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 $p_i$?

  • $\begingroup$ 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. $\endgroup$ – Jason Starr Feb 3 '14 at 19:40
  • $\begingroup$ 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 $a_{i,jk},p_i.$ $\endgroup$ – probably Feb 3 '14 at 21:06
  • $\begingroup$ 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. $\endgroup$ – Jason Starr Feb 3 '14 at 21:13
  • $\begingroup$ Yes, A_i and p_i are real-valued $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.