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Consider the mapping $$ \Psi: \mathbb R^2 \to \mathbb R^5, \\ \Psi(x) = \begin{pmatrix} x_1 \\ x_2 \\ x_1^2 \\ x_1 x_2 \\ x_2^2 \end{pmatrix}.$$ Which are the matrices $A \in \mathbb R^{m \times 5}$ such that $x \mapsto A \Psi(x)$ is injective?

Trivial special cases are of course when $A \in \mathbb R^{5 \times 5}$ is invertible, or when $A$ takes the special form $A = \begin{pmatrix} \tilde{A} & 0_{2 \times 3} \end{pmatrix}$ with $\tilde{A} \in \mathbb R^{2 \times 2}$ is invertible.

Are there, however, more general conditions that nicely generalize the above two special cases (which are too restrictive for my purposes) ?

Note also that $\Psi(x) = \begin{pmatrix} v_1(x) \\ v_2(x) \end{pmatrix}$, where $v_i$ is the $i$th Veronese embedding. So more generally, the question would be to characterize the matrices $A \in \mathbb R^N$ such that the mapping \begin{align*} x \mapsto A \begin{pmatrix} v_1(x) \\ \vdots \\ v_p(x) \end{pmatrix} \end{align*} is injective.

A simple scalar example: Consider $\Psi: x \mapsto \begin{pmatrix} x \\ x^2 \\ x^3 \end{pmatrix}$. Then, it can be easily seen by considering the vectors $(a_1, a_2, a_3)$ such that the polynomials $a \Psi(x)$ are strictly increasing or decreasing, that the problem is in fact feasible, i.e. one equation is sufficient for the unique solvability. And of course, the mapping $v \mapsto a_1 v_1 + a_2 v_2 + a_3 v_3$ has a two-dimensional kernel. Nevertheless, since this is not a linear algebra problem, the polynomial system of equations has a unique solution.

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  • $\begingroup$ The image of $\Psi$ are the rank one symmetric 2 by 2's. From that it should follow that unless A is invertible, the map isn't injective. $\endgroup$ – aginensky Sep 21 '15 at 21:04
  • $\begingroup$ Note that the mapping $\Psi$ also contains linear terms of $x$. Also, the example $\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{pmatrix} $ is clearly a counterexample to your statement. $\endgroup$ – user45183 Sep 21 '15 at 21:37
  • $\begingroup$ I think we have a terminology issue. His map is the 'affine' veronese map $A^2 \to A^5 $ . Properly interpreted , this is the map that takes a vector v to the tensor $v\otimes v$ . The image of the map are the rank one symmetric matrices. $\endgroup$ – aginensky Sep 21 '15 at 22:09
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Let $f(x)=A\psi(x)$.We assume that $A\in M_{m,5}$ is a generic matrix (in particular, $rank(A)=r=\inf(m,5)$ and each $r\times r$ submatrix of $A$ is invertible).

Proposition: If $m\geq 4$, then $f$ is generically injective, otherwise, it is not.

Proof. If $m\geq 5$, then $A$ contains an invertible $5\times 5$ matrix and we are done.

If $m=4$, then suppose that $f(x)=f(y)$, that is $A(\psi(x)-\psi(y))=0$. Then there are real $a,b,c,d$ s.t. $x_2-y_2=a(x_1-y_1),x_1^2-y_1^2=b(x_1-y_1),x_2^2-y_2^2=c(x_1-y_1),x_1x_2-y_1y_2=d(x_1-y_1)$

Since $a^2b-2ad+c\not=0$, there is only the trivial solution $x=y$.

If $m\leq 3$, it suffices to consider the case $m=3$.

There are real $a,b,c,d,e,f$ s.t. $x_1^2-y_1^2=a(x_1-y_1)+b(x_2-y_2),x_2^2-y_2^2=c(x_1-y_1)+d(x_2-y_2),x_1x_2-y_1y_2=e(x_1-y_1)+f(x_2-y_2)$

Take $S=\{a=3,b=4,c=7,d=-10,e=-2,f=6\}$; we obtain the non-trivial real solution: $x_1\approx -0.436,x_2\approx -8.564,y_1\approx -0.564,y_2\approx -8.436$ ($y_2=x_1-8,x_2=y_1-8$). Clearly, in a neighborhood of $S$, we obtain again a real non-trivial solution and the property is not generic.

EDIT. Answer to seno44. Linear algebra, my friend. One has a homogeneous system of $m$ equations in the 5 unknowns: $x_1-y_1,x_2-y_2,x_1^2-y_1^2,x_1x_2-y_1y_2,x_2^2-y_2^2$ which has rank $m$; since $A$ is generic, we may arbitrarily choose the $5-m$ parameters (firstly $x_1-y_1$ and secondly $x_1-y_1,x_2-y_2$); then we solve the system with respect to the other unknowns; we obtain these so called "principal unknowns" in the form of generic linear combination of the parameters.

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  • $\begingroup$ I am sorry, I don't understand what you wrote. Where does e.g. the relationship $x_2 - y_2 = a(x_1-y_1)$ come from? Where does the fact that each $r \times r$ submatrix of A is invertible come in? $\endgroup$ – user45183 Sep 23 '15 at 18:07

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