Skip to main content
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
edited body
Source Link
user45183
user45183

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.

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.

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.

edited body
Source Link
user45183
user45183

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 5} \end{pmatrix}$$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.

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 $A = \begin{pmatrix} \tilde{A} & 0_{2 \times 5} \end{pmatrix}$ 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.

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.

Source Link
user45183
user45183

Injectivity of a multivariate homogeneous polynomial mapping

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 $A = \begin{pmatrix} \tilde{A} & 0_{2 \times 5} \end{pmatrix}$ 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.