I am looking for a Jordan basis for $\phi: \mathbb{R}^3 \to \mathbb{R}^3, \phi((x_1,x_2,x_3))=(4x_1-x_2,4x_1,8x_1-4x_2+2x_3)$.
The matrix of the transformation in the standard basis is
$$A=M(\phi)_{\mathcal{st}}= \begin{bmatrix} 4 & -1 & 0 \\ 4 & 0 & 0 \\ 8 & -4 & 2 \\ \end{bmatrix}$$
The characteristic polynomial is $p_\phi(\lambda)=(2-\lambda)^3$, $r((A-2I)^0)-r((A-2I)^1)=3-1=2$ so the Jordan matrix must have two blocks of size $\ge 1$. I choose one of the possible Jordan forms $$M(\phi)_{\mathcal{A}}= \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}$$ From this, I know that for vectors from $\mathcal{A}$ we must have $$\phi(\alpha_1)=2\alpha_1$$ $$\phi(\alpha_2)=\alpha_1+2\alpha_2$$ $$\phi(\alpha_3)=2\alpha_3$$ This means that $\alpha_1$ and $\alpha_3$ are the eigenvectors, for instance $\alpha_1=(1,2,0),\alpha_3=(0,0,1)$. Now $$\phi(\alpha_2)=\alpha_1+2\alpha_2$$ $$\phi(\alpha_2)-2\alpha_2=\alpha_1$$ $$\phi(\alpha_2)-2\text{id}(\alpha_2)=\alpha_1$$ $$(\phi-2\text{id})(\alpha_2)=\alpha_1$$ so $$(A-2I)\alpha_2=\alpha_1$$ $$\begin{bmatrix} 2 & -1 & 0 \\ 4 & -2 & 0 \\ 8 & -4 & 0 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix}$$ This system doesn't have a solution and we are one vector short. What am I missing?