compute the Jordan matrix $ \ \mathcal{M}(T) \ $ If the characteristic and minimal polynomial of a linear transformation $ \ T: V \to V \ $ are given by
$ \chi(\lambda) =(x-\lambda)^4 \ , \ m(\lambda)=(x-\lambda)^4 $ ,where $ \ \lambda \ $ is an eigenvalue , then find the dimension of the eigenspace $ \ E(\lambda) \ $ i.e., $ \ dim \ E(\lambda) \ $  and compute the Jordan matrix $ \ \mathcal{M}(T) \ $.
Answer:
Here $ \ V \ $ is a vector space over $ \ \mathbb{C} \ $. 
$ \chi(\lambda) =(x-\lambda)^4 \ , \\ m(\lambda)=(x-\lambda)^4,  $ 
Since characteristics polynomial of $ \ T \ $  is $ \ (x-\lambda)^4  \ $ , we have
$dim \ V=4 \ $
algebraic multiplicity of $ \ \lambda \ $  is =$4 \ $
But geometric multiplicity of $ \lambda \ \ is=1 \ $
Thus $ \ dim  \ E(\lambda) =1 $
what would be the jordan matrix?
Am I right so far?
 A: If $\chi(\lambda)=(x-\lambda)^4$, the vector space $V$ has  dimension $4$, and its Jordan form is one of
$$\begin{bmatrix}\lambda& 1&0 & 0 \\ 0 &\lambda& 1&0  \\ 0 & 0 &\lambda & 1 \\ 0 & 0 &0 &\lambda\end{bmatrix},\quad\begin{bmatrix}\lambda& 1&0 & 0 \\0 &\lambda& 1&0  \\ 0 & 0 &\lambda & 0 \\ 0 & 0 &0 &\lambda \end{bmatrix},
\quad\begin{bmatrix}\lambda& 1&0 & 0 \\0 &\lambda& 0&0 \\ 0 & 0 &\lambda & 0 \\ 0 & 0 &0 &\lambda\end{bmatrix},\quad\begin{bmatrix}\lambda& 0&0 & 0 \\0 &\lambda& 0&0 \\ 0 & 0 &\lambda & 0 \\ 0 & 0 &0 &\lambda\end{bmatrix}.$$
Only the first form corresponds to $m(\lambda)=(x-\lambda)^4$. The other three have minimal polynomial equal to $(x-\lambda)^3$,  $(x-\lambda)^2$ and $x-\lambda$ respectively.
So there is a single Jordan block, and we know the dimension of the eigenspace $E_\lambda$ is the number of Jordan blocks corresponding to the eigenvalue $\lambda$, we have
$$\dim E_\lambda=1.$$
A: In this case we have one Jordan block of size $4$ for $\lambda$ that is
\begin{bmatrix}\lambda&1&0&0\\0&\lambda&1&0\\0&0&\lambda&1\\0&0&0&\lambda\end{bmatrix}
A: $\chi(\lambda) =(x-\lambda)^4$ implies that $\dim V = 4$ and $\sigma(T) = \{4\}$.
$m(\lambda)=(x-\lambda)^4$ implies that the size of the largest block with the eigenvalue $\lambda$ is precisely $4$.
Since the sum of the sizes of all blocks has to be $\dim V = 4$, and we already have a block of size $4$, we conclude that the Jordan form consists only of that block:
$$\begin{bmatrix}\lambda&1&0&0\\0&\lambda&1&0\\0&0&\lambda&1\\0&0&0&\lambda\end{bmatrix}$$
The dimension of the eigenspace $\ker (T - \lambda I)$ is precisely the number of blocks with the eigenvalue $\lambda$. In this case it is $\dim\ker (T - \lambda I) = 1$ because we only have one $\lambda$-block.
