Stable VAR($p$) procress: Is there an easy way to do this? Assume a $K$-dimensional VAR($p$) process given by
$$y_t=\nu+A_1y_{t-1}+\ldots+A_py_{t-p}+u_t$$
This process is called stable if the roots of the reverse characteristic polynomial are bigger than 1 in terms of the Euclidean norm. Therefore one has to check whether: 
$$\det(I_k-A_1z-\ldots-A_pz^p)\neq0 \text{ for } |z|\leq1.$$ 
Is there a way to find the values of $z$ without computing the polynomial $\det(I_k-A_1z-\ldots-A_pz^p)$ by hand? I see that there exist some implementations for the standard statistical programs, but I would like to understand how such a procedure works.
 A: The eigenvalues of
$$
    F = \left[\begin{array}{ccccc}
    A_1 & A_2 & \cdots & A_{p-1} & A_p\\
    I_k & O & \cdots & O & O\\
\vdots & \vdots & \ddots & \vdots &\vdots\\
    O & O & \cdots & I_k & O
    \end{array}\right]
$$
have to lie inside the unit circle, i.e.
$$
    \mathrm{det}(I_k\lambda^p - A_1\lambda^{p-1} - \cdots - A_p) = 0\qquad(\star)
$$
for a stable VAR(p) process.
Or equivalently:
All $z\in\mathbb{R}$ that satisfy
$$
    \mathrm{det}(I_k - A_1z - \cdots - A_pz^p) = 0\qquad\qquad(\star\star)
$$
lie outside the unit circle.
So, from my (basic) knowledge of Numerics, there are two possible ways of checking the stable condition: You either calculate the eigenvalues of $F$ or you solve the non-linear equation $(\star)$. I guess that most statistical software relies on the latter attempt since there are many algorithms that focus on solving non-linear equations as @Carlos Dutra already pointed out.
But to answer your question: In either way, there is no "simple" method to check the stable condition.
A: This is a root finding problem, and is independent of the AR study.
In practice, you will not calculate on hand the analytical solution, which is very complicated to higher orders. What algorithms do is search for a numerical approximation of the solution. To do this there are several methods, you can see the explanation of some of the most common, from the simplest to the most complex here and here. The Newton Method is a nice one to study.
