Is it possible to describe a AR(1) process as stable in terms of a stability triangle? 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 $\phi_2<1-\phi_1$.)


The second condition fails.
If you have an non-stable AR1 process then $\phi_1 = 1$. You could interpret it, if you like, as a AR2 process with $\phi_2 = 0$. Then you have that the second condition fails
$$0=\phi_2<1-\phi_1 = 0$$
The condition $0<0$ is false.

The characteristic equation of higher order but with coefficients zero will have the same roots but with some additional roots equal to zero. $\lambda^2-\lambda \theta_1- \theta_2$ and $\lambda-\theta_1$ are just the same when $\theta_2=0$ (except for that additional root $\lambda = 0$).
