Isn't this a dot product?

Question

The problem (Arfken, chapter 3, problem 3.10.26) begins with this and asks to verify it in Cartesian coordinates and in spherical coordinates.

$$\mathbf{A} \cdot \mathbf{\nabla}\mathbf{r} = \mathbf{A}$$

What troubles me is the dot relation, how this yields a vector not a scalar?

Answer 1

It would be easier to see how things going around with matrix notations, let the arbitrary vector $\mathbf{A} $ has components $a,b,c$: $$\mathbf{A} \cdot \mathbf{\nabla}\mathbf{r} = \mathbf{A} \implies\begin{bmatrix} a&b&c \end{bmatrix} \begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$ here it is obvious that the operator $\nabla$can't act on $\mathbf{r}$ directly. $$=(a\partial_x+b\partial_y+c\partial_z)\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

$$=\begin{bmatrix} (a\partial_x+b\partial_y+c\partial_z)x \\ (a\partial_x+b\partial_y+c\partial_z)y \\ (a\partial_x+b\partial_y+c\partial_z)z \end{bmatrix}$$

$$=\begin{bmatrix} a \\ b \\ c \end{bmatrix}$$

Q.E.D.

Answer 2

In $d$ dimensions we have $$ \mathbf{A} \cdot \mathbf{\nabla}\mathbf{r} = (\mathbf{A} \cdot \mathbf{\nabla})\mathbf{r} = (\sum_{k=1}^{d} A_k \partial_k)\mathbf{r} = \sum_{k=1}^{d} A_k (\partial_k \mathbf{r} ) = \sum_{k=1}^{d} A_k \mathbf{e}_k = \mathbf{A}, $$ where $\mathbf{e}_k$ is the unit vector in the $x_k$ direction.