Hello I am new to vector calculus and I have a basic question . The del operator which is defined as $\nabla = \Bigl(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\Bigr)$ which is basically a vector.

Let $a = (a_1,a_2,a_3)$ be a vector(or a vector function) . Then $a\cdot\nabla = a_1\frac{\partial }{\partial x}+a_2\frac{\partial }{\partial y}+a_3\frac{\partial }{\partial z}$ and $\nabla\cdot a =\frac{\partial a_1}{\partial x}+\frac{\partial a_2}{\partial y}+\frac{\partial a_3}{\partial z} $ obviously both are not equal i.e $\nabla\cdot a \neq a \cdot\nabla$ . But both $a,\nabla$ are vectors and for vectors dot product is commutative i.e $\nabla\cdot a = a \cdot\nabla$ should be true. I find this confusing , please explain.

  • $\begingroup$ you shall take the proper orientation for the vectors (column/row) and the proper convention on how the operator $\nabla$ will apply to vectors on the left or on the right. $\endgroup$
    – G Cab
    Feb 9, 2019 at 12:31

2 Answers 2


Yes as you noticed they are not equal.

It is true as you say that: $$a\cdot \nabla = a_1\cdot \frac{\partial}{\partial x}+a_2\cdot \frac{\partial}{\partial y}+a_3\cdot \frac{\partial}{\partial z}$$

And that $$\nabla \cdot a = \frac{\partial a_1}{\partial x}+\frac{\partial a_2}{\partial y}+\frac{\partial a_3}{\partial z}$$

$\nabla$ is an operator, not a vector. It works on functions, not vectors or scalars. It is still practical to use this notation because differentiation is a linear operation which means that almost always we can represent it with linear algebras by a matrix somehow.

The linearity how differentiation behaves on functions is somehow "compatible" with the language of linear algebra.

  • $\begingroup$ So you are saying that Del operator is not a vector , but in the book I am studying from Del was clearly given in terms of a vector each of whose components was the partial derivative operator .And in the question I forgot to mention 'a' can even be a vector function. I am changing that now. But even if 'a' is a vector function dot product should still work the same right ? $\endgroup$ Feb 9, 2019 at 13:09
  • $\begingroup$ It is a bit difficult to explain. Think of functions being special kinds of numbers. In higher algebra that is what functions are. Like really big complicated numbers. It is obvious that $f(x)\cdot g'(x) \neq g(x) \cdot f'(x)$, right? We should know this from calculus anyway. The nabla vector operates on something, changing what it was into something new. Also, the higher dimensions of $a$, the more careful we will need to be with our notation. For example if $a$ was a $2$-tensor it would not be completely obvious how to contract in a "dot" sense. $\endgroup$ Feb 9, 2019 at 13:28
  • $\begingroup$ Ohh I don't know much about n-dimensional algebra , then I shall take your word for it. Thanks !! $\endgroup$ Feb 9, 2019 at 13:38

The dot product is commutative if its arguments commute. But $\nabla$ is an operator and $a$ is a vector function; they do not commute.

This is however only one part of the explanation. The other part is that $\nabla \cdot a$ is defined by $\nabla$ only working on $a$. Had it been defined as an operator acting on a scalar function by $$(\nabla \cdot a) \phi := \sum_k \frac{\partial}{\partial x_k}(a_k \phi) \quad \quad\text{(NB: not the correct definition!)}$$ then the non-commutativity would have had more significance.


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