# Irreducibility of a resultant of real and imaginary parts of a characteristic polynomial

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 $p_i$?

• I am not sure this is a well-posed problem. Can you please explain how $R$ and $I$ are polynomials in $\omega$? Are you assuming some additional hypothesis in $A$? My suspicion is that, once properly formulated, the "generic" version of this question is a simple exercise using "incidence correspondences". However, I have trouble guessing what is really going on. – Jason Starr Feb 3 '14 at 19:40
• Let's we take a characteristic polynomial of A. This is a polynomial on $\lambda$: $(-1)^n\lambda^n+(-1)^{n-1}tr(A)\lambda^{n-1}+\ldots-\chi_{n-1}(A)\lambda+det(A).$ Thus $R= det(A) + (i\sqrt{\omega})^2\chi_2(A) +\ldots = det A -\omega\chi_2(A) + \omega^2\chi_4(A) - \ldots$ $I= \sqrt{\omega}^{-1}(-\sqrt{\omega}\chi_{n-1}(A)+\omega\sqrt{\omega}\chi_{n-3}(A)+\ldots)=-\chi_{n-1}(A)+\chi_{n-3}(A)\omega+\ldots)$ Therefore $R,I$ are polynomials in $\omega$ and $a_{i,jk},p_i$. We can take a resultant of $R,$ $I$ in $\omega$. It will be a polynomial in $(n^2+1)(k+1)$ variables $a_{i,jk},p_i.$ – probably Feb 3 '14 at 21:06
• Okay, that must mean that you are assuming that all of the matrices $A_i$ and the variables $p_i$ are real-valued. That is what I wanted to know. – Jason Starr Feb 3 '14 at 21:13
• Yes, A_i and p_i are real-valued – probably Feb 3 '14 at 21:20

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.