Finding the Jordan Canonical form of a $6 \times 6$ matrix Find the Jordan Canonical Form of the following matrix $$\begin{bmatrix} 
1 & 0 & 0 & 0 & 0 & 0\\
1 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0 & 0\\
1 & 0 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 0\\
1 & 1 & 1 & 1 & 1 & 1\\ \end{bmatrix}$$
My try: I go about finding the Jordan Basis for this matrix. It is clear that $1$ is the only eigenvalue of this matrix. So $$A-I=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0 & 0\\
1 & 1 & 1 & 1 & 1 & 0\\ \end{bmatrix}$$
Moreover Rank$(A-I)^2=1$. Moreover Rank$(A-I)^3=0$ . So We don't need to go further on evaluating the powers of matrices in our search for generalized eigenvectors. 
$$(A-I)^2=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
4 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$
Also $(A-I)^3=0$.
It is seen that the generalized eigenspace (say $U$) consists of 
$U$=span$\{v_1=(0,1,0,0,0,0)^t,v_2=(0,0,1,0,0,0)^t,v_3=(0,0,0,1,0,0)^t,v_4=(0,0,0,0,1,0)^t,v_5=(0,0,0,0,0,1)^t\}$
Here I run into a little problem. Since $(A-I)v_i=v_5$ for each $i=1,2,3,4$, I have only five vectors with me for the Jordan Canonical Basis. Moreover I can't choose any other arbitrary vector linearly independent to these four simply because if $v$ were such a vector , then there is no $i \gt 0$ (and integer) such that $(A-I)v^{i}=0$
I have also another question here. Since $(A-I)^3=0$, why should I not choose general eigen vectors corresponding to $(A-I)^3$??
I am a little stuck here.
Thanks for the help!!!

From rank computations you find: there are four linearly independent eigenvectors ($rank(I-A)=2$), hence four Jordan blocks. The nilpotence index of $I-A$ is three, hence the largest Jordan block has size $3\times 3$. 
Thus, the other blocks are $1\times 1$ blocks.
You need a vector $v_6$ that satisfies 
$$
(I-A)^3 v_6=0, \ (I-A)^2v_6\ne0.
$$
Take
$$
v_6=\pmatrix{1&0&0&0&0&0}.
$$
Then $(I-A)^2v_6, (I-A)v_6$ are in the Jordan basis too, which are
$$
(I-A)^2v_6 = \pmatrix{0&0&0&0&0&1}^T=v_5, \\
(I-A)v_6 = \pmatrix{0&1&1&1&1&1}^T=:v_7.
$$
Now you have three linearly independent vectors. Complement this set with three eigenvectors to form a basis, for instance 
$$
\pmatrix{0&1&-1&0&0&0}^T, 
\pmatrix{0&1&0&-1&0&0}^T, 
\pmatrix{0&1&0&0&-1&0}^T.
$$
Arrange them in the right order to obtain the Jordan basis.
