How to get Jordan Form above the diagonal and not below it? Let $ M=
  \begin{bmatrix}
    0 & 0 & 0 \\
    1 & 1 &1 \\
    -1 & -1 & -1\\
  \end{bmatrix}$ $\in M_{3}(\mathbb R^{3})$
I found that $\chi_{M}=X^{3}$ and $\mu_{M}=X^{2}$, since $\dim_{\mathbb R}\operatorname{Im}(M-0\cdot\operatorname{Id})=1$
So $$\ker(M-0\cdot\operatorname{Id})= \ker\begin{bmatrix}
    0 & 0 & 0 \\
    1 & 1 &1 \\
    -1 & -1 & -1\\
  \end{bmatrix}=\left\{\begin{bmatrix}
    -1\\
    1 \\
    0 \\
  \end{bmatrix}, \begin{bmatrix}
    -1\\
    0 \\
    1 \\
  \end{bmatrix} \right\}\quad(*)$$
Now define $(M-0\cdot\operatorname{Id})=N$.
It follows:
$0 \subset \ker N \subset \mathbb R^{3}$. We can find a basis of the complement to $(*)$ in $\mathbb R^{3}$, say $\begin{bmatrix}
    1\\
    0 \\
    0 \\
  \end{bmatrix}$. Now,
$N\begin{bmatrix}
    1\\
    0 \\
    0 \\
  \end{bmatrix}=\begin{bmatrix}
    0\\
    1 \\
    -1 \\
  \end{bmatrix} \in \ker N$. Now we find an element in $\ker N$ such that $\{\begin{bmatrix}
    0\\
    1 \\
    -1 \\
  \end{bmatrix}, v_{1}\}$ is a basis of $\ker N$. Let's choose $v_{1}=\begin{bmatrix}
    -2\\
    1 \\
    1 \\
  \end{bmatrix}$. So our Jordan basis is $$\left\{\begin{bmatrix}
    1\\
    0 \\
    0 \\
  \end{bmatrix},
\begin{bmatrix}
    0\\
    1 \\
    -1 \\
  \end{bmatrix},
\begin{bmatrix}
    -2\\
    1 \\
    1 \\
  \end{bmatrix}\right\}$$
Problem: For $S=\begin{bmatrix}
    1 & 0 & -2\\
    0 & 1 & 1\\
    0 & -1 & 1\\
  \end{bmatrix}$
and $J=S^{-1}MS$
, I only get a Jordan Matrix $J$ with ones below the diagonal, I am aware that this is also the Jordan Form, nonetheless in our lectures we state that $J$ has ones above the diagonal. For peace of mind, I would like to know I am not getting ones above the diagonal but rather below the diagonal?
 A: You must construct the Jordan basis more carefully.
You already determined that the Jordan form consists of one $2 \times 2$ block and one $1 \times 1$ block, i.e. $\dim \ker M = 2$ and $\dim \ker M^2 = 3$.
Write a table like this $$
\begin{array}{ccc}
\ker M^2 \,\dot-\,\ker M^1 & f_2\\
 \ker M^1 \,\dot-\,\ker M^0 & f_1 & f_3\\
\end{array}
$$
Pick a vector $f_2 \in \ker M^2 \setminus \ker M^1$, such as $f_2 = \pmatrix{1 \\ 0 \\ 0}$ and define $f_1 = Mf_2 = \pmatrix{0 \\ 1 \\ -1}$.
Pick a vector $f_3 \in \ker M$ linearly independent with $f_1$, such as $f_3 = \pmatrix{-1 \\ 1 \\ 0}$. 
Now $$\{f_1, f_2, f_3\} = \left\{\pmatrix{0 \\ 1 \\ -1}, \pmatrix{1 \\ 0 \\ 0}, \pmatrix{-1 \\ 1 \\ 0}\right\}$$ is a Jordan basis for $M$, with the Jordan form above the diagonal. Indeed: $$Mf_1 = 0$$ $$Mf_2 = f_1$$ $$Mf_3 = 0$$
Put those vectors in a matrix $S$ and we have $$J = S^{-1}MS = \pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0}^{-1}\pmatrix{0 & 0 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & -1}\pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0} = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$$
A: If your method has produced  an $r\times r$ Jordan block
$$\left[\matrix{\lambda&0&0&0&0\cr
1&\lambda&0&0&0\cr
0&1&\lambda&0&0\cr
0&0&1&\lambda&0\cr
0&0&0&1&\lambda\cr}\right]\ ,$$
valid for some basis vectors $e_1$, $\ldots$, $e_r$ (part of a larger basis), then replace the $e_i$ $(1\leq i\leq r)$ by
$$f_i:=e_{r+1-i}\qquad(1\leq i\leq r)\ ,$$
and you obtain a Jordan block with the $1$s above the diagonal.
