Finding Lag Polynomial Roots = Cayley-Hamilton?

by Syd Amerikaner   Last Updated September 14, 2018 15:19 PM - source

In my time series class we defined the lag polynomial by $\phi(L) = \sum_{i=0}^N\phi_i L^i$. It is well known that this polynomial can be factored as by $\prod_{i=0}^N(I - \lambda_i L)$ (note that $I$ denotes the "identity operator"). But then is the claim that we could "replace" $I$ by $1$ and $L$ by some some scalar $z$ and we look for the values $z_i$, $i=0,\dots, N$, such that $\prod_{i=0}^N(1 - \lambda_iz)$ is equal to zero.

I don't understand what "replace" means in context, or, in other words, why does it make sense to replace the operator by a scalar? This reminds me of the Caley-Hamilton theorem, where a scalar is "replaced" by a linear operator. Is this what is used to justify the above statement?

Cayley-Hamilton. Suppose that $T$ is a linear operator on a vector space $V$. Then $\chi_T(T) = \mathsf{null}$.

Here $V = \mathbb R$ (usually), $T = L$, $\chi_L(\lambda) = \prod_{i=0}^N(I - \lambda L)$, $\mathsf{null} = 0\ (\in\mathbb R)$. Correct? (assuming that there are $N$ distinct eigenvalues)

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