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.