largest open set such that $F(X) = (\operatorname{tr}(X),\det(X))$ is open

Question

I need to find the largest open set in $M_2 (\mathbb{R})$ such that the function: $F:M_2 (\mathbb{R}) \rightarrow \mathbb{R}^2$ given by $F(X) = (\operatorname{tr}(X), \det(X))$ is open.

I am lost with this question and don't really know where to start from.

Answer 1

$F$ is already an open map on $M_2(\mathbb R)$. Let $A$ be an arbitrary matrix in $M_2(\mathbb R)$. It suffices to show that there exists an arbitrarily small (in terms of diameter) set $S\ni A$ such that $F(S)$ is open. We consider two cases:

1. $A$ is not diagonalisable over $\mathbb R$. By a change of basis, we may assume that $A$ is a companion matrix of the form $\pmatrix{0&-d_0\\ 1&t_0}$. For any $\epsilon>0$, define $$ S=\left\{A+\pmatrix{0&-d\\ 0&t}:\|(t,d)\|_2<\epsilon\right\}. $$ Then $A\in S\subset B(A,\epsilon)$ (in Frobenius norm) and $F(S)$ is the open ball $B\left(F(A),\epsilon\right)$ (in Euclidean norm).
2. $A$ is diagonalisable over $\mathbb R$. By a change of basis, we may assume that $A$ is a diagonal matrix of the form $\operatorname{diag}(a,b)$. For any $\epsilon>0$, define $$ S=\left\{A+\pmatrix{0&-s\sqrt{d}\\ \sqrt{d}&t}:d\ge 0,\,s\in\{1,-1\},\,\|(t,d)\|_2<\epsilon\right\}. $$ Then $A\in S\subset B\left(A,\sqrt{\epsilon^2+2\epsilon}\right)$ and $F(S)$ is the open ellipse $F(A)+\pmatrix{1&0\\ a&1}B\left(0,\epsilon\right)$.