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 $\phi_2<1-\phi_1$.)


Indeed, since there is only one coefficient, there is no two-dimensional figure to plot (unlike the answer A proof for the stationarity of an AR(2) you cite, which discusses the AR(2) case). Stability means that all solutions to the characteristic polynomial are outside the unit circle.
In this case, the polynomial is just $1-\phi z=0$, such that the condition becomes
$$
|z|>1/|\phi|
$$
or $|\phi|<1$.
