Vectorial Identities Studying General Relativity I found, in a demonstration, this vectorial identity :

My question is, what does it mean
$(\textbf{A}\cdot \nabla) \textbf{A}$  ?
It can't be $\textbf{A}\cdot \nabla \textbf{A}$ because it would be a scalar. Sorry but it's the first time i saw that kind of product.
Thanks a lot in advance.
 A: Let $A=(a_1, a_2, a_3)$ and $\nabla = (\partial_1, \partial_2, \partial_3)$ then
$$A \cdot \nabla = a_1 \partial_1 + a_2 \partial_2+a_3 \partial_3$$
That's is, an operator.
A: Let $A_1,A_2,A_3$ denote the components of $\mathbf A$, and let $\partial_i$ denote the partial derivative with respect to $x_i$. Formally, $\mathbf A \cdot \nabla$ refers to the operator
$$
\mathbf A \cdot \nabla = \langle A_1,A_2,A_3\rangle \cdot \langle \partial_x,\partial_y,\partial_z \rangle = A_1 \partial_1 + A_2 \partial_2 + A_3 \partial_3.
$$
For any scalar function $f$, $(\mathbf A \cdot \nabla) f$ is the scalar function
$$
(\mathbf A \cdot \nabla) f = A_1 \partial_1f + A_2 \partial_2f + A_3 \partial_3f.
$$
Correspondingly, $(\mathbf A \cdot \nabla)\mathbf A$ is the vector function
$$
(\mathbf A \cdot \nabla)\mathbf A  = \langle (\mathbf A \cdot \nabla)A_1,(\mathbf A \cdot \nabla)A_2,(\mathbf A \cdot \nabla)A_3\rangle.
$$
If we take $\mathbf e_1,\mathbf e_2, \mathbf e_3$ to denote the standard basis vectors, then we can write
$$
(\mathbf A \cdot \nabla)\mathbf A = \sum_{i=1}^3 \sum_{j=1}^3 A_i \frac{\partial A_j}{\partial x_i} \mathbf e_j.
$$
