Is there a vector field $X$ on $\operatorname{M}_n(\mathbb{R})$ or $\operatorname{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \operatorname{trace}=\operatorname{Det} \\X\cdot \operatorname{Det}=-\operatorname{trace} \end{cases}$$ where $\operatorname{Det}$ is determinant?
A vector field $X$ on $\mathrm{GL}(n,\mathbb{R})$ with $\begin{cases} X.\mathrm{trace}=\mathrm{Det} \\X.\mathrm{Det}=-\mathrm{trace} \end{cases}$
Ali Taghavi
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