I want to figure out the cases when $\det A=0.$ I know then one or both of the eigen values are zero. Hence there are three corresponding Jordan Canonical Forms. In Exercise 4 of the same section the author asks to establish eight different types of qualitative behavior.
Any help in figuring this out is much appreciated. Thank you.
Each example will contain the matrix for $x' = A x, \delta$ and $\tau$, solution for $x_1(t), x_2(t)$, statement of stability and phase portrait.
Cases for $\det A \ne 0$:
Saddle Point
$$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \implies \delta = -1 \lt 0 \implies ~\mbox{Unstable Saddle}$$
The solution for the system is $x_1(t) = c_1e^t, x_2(t) = c_2 e^{-t}$.
Stable Node
$$A = \begin{bmatrix} -1 & 0 \\ 0 & -4 \\ \end{bmatrix} \implies \tau = -5 \lt 0, \delta = 4, \tau^2 - 4 \delta \ge 0 \implies ~\mbox{Stable Node}$$
The solution for the system is $x_1(t) = c_1e^{-t}, x_2(t) = c_2 e^{-4t}$.
Unstable Node
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \\ \end{bmatrix} \implies \tau = 5 \gt 0, \delta = 4, \tau^2 - 4 \delta \ge 0 \implies ~\mbox{Unstable Node}$$
The solution for the system is $x_1(t) = c_1e^{t}, x_2(t) = c_2 e^{4t}$.
Unstable Focus
$$A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \\ \end{bmatrix} \implies \tau = 2 \gt 0, \tau \ne 0, \delta = 5, \tau^2 - 4 \delta \lt 0 \implies ~\mbox{Unstable Focus}$$
The solution for the system is $x_1(t) = e^t(c_1 \cos 2t + c_2 \sin 2t), x_2(t) = e^t(-c_1 \cos 2t + c_2 \sin 2t)$.
Stable Focus
$$A = \begin{bmatrix} -1 & 2 \\ -2 & -1 \\ \end{bmatrix} \implies \tau = -2 \lt 0, \tau \ne 0, \delta = 5, \tau^2 - 4 \delta \lt 0 \implies ~\mbox{Stable Focus}$$
The solution for the system is $x_1(t) = e^{-t}(c_1 \cos 2t + c_2 \sin 2t), x_2(t) = e^{-t}(-c_1 \sin 2t + c_2 \cos 2t)$.
Center
$$A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix} \implies \tau = 0, \delta = 2 \gt 0 \implies ~\mbox{Marginally/Neutrally Stable Center}$$
The solution for the system is: $x_1(t) = c_1 \cos t + c_2 \sin t, x_2(t) = -c_1 \sin t + c_2 \cos t$
Cases for $\det A = 0$:
Case $\lambda \gt 0 = 1$:
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}$$
The solution for the system is (you can plot solution curves) $x_1(t) = c_1 e^t, x_2(t) = c_2$. The phase portrait is
Case $\lambda \lt 0 = -1$: Note: we could easily collapse this case with the previous, just look at the direction arrows (that is why the author says eight).
$$A = \begin{bmatrix} -1 & 0 \\ 0 & 0 \\ \end{bmatrix}$$
The solution for the system is (you can plot solution curves) $x_1(t) = c_1 e^{-t}, x_2(t) = c_2$. The phase portrait is
Case Linear Solution
$$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$
The solution for the system is (you can plot solution curves) $x_1(t) = c_1 + c_2 t, x_2(t) = c_2$. The phase portrait is
Case Degenerate: All Zero Matrix:
$$A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$$
The solution for the system is (you can plot solution curves) $x_1(t) = c_1, x_2(t) = c_2$.
There is no phase portrait for this case because every point is a constant solution, but it is neutrally stable.
Aside:
If you want to find eigenvalues and eigenvectors, some of the cases requires a Jordan Form and some do not because you can find two linearly independent eigenvectors for the eigenvalues. For example:
$$A = \begin{bmatrix} \lambda & 0 \\ 0 & 0 \\ \end{bmatrix} \implies \lambda_1 = 0, \lambda_2 = \lambda, v_1 = (0, 1), v_2 = (1, 0)$$
$$A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} \implies \lambda_{1,2} = 0, v_1 = (1, 0), v_2 = (0,1)$$
The second example is a generalized eigenvector and this has a Jordan Form
$$A = PJP^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$$