Jordan form of matrices So my professor gave me this question:
$A=\begin{pmatrix}
  0 & 2 & 5\\
  -5 & 5 & 10\\
   2 & -2 & -4 \\
  \end{pmatrix}$
I had to calculate $\forall 0 < i$ $kerA^{i}$ and $ImA^{i}$
So after calculating I reached those results:
$kerA$={$\begin{pmatrix}
  1 \\
  5 \\
   -2  \\
  \end{pmatrix}$}  
and $\forall 1<i$ 
$kerA^{i}$={$\begin{pmatrix}
  1 \\
  -1 \\
  0  \\
  \end{pmatrix}$,$\begin{pmatrix}
  -3 \\
  0 \\
  1  \\
  \end{pmatrix}$}
regarding the image it is just the span of the columns of $A^{i}$
Then he asked us to calculate $\forall 0 < i$ $ker(A-I)^{i}$ and $Im(A-I)^{i}$
So after calculating I reached those results:
$\forall 0<i$
$kerA^{i}=$ {$\begin{pmatrix}
  0 \\
  -5/2 \\
   1  \\
  \end{pmatrix}$}
regarding the image it is just the span of the columns of $(A-I)^{i}$
Then he asked us to show the Jordan form of $A$ 
How can I conclude that from what I proved. As far as I know the diagonal of jordan form contains the eigenvalues of $A$ (each eigenvalue repeat in the diagonal as the number of Algebraic multiplicities of the eigenvale ) and nothing more but how can I conclude from what I proved the eigenvalues of $A$.
Any help would be appreciated 
Thanks in advanced !!
 A: For a general method to compute the normal form in such case:
For each eigenvector, start with the power such that the kernel of $A-\lambda I$ does not change anymore (call that $m_\lambda$. Take a basis of a complement of $\operatorname{ker}(A-\lambda I)^{m_\lambda-1}$ in $\operatorname{ker}(A-\lambda I)^{m_\lambda}$. Now apply $(A-\lambda I)$ to that basis. The elements will lie in $\operatorname{ker}(A-\lambda I)^{m_\lambda -1}$. Now take a basis of the complement of the span of these vectors and $\operatorname{ker}(A-\lambda I)^{m_\lambda-2}$ in that space. Continue inductively. You will then get a basis for your whole space. If you change your matrix to that basis, you will get a matrix in Jordan normal form.
In your example (assuming your calculations are correct) take for example $b_0=(1,-1,0)^T$ and $Ab_0$ (for the eigenvalue $0$ and $b_1=(0,-5/2,1)^T$ for the eigenvalue $1$. Then your matrix represented in Jordan normal form will look like
$$\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix},$$
the first column corresponding to $Ab_0$, the second to $b_0$ and the third to $b_1$
