# 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 $$\lambda^p-\phi_1\lambda^{p-1}-\phi_2\lambda^{p-2}-...-\phi_{p-1}-\phi_p=0$$

• What happens to the polynomial and its roots when you plug in $1/u$ for $\lambda$? – whuber Oct 21 '13 at 15:41
• Could you ("hold my hand") and elaborate. I don't see it still even if i plug in 1/u. – user30490 Oct 21 '13 at 15:48

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 $\Phi$.

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.