Sea $T$ a linear operator on a subspace finite dimensional $V$ such that the Jordan Canonical Form is
$$\begin{pmatrix} 2 & 1 & 0 & 0 & 0 &0 & 0 &0 &0 & 0 \\ 0 & 2 & 1 & 0 & 0 &0 & 0 &0 &0 & 0 \\ 0 & 0 & 2 & 1 & 0 &0 & 0 &0 &0 & 0 \\ 0 & 0 & 0 & 2 & 0 &0 & 0 &0 &0 & 0 \\ 0 & 0 & 0 & 0 & 2 &0 & 0 &0 &0 & 0 \\ 0 & 0 & 0 & 0 & 0 &3 & 1 &0 &0 & 0 \\ 0 & 0 & 0 & 0 & 0 &0 & 3 &0 &0 & 0 \\ 0 & 0 & 0 & 0 & 0 &0 & 0 &3 &0 & 0 \\ 0 & 0 & 0 & 0 & 0 &0 & 0 &0 &0 & 1 \\ 0 & 0 & 0 & 0 & 0 &0 & 0 &0 &0 & 0 \\ \end{pmatrix}$$
i) Find the characteristic polinomyal
ii) For wich eingevalues the eigenspace is equal to the generalized eigenspaces?
iii) For each eigenvalue $\lambda_{i}$, find the minimal positive integer $p_{1}$ such that the generalized eigenspace $K_{\lambda_{i}}$ asociated to $\lambda_{1}$ match with the kernel of $(T-\lambda_{i}I)^{p_{1}}$, i.e $K_{\lambda_{i}}= Ker(T-\lambda_{i}I)^{p_{1}})$.
iv)Let $U_{i}$ the restriction of $T-\lambda_{i}I$ a $K_{\lambda_{i}}$ for each $i$. Find for each $i$, the dimension of $Ker(U_{i}),Ker(U_{i}^{2}),Ker(U_{i}^{3})$.
My solution
i) $p(x)=x^{2}(x-2)^{5}(x-3)^{3}$
ii) I think that for none of the three, because each one of them have a Jordan's block with size greater to $1$.
iii) The Jordan's basis of $\lambda_{1} ={0}$ only have two vector so $p_{1}=1$. For $\lambda_{2}=2$ we have that the jordan's block have size $4$ so the Jordan's basis is $\{v,(T-2I)v,(T-2I)^{2}v,(T-2I)^{3}v\}$, so $p_{2}=3$. By last $\lambda_{3}=3$ have $p_{3}=1$.
iv) I didn't understand how to do it, can you help me?
And can you recommend any lecture for thie theme?