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Consider the parametric complex map $f_{A,B,v}: \ \mathbb{C}^n \rightarrow \mathbb{C}$ defined as: $$ f_{A,B,v}(x) = Ax\cdot Bx \ |v \cdot x|^2, $$ where $A,B$ are $n \times n$ complex matrices, $v \in \mathbb{C}^n$, $\cdot$ is the standard Hermitian product and $| \ |$ is the complex absolute value.

Suppose that for some $A,B,v,A',B',v'$ it holds that $f_{A,B,v}(x) = f_{A',B',v'}(x)$ for all $x \in \mathbb{C}^n$. What can we say about the relationship between $A$ and $A'$, $B$ and $B'$, and $v$ and $v'$?

Thank you for your help.

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$\newcommand\ol\overline$For $x=(x_1,\dots,x_n)$, $$f_{A,B,v}(x)=\sum_{i,j,k,l,m}A_{ij}x_j\ol{B_{ik}}\ol{x_k}v_l\ol{x_l}\,\ol{v_m}x_m =\sum_{j,k,l,m}c_{j,k,l,m}x_j\ol{x_k}\,\ol{x_l}x_m,$$ where $\ol{\cdot}$ denotes the complex conjugation and $$c_{j,k,l,m}:=c_{j,k,l,m}(A,B,v):=v_l\ol{v_m}\sum_i A_{ij}\ol{B_{ik}}.$$ So, taking into account the symmetries $x_j\ol{x_k}\,\ol{x_l}x_m =x_m\ol{x_k}\,\ol{x_l}x_j =x_j\ol{x_l}\,\ol{x_k}x_m =x_m\ol{x_l}\,\ol{x_k}x_j$ and then equating the respective coeffcients, we see that $$f_{A,B,v}(x)=f_{A',B',v'}(x)$$ for all $x$ iff $$d_{j,k,l,m}(A,B,v)=d_{j,k,l,m}(A',B',v')$$ for all $j,k,l,m$, where $$d_{j,k,l,m}:=c_{j,k,l,m}+c_{m,k,l,j}+c_{j,l,k,m}+c_{m,l,k,j}.$$

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  • $\begingroup$ Thank you! I was also thinking of an argument via unique factorization of polynomials (with conjugate variables). Specifically, assume that $A,B,A',B'$ have rank $>1$. Then $Ax \cdot Bx$ is an irreducible quadric and therefore $f_{A,B,v}$ is a product of an irreducible quadric and a reducible one. Doesn't it follow from unique factorization that $|v \cdot x|^2 = |v' \cdot x|^2$ for all $x$, and therefore $v = \lambda v'$ for some $\lambda \in \mathbb{C}$? $\endgroup$
    – gm01
    Commented Jan 15 at 16:44
  • $\begingroup$ @gm01 : I don't think so. For instance, if $\sum_i A_{ij}\ol{B_{ik}}=0=\sum_i A'_{ij}\ol{B'_{ik}}$ for all $j,k$, then $v$ and $v'$ are arbitrary. More generally, what you are suggesting seems to hardly ever be true, given the symmetries $j\leftrightarrow m$ and $k\leftrightarrow l$, which "intermix" $v$ with $A$ and $B$. $\endgroup$
    – Iosif Pinelis
    Commented Jan 15 at 17:17
  • $\begingroup$ I am sorry, I meant to say that $A^*B$ and $A'^*B'$ have rank $>1$, where $*$ denotes the transpose conjugate. Does it work with this assumption? $\endgroup$
    – gm01
    Commented Jan 15 at 17:37
  • $\begingroup$ @gm01 : Even then, I don't think this will be true, because of the intermixing mentioned in the second sentence of my previous comment. It should be easy to find a counterexample to your suggestion, using the iff condition in the answer together with Mathematica or similar software. $\endgroup$
    – Iosif Pinelis
    Commented Jan 15 at 17:43

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