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$X=\text{set of all trace zero matrices with bounded entries in } M_2(\mathbb{R})$ 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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What do you mean by bounded entries? For a given matrix, the entries are bounded by the maximum of the absolute value of the entries, I think...? –  M Turgeon Apr 26 '12 at 21:07
    
Suppose you are given $|a_{ij}|\le2 \forall i,j$ –  Bunuelian Trick Apr 26 '12 at 21:09
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Ok, I see. $X$ is a set of matrices with bounded entries, and not the set. Sorry for the misunderstanding. –  M Turgeon Apr 26 '12 at 21:10
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up vote 3 down vote accepted

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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how to show that $X\mapsto Tr(X)$ is continous? –  Bunuelian Trick Apr 26 '12 at 21:13
    
thank you Davide!! so some how I was thinking in a right path isn't it?:) –  Bunuelian Trick Apr 26 '12 at 21:15
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What missed was the fact that the interval contains negative or positive values). –  Davide Giraudo Apr 26 '12 at 21:16
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It may be worth pointing out that your trick with $C$ also shows $Y$ is connected. Other wise, a priori, you could get something like $Y = [-2,-1]\cup[1,2]$. –  Jason DeVito Apr 26 '12 at 21:22
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