Tensor identity proof

Question

I have the following identity to prove:
$$\operatorname{grad} \operatorname{div}\textbf{A}=\operatorname{rot}\operatorname{rot}\textbf{A}+\operatorname{div}\operatorname{grad}\mathbf{A}$$
The part with $\operatorname{grad}\mathbf{A}$ is confusing to me as $\mathbf{A}$ is a vector so how do we calculate gradient of it? Could it be that my professor formulated the identity wrong?

Answer 1

They meant $(\text{div}\:\text{grad})\mathbf{A}$, i.e. $\Delta\mathbf{A}$, the Laplacian. This identity can be proven as follows:

$$(\nabla\times\nabla\times\mathbf{A})_i = \sum_{jk\ell m}\epsilon_{ijk}\epsilon_{k\ell m}\partial_j\partial_\ell A_m = \sum_{jk\ell m} \epsilon_{ijk}\epsilon_{\ell m k}\partial_j\partial_\ell A_m = \sum_{j\ell m}(\delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell})\partial_j\partial_\ell A_m$$ $$= \partial_i \sum_j\partial_j A_j - \sum_j\partial_j^2 A_i = \partial_i (\nabla\cdot\mathbf{A}) - (\nabla\cdot\nabla) A_i,$$ so that all in all $$ \nabla\times\nabla\times\mathbf{A} = \nabla(\nabla\cdot\mathbf{A}) - (\nabla\cdot\nabla)\cdot\mathbf{A}.$$