$\nabla(\vec{A}\cdot\vec{r})=\vec{A}.$ Using index notation for vector calculus. Practicing for exam

Question

I am learning the index notations in vector calculus and tried to prove the following identity:

Let $\vec{A}=(A_x,A_y,A_z)$ is a constant vector and let $\vec{r}=(x,y,z)$. Prove that $$\nabla(\vec{A}\cdot\vec{r})=\vec{A}.$$

So in the regular notation I have done like this: $$(\partial_x e_x+\partial_y e_y+\partial_z e_z)(A_x x+A_y y+A_z z)=A_xe_x+A_y e_y+A_z e_z=\vec{A}$$ But I would like to do in index notation as well.

Answer 1

Write $\vec{A}=a_i\hat{e_i}$. Then it follows that $$ \nabla(\vec{A} \cdot \vec{r}) = \partial_i (a_j r_j) \hat{e_i} = a_j (\partial_i r_j) \hat{e_i} = a_j\delta_{ij}\hat{e_i} = a_i\hat{e_i} = \vec{A}$$ where $\delta_{ij}$ is the Kronecker delta.