Question about a Jordan form matirx. Let $A = \begin{pmatrix}
\lambda_1 &  &  &  &  &  &  &  & \\ 
 1&  \lambda_1&  &  &  &  &  &  & \\ 
 &  1&  \lambda_1&  &  &  &  &  & \\ 
 &  & 1 &  \lambda_1&  &  &  &  & \\ 
 &  &  & 1 & \lambda_1&  &  &  & \\ 
 &  &  &  & 0 & \lambda_2 &  &  & \\ 
 &  &  &  &  &1  &\lambda_2  &  & \\ 
 &  &  &  &  &  &1  &\lambda_2  & \\ 
 &  &  &  &  &  &  &1  &\lambda_2 
\end{pmatrix} \in M_{9x9}(F)$ 
All of the empty spots are $0's$ and $\lambda_1 \neq \lambda_2$
This is obviously a Jordan's canonical form, therefore the characteristic polynomial of $A$ is $(x-\lambda_1)^{5}(x-\lambda_2)^{4}$
How do I show that $Ker(T-\lambda_1Id)^{5} = span(e_1,e_2,...e_5)$
and that $Ker(T-\lambda_1Id)^{5} = Ker(T-\lambda_1Id)^{3}$
 A: Actually, I think there's a mistake, you can't have $Ker(T-\lambda_1 I)^5=Ker(T- \lambda_1 I)^3$, otherwise the maximum size of the blocks regarding the eigenvalue $\lambda_1$ would be $3$. As far as your question is concerned, I think it's just a property of matrices with all zeroes except the sovra/under diagonal (don't know its name),and the fact that your $T-\lambda_1 I$ is a block matrix like:
$$
\begin{bmatrix}
 A&0 \\ 
0&B 
\end{bmatrix}$$
Just try to prove that given a matrix with all zeroes and ones over the diagonal, if you square that matrix you end up with one with all zeroes and ones two positions over the diagonal. Hope you get what I mean.
EDIT: Just seen your comment. I am not familiar with the notation your book uses, but I bet that means that you have a block of size 3 and one of size 2 for $\lambda_1$ and two blocks of size 2 for $\lambda_2$. The matrix you've written would be, using your notation:
$$ \begin{bmatrix}
J(\lambda_1)^{(5)}&0\\0&J(\lambda_2)^{(4)}
\end{bmatrix}$$
or at least, I think
