Formulas for partial composed product

Question

Let $A(x) = \prod\limits_i (x-\lambda_i)$ and $B(x) = \prod\limits_j (x-\mu_j)$. Then, their composed product is defined as $$ (A*B)(x) = \prod\limits_{i,j} (x-\lambda_i \mu_j). $$ Generally, we can show that if $A(x)$ and $B(x)$ have integer coefficients, so does $(A*B)(x)$. The reason for this is that if $A(x)$ and $B(x)$ are characteristic polynomials of matrices $A$ and $B$, then $(A*B)(x)$ is the characteristic polynomial of $A \otimes B$, their Kronecker product.

Now, let's focus on the composed product of $A(x)$ with itself. We can define polynomials $A_d(x)$ and $A_s(x)$ as

$$\begin{align} A_s(x) &= \prod\limits_{i < j} (x-\lambda_i \lambda_j), \\ A_d(x) &= \prod\limits_i (x-\lambda_i^2), \end{align}$$ in which case $(A*A)(x) = A_d(x) A_s^2(x)$. It appears that $A_s(x)$ and $A_d(x)$ should also have integer coefficients if $A(x)$ is as such, but I don't see any simple way to show it, or compute them without finding the roots, like it was possible with $(A*B)(x)$. Any ideas on how to proceed?

Ideally, I want to not only show that $A_s(x)$ and $A_d(x)$ have integer coefficients, but also provide some algorithmic way to find them if $A(x)$ is known, without finding all complex roots of $A(x)$.

The reason I ask is, assume that $A(x)$ is the characteristic polynomial of a linear recurrent sequence $u_n$. Then, $A_s(x)$ is the characteristic polynomial of the linear recurrent sequence $h_n = u_n u_{n+a+b} - u_{n+a} u_{n+b}$, and more generally, $A_s(x) A_d(x)$ appears to be the characteristic polynomial of $h_n = u_n u_{n+a}$, so it would be nice to have a general method to find it directly.

Answer 1

Thanks everyone for the comments! I will try to summarize what's going on a bit, and post it as an answer, as I was not really aware of how symmetric squares and exterior squares work, so I will provide somewhat more explicit constructions and explanations.

Tensor product of vector spaces $U$ and $V$ is the vector space $U \otimes V$ with a bilinear map $$\begin{gather} \otimes: U \times V \to U \otimes V, \\ (u, v) \mapsto u \otimes v, \end{gather}$$ such that $\{u \otimes v : u \in B_U, v \in B_V\}$ is a basis of $U \otimes V$, where $B_U, B_V$ are bases of $U$ and $V$.

Now, let $B_V = \{e_1,\dots,e_n\}$ and we will focus on the vector space $V$.

Tensor square $V^{\otimes 2}$ of a vector space $V$ is $V \otimes V$, it has a basis $$ \{e_i \otimes e_j : 1 \leq i, j \leq n\}. $$

Symmetric square $\operatorname{S}^2(V)$ of a vector space $V$ is the subspace of $V^{\otimes 2}$ with a basis $$ \{e_i \otimes e_j + e_j \otimes e_i : 1 \leq i \leq j \leq n\}. $$ Exterior square $\bigwedge^2(V)$ of a vector space $V$ is the subspace of $V^{\otimes 2}$ with a basis $$ \{e_i \otimes e_j - e_j \otimes e_i : 1 \leq i < j \leq n\}. $$

Generally, tensor square is isomorphic to the linear space of $n \times n$ matrices, while symmetric square and exterior square can be interpreted as subspaces of symmetric and antisymmetric matrices correspondingly.

For any matrix $A$, we can represent it as $A = \frac{A+A^\top}{2}+\frac{A-A^\top}{2}$, the sum of a symmetric and antisymmetric matrices, meaning that $V^{\otimes 2} = \operatorname{S}^2(V) \oplus \bigwedge^2(V)$, i.e. the tensor square is a direct sum of the symmetric square and the exterior square.

Now, if $A : V \to V$ is a linear transform on $V$, we can induce a linear transform $$\begin{gather} A^{\otimes 2}:V^{\otimes 2} \to V^{\otimes 2}, \\ u \otimes v \mapsto (Au) \otimes (Av), \end{gather}$$ which we will call its tensor square. Correspondingly, its symmetric square $\operatorname{S}^2(A)$ and exterior square $\bigwedge^2(A)$ are the reductions of the tensor square on corresponding subspaces. Since the tensor square of a space is the direct sum of the symmetric square and the exterior square, the tensor square of a linear map is the direct product of its symmetric square and exterior square.

Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $A$, and $e_1,\dots,e_n$ the eigenvectors. Then,

• The eigenvalues of $A^{\otimes 2}$ are $\lambda_i \lambda_j$ for all $1 \leq i,j \leq n$ and eigenvectors are $$e_{ij} = e_i \otimes e_j$$
• The eigenvalues of $\operatorname{S}^2(A)$ are $\lambda_i \lambda_j$ for $1 \leq i \leq j \leq n$ and eigenvectors are $$e_{ij} = e_i \otimes e_j + e_j \otimes e_i$$
• The eigenvalues of $\bigwedge^2(A)$ are $\lambda_i \lambda_j$ for $1 \leq i < j \leq n$ and eigenvectors are $$e_{ij} = e_i \otimes e_j - e_j \otimes e_i$$
• The eigenvalues of $A^2$ are $\lambda_i^2$ and eigenvectors are $e_i$.

These observations altogether pretty much explain how to represent

• $A_s(x) A_d(x)$ as the characteristic polynomial of $\operatorname{S}^2(A)$,
• $(A*A)(x)$ as the characteristic polynomial of $A^{\otimes 2}$,
• $A_s(x)$ as the characteristic polynomial of $\bigwedge^2(A)$,
• $A_d(x)$ as the characteristic polynomial of $A^2$.