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I'm stuck on this question:

Construct an element $x \in SO(1,2)$ which is not diagonalizable.

Any help would be appreciated

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    Im sorry it was no clear, it is the indefinite special orthogonal group – 2017-01-25
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    With $$I_{p,q} = \begin{pmatrix} I_p & 0 \\ 0 & -I_q \end{pmatrix}.$$ $$\mathfrak{so}(p,q) = \{X \in M_n(\mathbb{R}) : X^T I_{p,q} = -I_{p,q} X\}.$$ – 2017-01-25

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The group $SO(1,2)$ is isomorphic to $SL_2(\mathbb{R})$, which is perhaps more convenient to describe elements which are not diagonalisable over $\mathbb{R}$, such as parabolic or elliptic conjugacy classes, see here. For example, take $$ g=\begin{pmatrix} 1 & 1 \cr 0 & 1 \end{pmatrix}. $$

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    could you link me a proof of the isomorphism between $Sl_2(\mathbb R)$ and $SO(1,2)$ and what is the isomorphic equivalent of the matrix $g$ in $SL_2(\mathbb R)$ in $SO(1,2)$ – 2017-01-25
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    thanks for the answer anyway, im kind of lost in algebra – 2017-01-25
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    There are several links, e.g., [this one](http://math.stackexchange.com/questions/2081901/showing-that-so2-1-is-a-maximal-subgroup-of-sl3), or [here](http://www-users.math.umn.edu/~garrett/m/v/sporadic_isogenies.pdf). Of course, we could just write down all matrices in $SO(2,1)$ and then look at its Jordan form over the complex numbers. – 2017-01-25
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    I think this approach would be simpler to understand (for me) if anyone is patient enough to write it down – 2017-01-25
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    You could do the first step yourself, finding the explicit $3\times 3$-matrices for $SO(1,2)$. – 2017-01-25
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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.