The following question is motivated by the study of a stability border for a robust linear time-invariant control system.


Let us we have an affine family of $n\times n$ matrices with indeterminate ($\mathbb{R}$-valued) entries. 


$$
A=A\_0+p\_1A\_1+\ldots+p\_kA\_k,\qquad A\_i=\begin{pmatrix}a\_{i,11}\ldots a\_{i,1n}\\\ldots\\a\_{i,n1}\ldots a\_{i,nn}\end{pmatrix}
$$


Let $\chi\_A(\lambda)$ be a characteristic polynomial of $A$. Let us define $R= \Re( \chi\_A(i\sqrt{\omega})), I=\frac{\Im(\chi\_A(i\sqrt{\omega}))}{\sqrt{\omega}}.$ Both $R$ and $I$ are a polynomials depending on $\omega.$ Let $S$ be a resultant of $R$ and $I$ as $\omega$-polynomials. 


Is $S$ irreducible as a polynomial from $(k+1)(n^2+1)$ variables $a\_{i,jk}, p\_i$?
Is it true that if matrices $A\_i$ are, in some sense, in "general position" then $S$ is irreducible as a polynomial in $p\_i$?



Irreducibility of a resultant of real and imaginary parts of a characteristic polynomial
First there is a little fact about resultants. Let $(R(u),I(u))$ be a pair of polynomials in the variable $u$ of degrees $(m,m)$, resp. $(m,m-1)$. Define $x(v) = R(v^2) + vI(v^2)$, a polynomial in the variable $v$ of degree $n=2m+1$, resp. $n=2m$. Write, $$x(v) = c\_0v^n + c\_1v^{n-1} + \dots + c\_{n-1}v + c\_n.$$ Then we have the identity,
$$\text{Res}\_v(x(v),x(-v)) = \pm 2^nc\_0c\_n[\text{Res}\_u(R(u),I(u))]^2.$$ In particular, if $\chi(z)$ is a complex-valued polynomial of degree $n$, setting $$R(-u^2) = \frac{1}{2}(\chi(iu) + \chi(-iu))$$ and $$I(-u^2) = \frac{1}{2iu}(\chi(iu) - \chi(-iu)),$$ then $\chi(-iv)$ equals $R(v^2)+vI(v^2)$, and $\chi(iv)$ equals $R(v^2)-vI(v^2)$. Of course, up to multiplying by $i^a$ for some positive integer $a$ depending only on $n$, $\text{Res}\_z(\chi(z),\chi(-z))$ equals $\text{Res}\_v(\chi(-iv),\chi(iv))$. Thus, by the fact above, $$\text{Res}\_z(\chi(z),\chi(-z)) = i^a2^nc\_0c\_n[\text{Res}\_u(R(u),I(u))]^2.$$


Of course in the question, $c\_0$ equals $1$, and, up to a factor of $i$, $c\_n$ equals the determinant. I am going to restrict over the Zariski open subset of the parameter space where the determinant is nonzero, so that both $c\_0$ and $c\_n$ are nonzero. Then to prove that $\text{Res}\_u(R(u),I(u))$ is irreducible on this algebraic variety, i.e., the corresponding zero scheme is reduced and irreducible, it suffices to prove that the zero scheme of $\text{Res}\_z(\chi(z),\chi(-z)))$ is irreducible and has multiplicity $2$ at a generic point. Of course it suffices to check this after base change from $\mathbb{R}$ to $\mathbb{C}$.


