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I'm trying to understand vector notation, in particular when operators act on one another. I'm assuming the parenthesis are important to preserve the physical meaning of the operator. How do you know when to keep/introduce parenthesis? Can some one help expand the following expression?

$(\textbf{A} \cdot \nabla)(\textbf{A} \cdot \nabla)$

NOTE: $\textbf{A}$ is a vector in $\text{R}^3$ and I'm dealing with basic Euclidean space, where $\nabla$ is defined as the gradient in a conventional sense.

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In vector calculus it is common to interpret the differential operator $\nabla$ as if it were a vector itself, i.e. $ \nabla = (\partial_1, \partial_2, \partial_3)^T$. From this point of view it presents no problem to take the scalar product $A \cdot \nabla$, as $\nabla$ is just a vector: $$ A \cdot \nabla = A_1 \partial_1 + A_2\partial_2 + A_3\partial_3. $$ Note that this is giving you a new differential operator, which you could apply to either a scalar function $\alpha(x_1, x_2, x_3)$, i.e. $$ (A \cdot \nabla)\alpha = A_1 (\partial_1 \alpha) + A_2 (\partial_2 \alpha) + A_3(\partial_3 \alpha) \tag{$\ast$} $$ or componentwise to a vector.

Interpreting the differential operator $A \cdot \nabla$ as scalar, we can insert it into ($\ast$) and get $$ (A \cdot \nabla)\,(A \cdot \nabla) = A_1\partial_1 \big(A_1 \partial_1 + A_2 \partial_2 + A_3\partial_3 \big) + \dots = A_1^2 \partial_1^2 + A_1 A_2 \partial_1\partial_2 + A_1 A_3 \partial_1\partial_3 \cdots $$ If $A$ is a vector field you will have to take into account derivatives of $A$ as well (product rule).

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    I got that much and I believe it would be a tensor 2 operator...basically in the same way $(\textbf{A} \cdot \nabla)\textbf{A}$ works. But in the case I presented, this would result in the first tensor 2 term of the form $A_1\frac{\partial}{\partial x} (A_1\frac{\partial}{\partial x})$. For this term, do I have to apply a product rule? As a result, the first term would be $ A_1(\frac{\partial u}{\partial x} \frac{\partial}{\partial x} + A_1\frac{\partial^2 }{\partial x^2})$, where again, this is just the first term in the tensor 2. I'm not sure if this is correct. – 2017-02-14
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    Would you mind replacing the scalar in your solution to better reflect my question and carrying out the solution for the 1st term (I don't need the whole matrix, I can do that on my own). The part that is new to me is addressed in the comment above and I just want to be sure I get it right. – 2017-02-14
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    @ThatsRightJack: I don't understand. Why do you expect a matrix as outcome, if $A \cdot \nabla$ is clearly a scalar? – 2017-02-14
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    OK, I think I see what you're saying.... $(\textbf{A}\cdot \nabla)(\textbf{A}\cdot \nabla)$ would be a scalar operator. I guess the number of terms made me think matrix, but I forgot they are all summed. – 2017-02-14
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    And yes, A is a vector, so there would be terms like I mentioned 4 comments up in the summation? – 2017-02-14
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    I just noticed, in my first comment, I used "u" in the first term of the product rule. It should be "A1". I'm just use to working with velocity as my vector. – 2017-02-14
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    Just a side note. My original question (which I edited) was on the use of parenthesis. If the operator looked like this $\textbf{A}\cdot \nabla (\textbf{A}\cdot \nabla)$, would that change the story? You've answered the question as stated above and I acknowledge that, but could you just comment on the importance of the parenthesis and how that affects the operator interpretation. (I also had one other question 2 comments up in case you over look it). – 2017-02-15
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    I think the last line of my post already answers your question. For the operator $A \cdot \nabla(A \cdot \nabla)$ you can also interpret it as "vector $A$ scalar product with gradient of the scalar $A \cdot \nabla$." Not sure if it yields the same, but I'm confident you can perform the calculations yourself to check. – 2017-02-15