Let $U_1,\dots,U_k$ be $C^1$-vector fields on $\mathbb{R}^n$ and let $f:\mathbb{R}^n\to (0,\infty)$ be a $C^1$-function. Let $Lie(U_i)$ denote the Lie algebra generated by $\{U_i\}_i=1^k$. When is $dim(Lie(f\cdot U_i))=dim(Lie(U_i))$?
It's clear that, due to the assumption that $f$ is strictly positive, we have that $$ dim\left( \operatorname{span}\left\{ f U_i \right\}_{i=1}^k \right) = dim\left( \operatorname{span}\left\{ U_i \right\}_{i=1}^k \right). $$ When trying to compute the iterate Lie brackets, defined by: $$ [fU_i,fU_j] := D(f U_i)fU_j - D(fU_j)fU_i, \qquad (1) $$
Like in the above case, I'm thinking to reduce the iterated brackets with $f$ to those without.
What I've Worked out
From the matrix calculus, we have that $$ D(fU_i) = f D(U_i) + U_i\nabla f. \qquad (2) $$ Combining $(1)$ and $(2)$ we get that $$ \begin{aligned} [fU_i,fU_j] := & D(f U_i)fU_j - D(fU_j)fU_i \\ = & (f D(U_i) + U_i\nabla f) (fU_j) - (f D(U_j) + U_j\nabla f) (fU_i) \\ = & f^2 D(U_i)U_j + fU_i\nabla f U_j - f^2 D(U_j)U_i - fU_j\nabla f U_i \\ = & (f^2) \underbrace{\left(D(U_i)U_j - D(U_j)U_i \right)}_{Part of Lie(U_i)} + f\cdot \underbrace{\left( U_i\nabla f U_j - U_j\nabla f U_i \right)}_{???}, \end{aligned} $$ here I use $\cdot$ to emphasise that $f$ is multiplying the quantity on its right and not applied to it!
What can be done with this term: $\left( U_i\nabla f U_j - U_j\nabla f U_i \right)$?