Properties of AR(p) model to be invertible Suppose we are working with an $AR(p)$ model. We assume that $E( \epsilon^2 ) < \infty$.
We know that an $AR(P)$ model is invertible $\text{iff } \phi(z) \neq 0$ for $|z| \le 1$.
But why do we have to make this assumption? Why can we not have a zero inside the unit circle, but outside the unit circle it's fine?
 A: I like the material in the link that @whuber posted, but more often than not you see people just say what the eigenvalues have to be, without giving the logic behind it (this is taken from some notes by Eric Zivot, any errors are mine, of course). Re-writing an $AR(p)$ model (or $VAR(p)$) as an $AR(1)$ (or $VAR(1)$)
$$\left[\begin{array}{c}
y_{t}\\
y_{t-1}\\
\vdots\\
y_{t-p+1}
\end{array}\right]=\left[\begin{array}{cccc}
\beta_{1}' & \beta_{2}' & \beta_{\cdots}' & \beta_{p}'\\
I & 0 & 0 & 0\\
0 & I & 0 & 0\\
0 & 0 & I & 0
\end{array}\right]\left[\begin{array}{c}
y_{t-1}\\
y_{t-2}\\
\vdots\\
y_{t-p}
\end{array}\right]+\left[\begin{array}{c}
\varepsilon_{t}\\
0\\
\vdots\\
0
\end{array}\right]
 $$
For simplicity, we can re-write this as $Y_{t}=FY_{t-1}+\nu_{t}$. It can then be shown that $$Y_{t+j}=F^{j+1}Y_{t-1}+F^{j}\nu_{t}+\ldots +F\nu_{t+j-1}+\nu_{t}$$
If $\lim_{j\rightarrow\infty}F^{j}=0$, then it would be possible to say that $Y_{t}$ is stable since most of the above effectively cancels out and the initial value does not matter. To show that it is stable, one takes the eigendecomposition $F=T\Lambda T^{-1}$, where $\Lambda$ is a diagonal matrix of eigenvalues, and substitute that in to $F^{j}$ to get
$$\lim_{j\rightarrow\infty}F^{j}=T\Lambda^{j} T^{-1}$$ This implies that if $|\Lambda|$ has any elements that are larger than 1, then it will explode as $j\rightarrow\infty$. Hence, the requirement that the absolute value of the real component of the eigenvalues must be less than 1 to be stable. Complex eigenvalues lead to an oscillating behavior in the time series. If $Y_{t}$ is stable, then it is also stationary.
