To solve an automation engineering exercise, I need to find the Jordan normal form $J$ and a matrix $T$ so that $J=T^{-1}AT$, where $A$ is the initial matrix (given by the exercise). For example:
$$A = \begin{bmatrix}
-1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -4 & -4
\end{bmatrix} $$
To start, I easily found all the eigenvalues and eigenvectors of A:
$$λ_1=-1,\quad u_1=\begin{pmatrix}
1 \\
0 \\
0
\end{pmatrix},\quad λ_2=-2,\quad u_2=\begin{pmatrix}
0 \\
1 \\
-2
\end{pmatrix}$$
$λ_1$ has algebraic multiplicity 1, while $λ_2$ has algebraic multiplicity 2 and geometric multiplicity 1. I know that the first two columns of $T$ are $u_1$ and $u_2$, and that the third column must be a generalized eigenvector because there aren't any more eigenvectors. As far as I know, there are two methods to calculate them:
$$(A-λ_2 I)^2\cdot u_3=0\quad\textrm{or}\quad (A-λ_2 I)\cdot u_3=u_2$$
For simplicity's sake, I went with the second method:
$$\begin{pmatrix}
1&0&0 \\
0&2&1 \\
0&-4&-2
\end{pmatrix}\cdot\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=\begin{pmatrix}
0 \\
1 \\
-2
\end{pmatrix}\Rightarrow\begin{cases}
x=0\\
2y+z=1\\
-4y-2z=-2
\end{cases}$$
$$\Rightarrow\begin{cases}
x=0\\
y=1\\
z=1-2y=1-2\cdot1
\end{cases}\Rightarrow u_3=\begin{pmatrix}
0 \\
1 \\
-1
\end{pmatrix}$$
I chose $y=1$ because it's a valid choice, or so I understand. Now I should be able to calculate $T$, $T^{-1}$ and $J$:
$$T=\begin{pmatrix}
1&0&0 \\
0&1&1 \\
0&2&-1
\end{pmatrix},\quad T^{-1}=\frac{1}{3}\begin{pmatrix}
3&0&0 \\
0&1&1 \\
0&2&-1
\end{pmatrix}$$
$$\Rightarrow J=T^{-1}AT=\begin{pmatrix}
-1&0&0 \\
0&-10/3&-1/3 \\
0&16/3&-2/3
\end{pmatrix}$$
Which unfortunately isn't the Jordan normal form. It is a valid Jordan decomposition (from what I understand), but it's not what I'm looking for. The Jordan normal form should be:
$$J=\begin{pmatrix}
-1&0&0 \\
0&-2&1 \\
0&0&-2
\end{pmatrix}$$
With all the eigenvalues on the diagonal. If, however, I change my "valid choice" to $z=0$ and find $u_3=(0, -1/2, 0)^T$, then calculate $J=T^{-1}AT$, I get the Jordan normal form I need.
What am I doing wrong? Do I really need to pick an exact generalized eigenvector to make everything work? Please note that I also neeed $T$ and $T^{-1}$ (if I didn't need them, I would have just written the Jordan normal form directly, it's really easy). Are there better methods to calculate $J$ and $T$?
Extra context (not necessary for the question above)
$A$ is a matrix that represents how a dynamic system behaves. Without going into the details, the evolution in time of the system, given an initial state $x_0$, is:
$$x(t) = e^{A\cdot t}x_0 = Te^{J\cdot t}T^{-1}x_0$$
Where $J=T^{-1}AT$ is the Jordan normal form of $A$. Basically, the system is transformed using $T$ and $T^{-1}$ to make the matrix exponential $e^{A\cdot t}$ easier, since the exponential of a matrix in the Jordan normal form is the sum of the exponential of a diagonal matrix and the exponential of a nilpotent matrix, which are both trivial to calculate by hand.
Best Answer
You made a minor mistake: $$T=\begin{pmatrix} 1&0&0 \\ 0&1&1 \\ 0&\color{red}{-2}&-1 \end{pmatrix}\quad\Rightarrow\quad \color{red}{T^{-1}=\begin{pmatrix} 1&0&0 \\ 0&-1&-1 \\ 0&2&1 \end{pmatrix}}$$ and so $$J=T^{-1}AT=\color{red}{\begin{pmatrix} -1&0&0 \\ 0&-2&1 \\ 0&0&-2 \end{pmatrix}}.$$