# Regenrate a polynomial matrix from its determinant's roots

I explain the univariate case first. My question is about generalizing this to multivariate cases.

Univariate Case Consider the following polynomial:

$$f(L)=1-\phi_1L-\phi_2L^2-\ldots-\phi_pL^p$$

Assume that the roots of this polynomial are denoted by ($L_i$ for $i=1...p$) and might be smaller or larger than one (in or outside the unit circle).

Now, assume that we calculate the reciprocal of those roots that are inside the unit circle and create a new set of roots. It is easy to calculate the coefficients of a polynomial, by using these new roots (e.g., see this answer).

(Please note that the application is related to estimating ARMA models).

Multivariate case

Consider the following polynomial matrix:

$$f(L)=\mathbf{I}_n-\boldsymbol{\Phi}_1L-\boldsymbol{\Phi}_2L^2-\ldots-\boldsymbol{\Phi}_pL^p$$

Similar to the univariate case, some roots of this polynomial are in the unit circle and others are not. So the question is, is it possible to generalize the univariate approach and regenerate a polynomial matrix for which all roots are outside the unit circle?

I know that the eigenvalues of the following matrix:

$$\mathbf{F}=\begin{bmatrix} \boldsymbol{\Phi}_1 & \boldsymbol{\Phi}_2 & \boldsymbol{\Phi}_3& \dots & \boldsymbol{\Phi}_{p-1}& \boldsymbol{\Phi}_p\\ \mathbf{I}_n& \mathbf{0}&\mathbf{0}&\ldots&\mathbf{0}&\mathbf{0}\\ \mathbf{0}& \mathbf{I}_n&\mathbf{0}&\ldots&\mathbf{0}&\mathbf{0}\\ \vdots& \vdots&\vdots&\ldots&\vdots&\vdots\\ \mathbf{0}& \mathbf{0}&\mathbf{0}&\ldots&\mathbf{I}_n&\mathbf{0}\\ \end{bmatrix} : np\times np$$

(with $\boldsymbol{\Phi}_i : n \times n$), satisfy:

$$|\mathbf{I}_n\lambda^p-\boldsymbol{\Phi}_1\lambda^{p-1}-\boldsymbol{\Phi}_2\lambda^{p-2}-\ldots-\boldsymbol{\Phi}_p|=0$$

(e.g., See Hamilton (1994) p. 259)

Therefore, I think maybe I can use the eigenvalues and eigenvectors of such matrix and calculate the polynomial. However, it is not clear how.

Thanks.

• If $D$ is a diagonal matrix and $V$ is an invertible matrix, then $F = V D V^{-1}$ will be a matrix whose eigenvalues and eigenvectors will be the diagonal elements of $D$ and the columns of $V$, coupled in pairwise order of the columns of the two matrices. I'm not sure if that answers your question, because the details of what you are asking are a bit unclear. – Igor Khavkine Apr 9 '17 at 13:40
• +1 for "it's a bit unclear what you are asking". What is $r$ for? What is this $F^*$ that you want to recreate? That $^*$ does not stand for a conjugate transpose, does it? – Federico Poloni Apr 9 '17 at 16:47
• In any case, I suspect that the answer to what you are asking is either on page 3 or page 20 of Gohberg-Lancaster-Rodman's book Matrix Polynomials. :) – Federico Poloni Apr 9 '17 at 16:50
• I edited the entire question. Please let me know if anything is still unclear. Thanks. – Ron Apr 9 '17 at 20:41
• OK, now that the question is more clear, there is a specific result that helps from the very useful book suggested by Federico Poloni. See my answer below. – Igor Khavkine Apr 9 '17 at 21:28

## 1 Answer

Federico Poloni suggested the right reference, Matrix Polynomials by Gohberg-Lancaster-Rodman. Though, the more precise point would be to Theorem 2.4 and other results in Chapter 2.