Worked-out solution

$\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{-1 & 0\\1 & -1\\}$

1. $\quad$ The characteristic polynomial of $A$ is because of its lower triangle form
   $\quad$equal to $(\lambda+1)^2$
   $\quad$The eigenvalues of $A$ are zeros of the characteristic polynomial: $-1$.
   
   $\quad$ Let $\cv{v_1\\ v_2}$ be an eigenvector corresponding to the eigenvalue $-1$:
   $\quad$$\matrix{-1 & 0\\1 & -1\\}\cdot \cv{v_1\\ v_2}=-\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} -v_1 &= -v_1 \\[0.2cm] v_1-v_2 &= -v_2\end{aligned}\right.$
   $\quad$So: $v_1=0$ and an eigenvector corresponding to the eigenvalue $-1$ (with integral coefficients)
   $\quad$is $\vec{v}=\cv{0\\1}$.
2. $\quad$A generalised eigenvector $\vec{w}=\cv{w_1\\ w_2}$ corresponding to the eigenvalue $-1$ satisfies
   $\quad$$(A-\lambda\cdot I)\vec{w}=\vec{v}$ with $\lambda=-1$ and $\vec{v}=\cv{0\\ 1}$
   $\quad$So: $\matrix{0 & 0\\1 & 0\\}\cdot \cv{w_1\\ w_2}=\cv{0\\ 1}$.
   $\quad$Thus: $w_1=1$ and $w_2$ can be freely chosen.
   $\quad$Take $w_2=0$, then $\vec{w}=\cv{1\\ 0}$ is a generalised eigenvector.
   $\quad$The general solution is $$\begin{aligned}\vec{x}(t)&=\beta\, e^{-t}\cdot\vec{v}+ \alpha\, e^{-t}\cdot \left(t\cdot \vec{v}+\vec{w}\right) \\ \\ &= \beta\, e^{-t}\cdot\cv{0\\1}+ \alpha\, e^{-t}\cdot \left(t\cdot \cv{0\\1}+ \cv{1\\ 0}\right)\\ \\ &= \cv{\alpha\, e^{-t}\\ \beta\, e^{-t}+\alpha\, t\, e^{-t}}\end{aligned}$$ $\quad$ with constants $\alpha$ and $\beta$.
   
   
3. $\quad\vec{x}(0)=\cv{3\\2}$ means that $x_1(0)=3$ and $x_2(0)=2$.
   $\quad$Substitution in the general solution (b) gives $\alpha=3$ and $\beta=2$,
   $\quad$and thus
   $\vec{x}(t)=\cv{3e^{-t}\\ 2e^{-t}+3t e^{-t}}$.
   
   $\quad$Then: $\vec{x}(1)=\cv{3/e\\ 5/e},\quad \vec{x}(2)=\cv{3/e^2\\ 8/e^2}, \quad \vec{x}(4)=\cv{3/e^4\\ 14/e^4}, \quad \vec{x}(8)=\cv{3/e^8\\ 18/e^8}$.
   
   
4. $\quad\vec{x}(0)=\cv{1\\0}$ means that $x_1(0)=1$ and $x_2(0)=0$.
   $\quad$Substitution in the general solution (b) gives $\alpha=1$ and $\beta=0$,
   $\quad$and thus $\vec{x}(t)=\cv{e^{-t}\\ t e^{-t}}$.
   
   $\quad$Then: $\vec{x}(1)=\cv{1/e\\ 1/e},\quad \vec{x}(2)=\cv{1/e^2\\ 2/e^2}, \quad \vec{x}(4)=\cv{1/e\\ 4/e^4}, \quad \vec{x}(8)=\cv{1/e^8\\ 8/e^8}$.
   
   $\quad$When $\left\{\begin{aligned} x_1 &= e^{-t}\\ x_2 &= te^{-t}\end{aligned}\right.$ then $\ln(x_1)=-t$ and $x_2=t\cdot x_1$ So: $x_2=-x_1\ln(x_1)$
   
   
5. $\quad\vec{x}(0)=\cv{0\\1}$ means that $x_1(0)=0$ and $x_2(0)=1$.
   $\quad$Substitution in the general solution (b) gives $\alpha=0$ and $\beta=1$. $\quad$and thus $\vec{x}(t)=\cv{0\\ e^{-t}}$.
   
   $\quad$Then: $\vec{x}(1)=\cv{0\\ 1/e},\quad \vec{x}(2)=\cv{0\\ 1/e^2}, \quad \vec{x}(4)=\cv{0\\ 1/e^4}, \quad \vec{x}(8)=\cv{0\\ 1/e^8}$.
   
   $\quad$The orbit is $\left\{\cv{0\\y}\,\middle\vert\, y>0\right\}$.
   
   
6. $\quad\vec{x}(0)=\cv{\alpha\\ \beta}$ means $x_1(0)=\alpha$ and $x_2(0)=\beta$.
   $\quad$Substitution in the general solution (b) gives $\vec{x}(t)=\cv{\alpha\, e^{-t}\\ \beta\, e^{-t}+\alpha\, t\, e^{-t}}$.
   
   $\quad$When $\left\{\begin{aligned} x_1 &= \alpha\, e^{-t}\\ x_2 &= \beta\, e^{-t}+\alpha\, t\, e^{-t}\end{aligned}\right.$, then generically $t=\ln\left(\frac{\alpha}{x_1}\right)$ and $\alpha\cdot x_2=\beta\cdot x_1 + \alpha\cdot t\cdot x_1$.
   $\quad$The orbit has generically the equation $\alpha \cdot x_2 =\beta\cdot x_1+\alpha\cdot x_1\cdot \ln\left(\frac{\alpha}{x_1}\right)$
   $\quad$ 'Generically' means that we assume logarithms with a positive argument.
   $\quad$So we consider the case $\alpha>0$ and $x_1>0$ and the case $\alpha<0$ and $x_1<0$.
   $\quad$The case that $\alpha=0, \beta>0$ is dealt with in (e):
   $\quad$The trajectory is the positive vertical axis.
   $\quad$This is illustrated in the phase portrait below.
   $\quad$The equilibrium $(0,0)$ is attracting.
   $\quad$