Is there a vector field $X$ on $\mathrm{M}_n(\mathbb{R})$ or $\mathrm{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \mathrm{trace}=\mathrm{Det} \\X\cdot \mathrm{Det}=-trace \end{cases}$$ where $\mathrm{Det}$ is determinant?
A vector field $X$ on $GL(n,\mathbb{R})$ with $\begin{cases} X.\mathrm{trace}=\mathrm{Det} \\X.\mathrm{Det}=-\mathrm{trace} \end{cases}$
Ali Taghavi
- 356
- 8
- 31
- 123