$X=\text{set of all trace zero matrices with bounded entries in } M_2(\mathbb{R})$ and $Y=\{detA: A\in X\}\subseteq\mathbb{R}$ Does there exist $\alpha<0$ and $\beta>0$ such that $Y=[\alpha,\beta]$? So far I think they are asking whether $Y$ is compact or not, for that enough to show $X$ is compact as det is continuous so we will be done, the set $X$ is bounded clearly, but to show closed I hope it is a subset of kerT where $T:M_2(\mathbb{R})\rightarrow \mathbb{R}$ given by $T(A)=trace(A)$, nothing more I can conclude now. Please help.
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The map $X\mapsto \operatorname{Tr}(X)$ is continuous (as a linear map over a finite dimensional vector space), so $X$ is closed and bounded, hence compact. The map $A\mapsto \det A$ is continuous so $Y$ compact. Now, what we have to see is whether we can have negative or positive determinant. Taking $A:=\pm C\pmatrix{-1&0\\0&1}$, we can have negative or positive values (we choose $0\leq C<M$, where $M$ is such that $|a_{ij}|\leq M, i,j\in\{1,2\}, A\in X)$. |
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