Consider the nonlinear mapping $\phi: \mathbb R^{2 \times 2} \to \mathbb R^3$ given by $X \mapsto \begin{pmatrix} x\_{11} x\_{21} \\ x\_{11} x\_{22} + x\_{21} x\_{12} \\ x\_{12}x\_{22} \end{pmatrix}$.


I think that the image $\phi(\mathbb R^{2 \times 2})$ is a convex cone, i.e. in particular: for all $X', X''$ there exists $H$ so that $\phi(X') - \phi(X'') = \phi(H)$.


I do not see how I can prove this elegantly (!) though. I am looking forward for a general method that can be also applied for other cases. Thank you in advance.



Proof that image of a polynomial map is a cone
A condition for $y = (y\_1,y\_2,y\_3)$ to be in the image is that the discriminant of $t^2 y\_3 - t y\_2 + y\_1$, namely $y\_2^2 - 4 y\_1 y\_3$, is nonnegative. This is **not** a convex constraint. For example, $(1,0,1)$ is not in the image, but
$(1,2,1) = \phi\pmatrix{1 & 1\cr 1 & 1\cr}$ amd $(1,-2,1) = \phi(\pmatrix{-1 & 1\cr -1 & 1\cr}$ are. Thus the image is not a convex cone.


