Curl of a linear operator applied to a vector field

Let $A$ be a $3\times 3$ matrix such that $A_{ij}$ is a smooth real valued function on $\mathbb{R}^3$ and let $v$ be a smooth vector field in $\mathbb{R}^3$. My problem is to simplify the expression: $$\text{curl}(Av)\, .$$ Here is my initial attempt. Using the Einstein summation convention, we have $(Av)_i = A_{ij}v_j$ and $(\text{curl}(B))_i = \epsilon_{ijk}\partial_jB_k$. Then, \begin{align*} (\text{curl}(Av))_i &= \epsilon_{ijk}\partial_j(A_{kl}v_l)\\ &=\epsilon_{ijk}v_l\partial_jA_{kl}+ \epsilon_{ijk}A_{kl}\partial_jv_l\, . \end{align*} How can I simplify this further to get an expression involving the standard vector differential operators?

I don't know if the following could be of help: If $f$ is a scalar field and $\vec a$ a vector field then $${\rm curl}(f\,\vec a)=f\,{\rm curl}(a)+\nabla f\times\vec a\ .\tag{1}$$ Denote the $j^{\rm\, th}$ column of your matrix $A$ by $\vec A_j$. Then you can write the field $Av$ whose ${\rm curl}$ has to be calculated in the form $$Av=\sum_{j=1}^3 v_j\,\vec A_j\ ,\tag{2}$$ where now the components of the field $v$ appear as scalar functions. Applying $(1)$ to the three summands in $(2)$ individually we obtain $${\rm curl}(Av)=\sum_{j=1}^3\bigl(v_j\,{\rm curl}(\vec A_j)+\nabla v_j\times \vec A_j\bigr)\ .$$ But I don't see a way to encode each of the sums $$\sum_{j=1}^3 v_j\,{\rm curl}(\vec A_j)\>,\qquad\sum_{j=1}^3\nabla v_j\times \vec A_j$$ in a coordinate free way as "vectorial objects".
• For the case where $v=\nabla u$ ($u$ is a scalar function), can we find a formula for $curl (A\nabla u)$? May 4, 2021 at 12:54