How do I find Jordan basis? I have a matrix:
$$A=\begin{pmatrix}0&1&0\\-4&4&0\\-2&1&2\end{pmatrix}$$
solving $\det|A-\lambda{I}|$ I got characteristic polynom that equals to $(2-\lambda)^3 = 0$ for eigenvalue found two eigenvectors and one generalized eigenvector: $v_1=(1,2,0)\quad v_2=(0,0,1) \quad v_3=(1,0,0)$
What do I have to do to find Jordan basis here? (and what do I need to find Jordan basis in general, I mean is there appropriate alghoritm?, What I read did not make things more clear).
 A: Here is the way to go: consider the sequence of kernels:
$$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$
The sequence stops after step $2$ since
$$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix}
0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.
$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say
$$e_3=(0,1,0). $$
Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say
$$e'_1=(1,2,0).$$
The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the linear map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form:
$$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$
