Reciprocal roots and eigenvalues relationship in time series I came across a result in a time series textbook the other day and have not been able to understand why it is true (the authors don't give a proof but just state it as true).  I want to show that the eigenvalues of the matrix $\mathbf{G}$ given by
$$G=
\begin{pmatrix}
\phi_1&\phi_2 &\phi_3 &...&\phi_{p-1} & \phi_p\\
1 & 0 &0 &...& 0 &0\\
0 & 1 & 0 &... &0 &0\\
\vdots & & & \ddots&0&0\\
0 & 0 &...&...&1 &0
\end{pmatrix}
$$
correspond to the reciprocal roots of the $AR(p)$ characteristic polynomial
$$\Phi(u)=1-\phi_1u-\phi_2u^2-...-\phi_pu^p$$
The one thing i was able to deduce is that the eigenvalues of $\mathbf{G}$ must satisfy
$$\lambda^p-\phi_1\lambda^{p-1}-\phi_2\lambda^{p-2}-...-\phi_{p-1}-\phi_p=0$$
 A: An eigenvalue of any matrix $\mathbb{G}$ must be a root of its characteristic polynomial $p_G(\lambda) = \det(\lambda - \mathbb{G}).$  By row-reducing the latter we readily find that
$$p_G(\lambda) = \lambda^p-\phi_1\lambda^{p-1}-\phi_2\lambda^{p-2}-...-\phi_{p-1}\lambda-\phi_p.$$
If $u = 1/\lambda$ is the reciprocal of an eigenvalue, then $1/u = \lambda,$ whence
$$0 = p_G(\lambda) = p_G\left(\frac{1}{u}\right) = \left(\frac{1}{u}\right)^p-\phi_1\left(\frac{1}{u}\right)^{p-1}-\phi_2\left(\frac{1}{u}\right)^{p-2}-...-\phi_{p-1}\left(\frac{1}{u}\right)-\phi_p \\
= u^{-p}\left(1-\phi_1u^{1}-\phi_2u^{2}-...-\phi_{p-1}u^{p-1}-\phi_pu^{p}\right) = u^{-p}\Phi(u).$$
Since $u$ must be nonzero (it's the reciprocal of a number), multiplying both sides by $u^p$ does not change the roots: $u$ must therefore be a root of $\Phi$.
A: For any polynomial $p(x)$, we can define a reciprocal polynomial of the form $x^n p(1/x)$ where the roots of this reciprocal function are the reciprocal roots of the original polynomial.
In the case of $\phi(u)$, the reciprocal polynomial would look like:
\begin{align}
u^p p(1/u) &= u^p - \phi_1 u^p/u - \phi_2 u^p/u^2  - \ldots - \phi_p u^p/u^p   \\
             &= u^p - \phi_1 u^{p-1}  - \phi_2 u^{p-2}  - \ldots - \phi_p.
\end{align}
A reordering of this equation reveals the exact characteristic eigenvalue equation that you have found above, only now the $u$ values have become constant eigenvalues.  Thus, solutions to the eigenvalue equation will be reciprocal roots of the ${\rm AR}(p)$ characteristic equation.
A: It is possible to perform the proof with Rouché's theorem. This appears in:
Existence and Stochastic Structure of a Non-negative Integer-valued Autoregressive Process by Alain Latour
a place where to read the paper:
https://onlinelibrary.wiley.com/doi/abs/10.1111/1467-9892.00102?casa_token=Deqs3ZkGdZoAAAAA:PLGUPa404nSVr9KO5T52lSm1KJCvyQzjGF78i7_BeWk_jqNm2aU2bUzPJEN9_y8kdiBkZWdNPBxbtVYM
