<s>[INST] Finding the Jordan Basis and the Canonical form corresponding to the Jordan basis \begin{pmatrix} 4 & 0 &0 \\ 2 &1 &3 \\ 5& 0 &4 \end{pmatrix}
I know that the Characteristic polynomial is : $$(t-4)^2(t-1)$$
I started with eigenvalues $λ=1$
 and got in the null space:
\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}
$λ=4$
 and got :
\begin{pmatrix} 0\\ -1 \\ 1 \end{pmatrix},
in $A- 4I$ 
and got 
\begin{pmatrix} -1\\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix},
in $(A- 4I)^2$
Which vector(s) do I choose correctly  for the Jordan Basis?
I chose $[0,1,0],[1,0,1]$ and got $b_3$ as $A-4I[b_2] = [0,5,5]$
What procedures should I use for choosing the correct Jordan basis vectors?
Secondly am I right?
Lastly, How do I get the Canonical form "corresponding to the basis"?
is it \begin{pmatrix} 1 & 0 &0 \\ 0 &4 &1 \\ 0& 0 &4 \end{pmatrix}
or  \begin{pmatrix} 4 & 1 &0 \\ 0 &4 &0 \\ 0& 0 &1 \end{pmatrix}
How do I set it up correctly?
 [/INST] You have the right idea, but you made a slight computational error: the $4$-eigenspace of $A$ is spanned by $(0,1,1)$, not $(0,-1,1)$. You correctly picked a vector $(1,0,1) \in \ker (A-4I)^2$ that doesn't belong to $\ker(A-4I)$, which let you construct the Jordan chain $((1,0,1),(A-4I)(1,0,1)) = ((1,0,1),(0,5,5))$. (This now works because $(0,5,5)$ is an eigenvector for the eigenvalue $4$.)
Then you just place the Jordan chains as columns of the change of basis matrix (reading a chain backwards if you want $1$s on the superdiagonal):
$$P = \begin{pmatrix} 0 & 0&1\\ 1&5&0 \\ 0&5&1 \end{pmatrix}.$$
And then we get
$$ P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{pmatrix}.$$
If instead you chose 
$$P = \begin{pmatrix} 0&1&0\\ 5&0&1 \\ 5&1&0 \end{pmatrix},$$
you'd get 
$$ P^{-1}AP = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
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