find the Jordan form and $P$ such that $P^{-1}AP = J$. Consider the matrix $$A = \left(\begin{array}{cccc} -11&0&-9\\32&1&24\\16&0&13 \end{array}\right)$$
I want to find the Jordan form of $A$, with $1$-s at the bottom and the jordan basis, which is $P$ columns such that $P^{-1}AP = J$.
I evaluated the charechteristic polynomial which is $f_A(x) = -(x-1)^3$. So we have one eigenvalue which is $\lambda =1$.
Now, 
$$A-I = \left(\begin{array}{cccc} -12&0&-9\\32&0&24\\16&0&12 \end{array}\right)$$
$$\ker (A-I) = \text{span}\{(0,1,0),(1,0,3/4)\}$$
$$\ker (A-I)^2 =\mathbb{R}^3$$
As I understand I need to find a vector, $v_3$ such that $\ker (A-I) \cup v_3 = \ker (A-I)^2 = \mathbb{R}^3$.
Then, $$v_2 = (A-I)v_3 \\ v_1 = (A-I)v_2$$
And we (suppose to) get $P = (v_1,v_2,v_3)$ such that $P^{-1}AP = J$.
I chose $v_3 = (1,0,0)$, but it doesn't work, and for the matter of fact I'm confused by all this.
I'd be glad if you could explain to me how to correct this.
Thanks.

Saying that 1 is an eigenvalue for matrix $A$ means that there exist non-zero vectors, $v$, such that $Av= v$ or $(A- I)v= 0$.
Here, that means $$\begin{bmatrix}-12 & 0 & -9 \\ 32 & 0 & 24 \\ 16 & 0 & 12\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}\begin{bmatrix}-12x- 9z \\ 32x+ 24z \\ 16x+ 12z \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$
You basically have that. Now note that this gives you the three equations $$-12x- 9z= 0,$$ $$32x+ 24z= 0,$$ and $$16x+ 12z= 0.$$  From the first equation, $9z= -12x$ so $z= (-12/9)x= (-4/3)x.$  Setting $z= (-4/3)x$ in the second and third equations gives $$32x+ 24(-4/3)x= 32x- 32x= 0$$ and $$16x+ 12z= 16x+ 12(-4/3)x= 16x- 16x= 0.$$
That is, for any value of $x, z= (-4/3)x$ will satisfy all those equations.  Since $y$ did not enter into this, $y$ can be anything.  That is, the eigenvectors must be of the  form $$\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix} x \\ y \\ (-4/3)x \end{pmatrix}= x\begin{pmatrix}1 \\ 0 \\ -4/3\end{pmatrix}+ y\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}.$$  
We can use $\begin{pmatrix}3\\ 0\\ -4\end{pmatrix}=3\cdot \begin{pmatrix}1\\ 0\\ -4/3\end{pmatrix}$ and $\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}$ form a basis for the Eigen space.  since that is a two dimensional subspace, the "Jordan Normal Form" will be of the form $$J= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 0 \\0 & 0 & 3\end{bmatrix}$$ and we can form the matrix $$P^{-1}$$ using the vectors $$\begin{pmatrix}3 \\ 0\\ -4\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}$$ as columns.  
