# Divergence of a vector-tensor multiplication

I have the following expression:

$$\nabla \cdot (\vec{\mathbf{v}} \mathbf{A})$$.

Where: $$\vec{\mathbf{v}}$$ is vector [1x3] and $$\mathbf{A}$$ a tensor [3x3]

I would like to know if the following identity holds:

$$\nabla \cdot (\vec{\mathbf{v}} \mathbf{A}) = \vec{\mathbf{v}} \cdot \nabla \mathbf{A} + (\nabla \cdot \vec{\mathbf{v}} )\mathbf{A}$$

If not, is there any identity for the divergence of a vector multiplied by a tensor?

Best regards

You can be sure that your answer's not quite correct because the second term is not a scalar. Let's compute it out in cartesian coordinates: \begin{align} \nabla\cdot(\mathbf{v}\mathbf{A}) &= \nabla\cdot(v_i A_{ij}\mathbf{e_j})\\ &= (v_i A_{ij})_{,j}\\ &= v_{i,j} A_{ij} + v_i A_{ij,j}\\ &= (v_{i,j}\ \mathbf{e_i} \otimes \mathbf{e_j}):\mathbf{A} + \mathbf{v} \cdot (A_{ij,j} \mathbf{e_i})\\ &= \nabla\mathbf{v} : \mathbf{A} + \mathbf{v} \cdot (\nabla\cdot\mathbf{A}) \end{align} Note that $$\mathbf{e_i}$$ are the $$3 \times 1$$ basis vectors for $$\mathbb{R}^3$$. In light of this, $$\mathbf{v}\mathbf{e_i}$$ is already an inner product, but I wrote $$\mathbf{v} \cdot \mathbf{e_i}$$ just to be extra clear there.