Find Jordan normal form and basis 
Let $$A=M(\varphi)^{st}_{st}={\begin{bmatrix}0&1&1\\-4&-4&-2\\0&0&-2\end{bmatrix}}$$ and $ \varphi: \mathbb R^{3} \rightarrow \mathbb R^{3}$. Find the Jordan normal form $J_{A}$ for the matrix $A$ and a basis $X$ for the endomorphism $\varphi$ such that $J_{A}=M(\varphi)^{X}_{X}$.

I did this task but unfortunately my basis is not correct and I do not know where I'm making a mistake.
My try:
The appointment of $ J_ {A} $ is clear to me, so I will write only the result:
$$J_{A}={\begin{bmatrix}-2&1&0\\0&-2&0\\0&0&-2\end{bmatrix}}$$Then I am trying to find basis:
$$(A+2I)^{2}={\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}$$
$$\alpha_{2} \in \ker(\varphi+2id)^{2}- \ker(\varphi +2id)$$
$$\ker(\varphi+2id)^{2}=\mathbb R^{3}$$
$$\ker(\varphi+2id)=lin\left\{(1,-2,0),(0,-1,1)\right\}$$
From the above conclusions I think that I can take: 
$$\alpha_{2}=(0,1,0)$$
$$\alpha_{1}=(\varphi+2id)(\alpha_{2})=(1,-2,0)$$
As $\alpha_{3}$ I choose a linearly independent vector and I have for example:
$$\alpha_{3}=(0,0,1)$$
So my basis is:
$$X=\left\{(1,-2,0),(0,1,0),(0,0,1)\right\}$$
I know that basis can be different. That's why I checked my answer: $$J_{A}=M(\varphi)_{X}^{X}=M(id)^{X}_{st}M(\varphi)^{st}_{st}M(id)^{st}_{X}$$ From my basis I have: $$M(id)^{st}_{X}={\begin{bmatrix}1&0&0\\-2&0&1\\0&1&0\end{bmatrix}}$$Then I made a multiplication: $$M(id)^{X}_{st}M(\varphi)^{st}_{st}M(id)^{st}_{X}$$ and I get: $${\begin{bmatrix}-2&1&1\\0&-2&0\\0&0&-2\end{bmatrix}}$$Of course this is not equal $J_{A}$ so I know that I have a mistake.I would like to add that it is very strange for me that when I was curiosity I changed the vector $\alpha_{1}$ and $\alpha_{2}$ then I get $J_{A}$. However by my calculation I can't do it.
Can you help me and say where I am doing something wrong? I suspect that I make some mistake in determining $ \alpha_{2} $ but I still do not realize what.
 A: It looks like you've successfully obtained the data of $A$ and its Jordan form $J$ as
\begin{align*}
A &= \left[\begin{array}{rrr}
0 & 1 & 1 \\
-4 & -4 & -2 \\
0 & 0 & -2
\end{array}\right] & J &= \left[\begin{array}{rr|r}
-2 & 1 & 0 \\
0 & -2 & 0 \\
\hline
 0 & 0 & -2
\end{array}\right]
\end{align*}
Note that our table of eigenvalues is
$$
\begin{array}{c|c|c}
\lambda & \operatorname{am}_A(\lambda) & \operatorname{gm}_A(\lambda) \\ \hline
-2 & 3 & 2
\end{array}
$$
Here, $\operatorname{am}_A(-2)$ is the algebraic multiplicity of $-2$ as an eigenvalue of $A$ (the number of times $-2$ appears on the diagonal of $J$) and $\operatorname{gm}_A(-2)$ is the geometric multiplicity (the number of Jordan blocks in $J$ corresponding to $-2$).
Let's find the change of basis matrix using this algorithm described by Stefan Friedl.
First, we compute the numbers
$$
d_k = \operatorname{nullity}((A+2\cdot I)^k)-\operatorname{nullity}((A+2\cdot I)^{k-1})
$$
for $1\leq k\leq\operatorname{gm}_A(-2)=2$. Computing these numbers gives
\begin{align*}
d_1 &= 2 & d_2 &= 1
\end{align*}
We use these numbers to construct a diagram 
$$
\begin{array}{cc}
\Box & \Box \\
\Box 
\end{array}
$$
According to our algorithm, we start at the bottom of this diagram and fill the boxes in row $k$ with linearly independent vectors that belong to $\operatorname{Null}((A+2\,I)^k)$ but not $\operatorname{Null}((A+2\,I)^{k-1})$. As soon as a box is filled with a vector $\vec{v}$, we fill the box above with $(A+2\,I)\vec{v}$.
By noting that 
\begin{align*}
A+2\,I &= \left[\begin{array}{rrr}
2 & 1 & 1 \\
-4 & -2 & -2 \\
0 & 0 & 0
\end{array}\right] & (A+2\,I)^2 &= \left[\begin{array}{rrr}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{align*}
we see that our diagram can take the form
$$
\begin{array}{cc}
\fbox{$\left\langle2,\,-4,\,0\right\rangle$} & \fbox{$\left\langle0,\,1,\,-1\right\rangle$}\\
\fbox{$\left\langle1,\,0,\,0\right\rangle$}
\end{array}
$$
This defines our change of basis matrix $P$ as
$$
P=\left[\begin{array}{rrr}
2 & 1 & 0 \\
-4 & 0 & 1 \\
0 & 0 & -1
\end{array}\right]
$$
We can check our work by verifying that $A=PJP^{-1}$. Indeed, we have
$$
\overset{A}{\left[\begin{array}{rrr}
0 & 1 & 1 \\
-4 & -4 & -2 \\
0 & 0 & -2
\end{array}\right]}
=
\overset{P}{\left[\begin{array}{rrr}
2 & 1 & 0 \\
-4 & 0 & 1 \\
0 & 0 & -1
\end{array}\right]}
\overset{J}{\left[\begin{array}{rr|r}
-2 & 1 & 0 \\
0 & -2 & 0 \\
\hline
 0 & 0 & -2
\end{array}\right]}
\overset{P^{-1}}{\left[\begin{array}{rrr}
0 & -\frac{1}{4} & -\frac{1}{4} \\
1 & \frac{1}{2} & \frac{1}{2} \\
0 & 0 & -1
\end{array}\right]}
$$
