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I have a question on the decomposition of polynomial matrices.

Suppose $A(\lambda) = \sum_{j=0}^L \lambda^j A_j$ is an $n \times n$ matrix of polynomials, which is Hermitian on the real axis $\lambda \in \mathbb{R}$. What is the necessary and sufficient condition to decompose $A(\lambda)$ into the following form: $$ A(\lambda) = U(\lambda) \begin{bmatrix}d_1(\lambda) & & \\ & \ddots & \\ & & d_n(\lambda) \end{bmatrix} U^\star(\lambda) $$ where $U(\lambda)$ is an $n\times n$ matrix of polynomials, and $D(\lambda) = \begin{bmatrix}d_1(\lambda) & & \\ & \ddots & \\ & & d_n(\lambda) \end{bmatrix} $ is a diagonal matrix of polynomials.

An easy necessary condition for such a decomposition to exist is that the number of positive/negative eigenvalues of $D(\lambda)$ has to be greater than or equal to that of $A(\lambda)$ for all $\lambda \in \mathbb{R}$. But is this also a sufficient condition?

A special case for such problems is the J-spectral decomposition, which considers the case that $D(\lambda)$ is a constant diagonal matrix $J = \begin{bmatrix}I_{p} & \\ & -I_{q}\end{bmatrix}$. According to the beautiful result from GohbergLancasterRodman, the necessary and sufficient condition for such a decomposition to exist is that the signature of $A(\lambda)$ is constant for all $\lambda \in \mathbb{R}\setminus\sigma(A)$, where $\sigma(A):=\{\lambda\in\mathbb{R}: det(A(\lambda))=0\}$. This condition is exactly the same as the necessary condition we stated above in the general case, which shows that the necessary condition is also sufficient for this special case.

Thanks!

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    $\begingroup$ This type of diagonalization is extremely restrictive. Consider $A$ as a matrix with entries in the polynomial ring. Necessary is that the eigenvalues all be polynomial (in general, they are analytic with branch points in general, no branches if they are well separated). [e.g., take the companion matrix of $x^2 - f$ where $f $ is a polynomial in $\lambda$ but not a square; this is not symmetric, but can probably be made symmetric for suitable such $f$. A condition that is close to necessary is that for each $\lambda$, $A(\lambda)$ has no multiple roots (this avoids branch points). $\endgroup$
    – David Handelman
    Dec 1, 2017 at 17:09
  • $\begingroup$ This reminds me a lot of Wiener-Hopf factorization --- for instance, this may be relevant: doi.org/10.1016/0024-3795(83)80022-6. $\endgroup$
    – Federico Poloni
    Dec 1, 2017 at 18:34
  • $\begingroup$ @DavidHandelman Thanks for your reply. I don't need $U(\lambda)$ to be unitary, so the eigenvalues need not be polynomials. For example $$A(\lambda)=\begin{bmatrix}1&\lambda\\ \lambda&\lambda\end{bmatrix}$$ $$U(\lambda)=\begin{bmatrix}1&0\\ \lambda&1\end{bmatrix}$$ $$D(\lambda)=\begin{bmatrix}1&\\ &\lambda-\lambda^2\end{bmatrix}$$ Then $A(\lambda)=U(\lambda)D(\lambda)U^\star(\lambda)$. The eigenvalues are not polynomials. I actually have some $d_j(\lambda)=(\lambda-\alpha_j)(\lambda-\beta_j)$ in mind. When can I reduce $A(\lambda)$ to such $D(\lambda)$ with polynomial matrix $U(\lambda)$? $\endgroup$
    – William
    Jan 12, 2018 at 0:42
  • $\begingroup$ @FedericoPoloni Thanks for your reply. I briefly read the paper and some relevant papers you mentioned. If I understand correctly, it seems that the Wiener-Hopf factorization is not talking about factorizations of Hermitian matrices $A^\star(\lambda)=A(\lambda)$. And the factorization in the paper $H_-(\lambda)\neq H_+^\star(\lambda)$ in general. $\endgroup$
    – William
    Jan 12, 2018 at 1:02

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