<s>[INST] Finding a Jordan Basis after finding the Jordan Canonical Form The question asked to find the Jordan Canonical Form and Jordan Basis of $\begin{bmatrix}1 & 1 & 0 & -1\\0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$, and after finding the characteristic polynomial, I used its eigenvalues and their multiplicities to find the JCF.
I found the JCF to be $\begin{bmatrix}0 & 0 & 0 & 0\\0 & 1 & 1 & 0\\0&0&1&1\\ 0 & 0 & 0 & 1\end{bmatrix}$, but I am unsure of how to find a Jordan Basis for this. 
Intuitively, I'd think the Jordan Basis for this would be the vectors that make up the three linearly independent columns, but my textbook doesn't really explicitly state what the basis is when they define the JCF. Is my intuition correct or am I way off? If so how can I find the Jordan Basis for this JCF?
 [/INST] let your matrix be $A$ and name $B = A - I.$ the characteristic polynomial says $A B^3 = 0.$ This is also the minimal polynomial. 
The method I like is to find a column vector $w$ with $B^3 w = 0$ but $B^2 w \neq 0.$ I like zeros and ones, so I am choosing
$$
w = 
\left(
\begin{array}{r}
0 \\
0 \\
0 \\
1 \\
\end{array}
\right)
$$
This is going to be the far right column, number 4. Next $v = B w$ with
$$
v = 
\left(
\begin{array}{r}
-1 \\
1 \\
0 \\
0 \\
\end{array}
\right)
$$
We reach a genuine eigenvector finally with $u = Bv$
$$
u = 
\left(
\begin{array}{r}
1 \\
0 \\
0 \\
0 \\
\end{array}
\right)
$$
because $Bu = B^2 v = B^3 w = 0,$ so $Au = u.$
The 0 eigenvector is the first column. Call the whole thing $R$
$$
R = 
\left(
\begin{array}{rrrr}
0&1&-1&0 \\
0&0&1&0 \\
1&0&0&0 \\
0&0&0&1 \\
\end{array}
\right)
$$
Next calculate $\det R = 1$ and
$$
R^{-1} = 
\left(
\begin{array}{rrrr}
0&0&1&0 \\
1&1&0&0 \\
0&1&0&0 \\
0&0&0&1 \\
\end{array}
\right)
$$
finally $J = R^{-1} A R$
</s>