Find a matrix $P$ such that $P^{-1}AP$ is in Jordan Normal Form. The matrix in question is:
$$A = \begin{bmatrix}
     2 && -1 \\
     4 && 6
\end{bmatrix}$$
This has characteristic and minimal polynomial $p(\lambda) = (\lambda - 4)^2$
An eigenvector is $(1,-2)$, from a one dimensional  eigenspace so not diagonalisable.
Don't I now have to find an eigenvector $(A-4\mathbb{I})^2v_2 = 0$?
but $(A-4\mathbb{I})^2 = 0$.
Now I am pretty stumped. I can not seem to find some good material on this.
 A: You have two options. 


*

*Since you already have an eigenvector $v_1=(1; -2)^\top$, just find a non-zero vector $v_2$ such that $(A-4 I)v_2 = v_1$. Then $v_1,v_2$ are $1$st and $2$nd columns of matrix $P$ respectively. 

*Or, since $(A-4 I)^2$ is the zero matrix, you could take the standard impulse basis $e_1=(1,0)^\top$, $e_2=(0,1)^\top$ and try to find $w=(A-4I)e_1$. If $w \neq \vec 0$ then $w,e_1$ is your basis, otherwise $v=(A-4I)e_2$ should be non-zero and the basis is $v,e_2$.
A: The only possible Jordan normal form for a non-diagonalizable 2x2 matrix is $${\bf A} = {\bf SJS}^{-1}\hspace{1cm}\textrm{if}\hspace{1cm} {\bf J} = \left[\begin{array}{cc}\lambda & 1\\ 0&\lambda\end{array}\right]$$ So you see by inspection $${\bf P} = {\bf S}^{-1}$$
So we now search for a vector $\bf v$ is so that $${\bf Av} = \lambda {\bf v} + 1{\bf w}, \textrm{   if  }  {\bf w} = [1,-2]^T$$
( The numbers 1 and $\lambda$ come from the positions in the $\bf J$ matrix above ). You can rewrite this as a linear equation system which you can solve. Then $\bf v$ and $\bf w$ will be the columns in the base change matrix.
