Is it possible to describe AR(1) as stable in terms of a unit triangle as in the explanation below or not because a lack of a second degree characteristic polynomial?
Consider the equation $$ \lambda^2-\phi_1\lambda-\phi_2=0 $$
If $z$ is a root of the "standard" characteristic equation $1-\phi_1 > z-\phi_2 z^2=0$ and setting $z^{-1}=\lambda$, the display obtains from rewriting the standard one as follows: \begin{eqnarray*} 1-\phi_1 > z-\phi_2 z^2&=&0\\ \Rightarrow z^{-2}-\phi_1 z^{-1}-\phi_2 &=&0\\ > \Rightarrow \lambda^2-\phi_1\lambda -\phi_2 &=&0 \end{eqnarray*} Hence, an alternative condition for stability of an $AR(2)$ is that all roots of the first display are inside the unit circle, $|z|>1 > \Leftrightarrow |\lambda|=|z^{-1}|<1$.
We use this representation to derive the stationarity triangle of an $AR(2)$ process, that is that an $AR(2)$ is stable if the following three conditions are met:
- $\phi_2<1+\phi_1$
- $\phi_2<1-\phi_1$
- $\phi_2>-1$
Recall that you can write the roots of the first display (if real) as $$ > \lambda_{1,2}=\frac{\phi_1\pm\sqrt{\phi_1^2+4\phi_2}}{2} > $$ to find the first two conditions.
Then, the $AR(2)$ is stationary iff $|\lambda|<1$, hence (if the $\lambda_i$ are real): \begin{eqnarray*} > -1<\frac{\phi_1\pm\sqrt{\phi_1^2+4\phi_2}}{2}&<&1\\ \Rightarrow -2<\phi_1\pm\sqrt{\phi_1^2+4\phi_2}&<&2 > \end{eqnarray*} The larger of the two $\lambda_i$ is bounded by $\phi_1+\sqrt{\phi_1^2+4\phi_2}<2$, or: \begin{eqnarray*} > \phi_1+\sqrt{\phi_1^2+4\phi_2}&<&2\\ \Rightarrow > \sqrt{\phi_1^2+4\phi_2}&<&2 - \phi_1\\ \Rightarrow > \phi_1^2+4\phi_2&<&(2 - \phi_1)^2\\ \Rightarrow \phi_1^2+4\phi_2&<&4 - > 4\phi_1+\phi_1^2\\ \Rightarrow \phi_2&<&1 - \phi_1 \end{eqnarray*} Analogously, we find that $\phi_2<1 + \phi_1$.
If $\lambda_i$ is complex, then $\phi_1^2<-4\phi_2$ and so $$\lambda_{1,2} = \phi_1/2\pm i\sqrt{-(\phi_1^2+4\phi_2)}/2.$$ The squared modulus of a complex number is the square of the real plus the square of the imaginary part. Hence, $$ \lambda^2 = (\phi_1/2)^2 + > \left(\sqrt{-(\phi_1^2+4\phi_2)}/2\right)^2 = > \phi_1^2/4-(\phi_1^2+4\phi_2)/4 = -\phi_2. $$ This is stable if $|\lambda|<1$, hence if $-\phi_2<1$ or $\phi_2>-1$, as was to be shown. (The restriction $\phi_2<1$ resulting from $\phi_2^2<1$ is redundant in view of $\phi_2<1+\phi_1$ and $\phi_2<1-\phi_1$.)