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Will this be true in vector calculus? Given are the vector fields ${\mathbf A}$ and ${\mathbf B}$ such that ${\mathbf A}=\lambda {\mathbf B}$. $\lambda$ is a scalar field. Also, $\nabla \cdot {\mathbf A}=0$.

Then I do: $$\nabla \cdot {\mathbf A}=\nabla \cdot (\lambda {\mathbf B}) = 0$$. $$\nabla \cdot {\mathbf A}=\lambda (\nabla \cdot {\mathbf B}) + \nabla\lambda \cdot {\mathbf B} = 0$$ This means $$\nabla\lambda \cdot {\mathbf B} = -\lambda (\nabla \cdot {\mathbf B})$$ Then $$\nabla\lambda = -\lambda \nabla $$

Is this last expression true? I mean, will it be true if used in other expressions to substitute the Lhs with the Rhs or vice versa? Thanks!

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    No. E.g. if $\mathbf{A}=\mathbf{B}=0$. Even otherwise, $\mathbf{B}$ is not arbitrary, so you can't just "divide" it away. – 2017-02-27
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    You mean, only this is true: If $\mathbf{U} \cdot \mathbf{r} = \mathbf{V} \cdot \mathbf{r}$ then $\mathbf{U}=\mathbf{V}$, because $\mathbf{r}$ is a radius vector (and thus, arbitrary)? – 2017-02-27
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    I guess, somehow one needs to prove arbitrarity of $\mathbf{B}$ or else is not true. Thanks! – 2017-02-27

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