So now let $\textbf{GL}\_n$ be the complex affine variety parameterizing invertible matrices $B$ over $\mathbb{C}$. Let $Y$ be the zero scheme of $\text{Res}\_z(\chi\_B(z),\chi\_B(-z))$. Of course $\chi\_B(-z)$ equals $(-1)^n\chi\_{-B}(z)$. Thus $Y$ is also the zero scheme of $\text{Res}\_z(\chi\_B(z),\chi\_{-B}(z))$. As a Zariski closed subset, $Y$ is the set of invertible matrices $B$ that have two opposite eigenvalues, $\lambda$ and $-\lambda$. Of course $\lambda$ is necessarily nonzero since $\text{det}(B)$ is nonzero. Denote by $\mathbb{G}\_m$ the complex affine variety parameterizing nonzero scalars $\lambda$. Denote by $X$ the Zariski closed subset of $\textbf{GL}\_n\times \mathbb{G}\_m$ parameterizing pairs $(B,\lambda)$ such that both $\lambda$ and $-\lambda$ are eigenvalues of $B$. Finally, let $W$ be the Zariski closed subset of $$\textbf{GL}\_n\times \mathbb{G}\_m\times [(\mathbb{P}^{n-1}\times \mathbb{P}^{n-1})\setminus \Delta(\mathbb{P}^{n-1})]$$ parameterizing data $(B,\lambda,[v],[w])$ such that $Bv=\lambda v$ and $Bw=-\lambda w$. Consider the projection,
$$\pi:W \to \mathbb{G}\_m\times [(\mathbb{P}^{n-1}\times \mathbb{P}^{n-1})\setminus \Delta(\mathbb{P}^{n-1})].$$ 
This is a Zariski local fiber bundle whose fiber over $(\lambda,[v],[w])$ is isomorphic to a dense Zariski open subset of the affine linear space parameterizing matrices $B$ satisfying the two affine linear equations, $$Bv=\lambda v, \ \ Bw=-\lambda w.$$ Therefore, since the target of $\pi$ is irreducible and since $\pi$ is a Zariski local fiber bundle with irreducible fiber, the domain $W$ of $\pi$ is irreducible. In fact $W$ is smooth (hence reduced, i.e., of "multiplicity one"). 


By parameter counts, the projection $W\to X$ is surjective and birational. Thus $X$ is also irreducible and reduced. On the other hand, $X\to Y$ is surjective and generically $2$-to-$1$. Thus $Y$ is irreducible and has multiplicity $2$ as a Cartier divisor in $\textbf{GL}\_n$. Therefore the zero scheme of $\text{Res}\_u(R(u),I(u))$ is irreducible and reduced on the Zariski open subset $\textbf{GL}\_n\subset \textbf{Mat}\_{n\times n}$ of the space of *all* $n\times n$ matrices. 


To complete the analysis, it suffices to determine the multiplicity of $\text{det}(B)$ as a factor of $\text{Res}\_u(R(u),I(u))$, which is equivalent to determining the multiplicity of $\text{det}(B)$ as a factor of $\text{Res}\_z(\chi\_B(z),\chi\_{-B}(z))$ via the resultant identity above. But for the parameter $t$ and for the diagonal matrix $B$ with diagonal entries $(1,1,\dots,1,t)$, it is straightforward to compute $$\text{Res}\_z(\chi\_B(z),\chi\_{-B}(z)) = \pm 2^{n^2-2n+1}(1+t)^{2n-1}(2s).$$
Since the multiplicity of $s$ equals $1$, it follows that the multiplicity of $\text{det}(B)$ as a factor of $\text{Res}\_u(R(u),I(u))$ is $0$. Therefore $\text{Res}\_u(R(u),I(u))$ is irreducible as a polynomial on $\text{Mat}\_{n\times n}$.


Now, denote by $(\text{Mat}\_{n\times n})^{k+1}$ the affine space of $(k+1)$-tuples of $n\times n$ matrices, $(A\_0,A\_1,\dots,A\_k)$. Also denote by $\mathbb{A}^k$ the affine space of $k$-tuples of scalars, $(p\_1,\dots,p\_k)$. Then the morphism,
$$ B : (\text{Mat}\_{n\times n})^{k+1}\times \mathbb{A}^k \to \textbf{Mat}\_{n\times n}, \ \ ((A\_0,A\_1,\dots,A\_k),(p\_1,\dots,p\_k)) \to A\_0 + p\_0 A\_1 + \dots + p\_kA\_k, $$
is a Zariski local fiber bundle with fiber isomorphic to $(\textbf{Mat}\_{n\times n})^k\times \mathbb{A}^k$. Indeed, there is an automorphism,
$$ g : \text{Mat}\_{n\times n})^{k+1}\times \mathbb{A}^k \to \text{Mat}\_{n\times n})^{k+1}\times \mathbb{A}^k , \ \ ((A\_0,A\_1,\dots,A\_k),(p\_1,\dots,p\_k)) \mapsto ((B,A\_1,\dots,A\_k),(p\_1,\dots,p\_k)), $$ that makes this manifest. Since $B$ is a Zariski local fiber bundle whose fiber is smooth and irreducible, the inverse image of the reduced, irreducible Cartier divisor $\text{Zero}(\text{Res}\_u(R(u),I(u)))$ is still reduced and irreducible. 


