A question on determinant of a matrix polynomial

Question

Let

• ${A_j} \in {\mathbb{C}^{n \times n}},0<{w_j}\in \mathbb{R} (j = 0,1,2....m)$ and $\lambda $ is a complex variable such that $\lambda=x+iy$ and $x,y\in \mathbb{R}$.

• ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + .....{A_1}\lambda + {A_0}$ is a matrix polynomial.

• ${\rm{Q(}}\lambda {\rm{) = }}{{\rm{w}}_m}{\lambda ^m} + .....{w_1}\lambda + {w_0}$
• t=$Q{(\left| \lambda \right|)^2}$

Why does $$D(x,y)=\det (tI - P{(\lambda )^*}P(\lambda )) = \sqrt {{x^2} + {y^2}} p(x,y) + q(x,y)$$ where $p(x, y)$ and $q(x, y)$ are real polynomials in $x, y$?

Furthermore, if $Q(x)$ is even function, then $p(x,y)=0 $?

Answer 1

The statement is false if $P^*$ is taken to mean the element by element complex conjugate of $P(\lambda)$. A counterexample: let $m=1$, $\omega_0 = \omega_1 = 1$, and $$A_0 = \left( \begin{array}{cc} 0&i\\2i&0 \end{array} \right) \\ A_1 = \left( \begin{array}{cc} 1&i\\0&2 \end{array} \right) $$ Then coefficients in $p(x,y)$ come out to be complex non-real.

You must have meant $P^\dagger(\lambda)P(\lambda)$, the Hermitian conjugate of the matrix. With that change:

$t$ is a real polynomial in the variable $\lambda = \sqrt{x^2+y^2}$.

Proposition 1: $\forall n \in \Bbb{N} : \lambda^n $ is either a polynomial in $x^2$ and $y^2$ or (if $n$ is odd) $\sqrt{x^2+y^2}$ times a polynomial in $x^2$ and $y^2$.

Since all the $\omega_m$ are real, we have by proposition 1 that $t = Q(|\lambda| = Q(\sqrt{x^2+y^2}$ is of the form $$ t = P_1(x,y) + \sqrt{x^2+y^2}P_2(x,y) $$ where $P_i(x,y)$ are both real polynomials.

Now for a given set of $A_m$, each element of $P(\lambda)$ is a (possibly complex) polynomial in $(x,y)$. But each element of $P^\dagger(\lambda)P(\lambda)$ is a real-valued, thus it is a real polynomial in $(x,y)$.

Then each element of $tI - P^\dagger(\lambda)P(\lambda)$ is a real polynomial in $(x,y)$ plus, for diagonal elements, an expression of the form $P_1(x,y) + \sqrt{x^2+y^2}P_2(x,y)$.

So each element of $tI - P^\dagger(\lambda)P(\lambda)$ is of the form $P_1(x,y) + \sqrt{x^2+y^2}P_2(x,y)$

Finally, the determinant of a matrix is a polnomial function of all of its elements. This brings us home, because any polynomial function of elements of the form $P_1(x,y) + \sqrt{x^2+y^2}P_2(x,y)$ is itself of the form $P_3(x,y) + \sqrt{x^2+y^2}P_4(x,y)$. Identify in your problem $q(x,y)$ with $P_3$ and p(x,y) with $P_4$.

By the way, if only even powers appear in $Q$, then $P_2(x,y) = 0$ since every term ins itself a polynomial in $x^2+y^2)$. Since the off-diagonal elements are also pure polynomials in $x$ and $y$, in that situation, $p(x,y) = 0$.

Answer 2

Variable $t$ is a polynomial in $|\lambda|$. The even powers are real polynomials in $x,y$. The odd powers are real polynomials in $x,y$ multiplied by $|\lambda|=\sqrt{x^2+y^2}$. The quantity $P^*P$ also produces real polynomials in $x,y$. The result follows.