$T$ is the linear operator on $P_2(\mathbb{R})$ defined by $T(f(x))=2f(x)-f'(x)$
1) Find a basis for a generalized eigenspace of $T$ consisting of a union of disjoint cycles of generalized eigenvectors
2) Find a Jordan Canonical Form for T
Let $\alpha = \{1, x, x^2\}$ be the standard ordered basis
$A= [T]_{\alpha}= \left[ {\begin{array}{ccc} 2&-1&0\\ 0&2&-2\\ 0&0&2\\ \end{array} } \right]$
$det\left(A-\lambda I\right) = \left(2-\lambda\right)^3$.
Thus, $\lambda=2$ and has a multiplicity of $m=3$
I know my dot diagram (one cycle of length 3) and thus was able to find the Jordan Canonical form easily: $ J= \left[ {\begin{array}{ccc} 2&1&0\\ 0&2&1\\ 0&0&2\\ \end{array} } \right]$
What I need help with is finding the basis for each generalized eigenspace... I found the answer in the book: $\beta=\{2,-2x,x^2\}$, but am not sure how to come to that
Since the length of the cycle is $3$, you start with an arbitrary non-zero vector $z_3\in\ker (A - 2I)^3 \setminus \ker (A - 2I)^2$.
The solutions says $z_3 = x^2$ (probably this is the "simplest" one). Then, you must take $z_2 = (A- 2I)z_3$ and $z_1 = (A - 2I)z_2$.
But! It is more important to understand, why this works.