# 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

There seems to be a typo. Judging from your computed kernel it should be $-4$ in the matrix. –  Julian Kuelshammer Jun 23 '13 at 8:58
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 $b_1$