need to worry about what the full set of solutions may be.
