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 $\begin{cases} X.\mathrm{trace}=\mathrm{Det} \\X.\mathrm{Det}=-\mathrm{trace} \end{cases}$
$\DeclareMathOperator\trace{trace}\DeclareMathOperator\Det{Det}$**No**. Let $\sum\_{i,j}x\_{ij}(M)\frac\partial{\partial m\_{ij}}$ be this vector field. These conditions amount to writing
$$\sum\_ix\_{ii}(M)=\det M,\qquad\sum\_{ij}x\_{ij}(M)\hat m\_{ij}=-{\rm Tr}M,$$
where $\hat M$ is the cofactor matrix. Take $M=a I\_n$, for which $\hat M=a^{n-1}I\_n$. One must have
$$\sum\_ix\_{ii}(a I\_n)=a^n,\qquad a^{n-1}\sum\_{i}x\_{ii}(a I\_n)=-na,$$
which is inconsistent when $a^{2n-1}\ne-an$.


**Edit**. Since both the trace and the determinant are homogeneous polynomial, you could be interested in a vector field in which the conditions are consistent with homogeneity. Then its coordinates would be themselves homogeneous polynomials in the entries of $M$. For instance, looking for a vector field $X$ such that
$$X\cdot \trace=\Det,\qquad X\cdot \Det=-(\trace)^{2n-1}$$
makes sense. The last righ-hand side could be replaced by $- \trace^{n-1}\Det$.


