<s>[INST] Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$ I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan normal form).
I could find that $\lambda=3$ is $A$'s only eigenvalue and its algebraic multiplicity is $3$.
Afterwards, I solved the system $(A-3I_3)X=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and I got that $x=\begin{pmatrix} -a-b \\ a \\ b \end{pmatrix}$ for some parametres $a, b\in \mathbb{C}$.
I picked the eigenvectors $X_1=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $X_2=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ and then I got stuck. I tried solving both the equations $(A-3I_3)X'=X_1$ and $(A-3I_3)X''=X_2$, but these have no solutions, so I can't find any generalised eigenvectors. What should I do?
 [/INST] After you get the Jordan form (as psidaga's answer did)
\begin{align*}
J = \begin{pmatrix}
3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix},
\end{align*}
in order to determine the similarity matrix $P$ such that $P^{-1}AP = J$, it is is equivalent to solve the system
\begin{align*}
& A\alpha_1 = 3\alpha_1, \tag{1} \\
& A\alpha_2 = \alpha_1 + 3\alpha_2, \tag{2} \\
& A\alpha_3 = 3\alpha_3,
\end{align*}
where $\alpha_1, \alpha_2, \alpha_3$ are three columns of $P$. As you did, you can fix $\alpha_3$ to be $(-1, 0, 1)^T$. To determine $\alpha_1$ and $\alpha_2$, note that $A - 3I_{(3)}$ is nilpotent with nilpotent index $2$, i.e., $(A - 3I_{(3)})^2 = 0$, hence any non-zero vector $\alpha_2$ that is not already an eigenvector and $\alpha_1 := (A - 3I_{(3)})\alpha_2$ would satisfy equations $(1)$ and $(2)$, since $(A - 3I_{(3)})\alpha_1 = (A - 3I_{(3)})^2\alpha_2 = 0$. For example, let $\alpha_2 = (1, 0, 0)^T$, then $\alpha_1 = (1, -2, 1)^T$. It can be easily verified that
\begin{align*}
P = \begin{pmatrix}
1 & 1 & -1 \\
-2 & 0 & 0 \\
1 & 0 & 1 
\end{pmatrix}
\end{align*}
is invertible and satisfies $A = PJP^{-1}$.
Of course, $P$ is not unique, depending on what $\alpha_2$ you are choosing.
</s>