0
$\begingroup$

I'm doing a problem where I'm taking the divergence of a matrix. If the dot product $A\cdot B$ is essentially $A^{T}B$ then surely $$ \nabla \cdot A =\nabla^TA$$

where A is a matrix. If $\nabla^T$ is 1x3 and $A$ is a 3x3, then matrix multiplication should yield a 1x3 too. Thus it is not a vector, but a vector transposed. However say $A$ is an outer product. Then $$\nabla \bullet (a\otimes b ) = (\nabla \cdot a) \otimes b $$ where $\nabla \cdot a$ is a scalar, and a scalar tensor product a vector is just the scalar times the vector. Whats with this contradiction?

$\endgroup$
2
  • $\begingroup$ Nabla, not niabla. $\endgroup$
    – user65203
    Jun 12, 2018 at 5:08
  • $\begingroup$ Where do you draw $\nabla\cdot (a\otimes b)=(\nabla\cdot a)\otimes b$ from ? $\endgroup$
    – user65203
    Jun 12, 2018 at 5:12

1 Answer 1

Reset to default
1
$\begingroup$

The contradiction between your results (that $\nabla \cdot A$ is a vector and its transponse) arises because $\nabla$ is not truly a vector, it is just commonly referred to and represented as such for mathematical convenience. (See this post on the Physics StackExchange for some discussion on this.) In my experience, this is in some sense because we can express each partial derivative in terms of components just like we can do for a vector, and because these transform w.r.t. coordinate systems just like typical vectors do. It is in some heuristic sense an "effective" vector.

That being said, there are some noteworthy things to observe about $\nabla^T$, particularly that it is correct to identify it as $\nabla \cdot$ in some cases (see this post). Where it fails is precisely when trying to use these operators on tensors, in which case the component-wise description of the operation is correct (try it using index notation!) but an intuitive linear algebra conceptualization of the result fails, as you've noted.

Hopefully this clears things up!

$\endgroup$
2
  • 1
    $\begingroup$ Isn't $\nabla^T$ equal to $-\text{div}$ rather than $\text{div}$? (Under suitable assumptions, of course.) $\endgroup$
    – littleO
    Jun 12, 2018 at 6:06
  • $\begingroup$ I -think- the negative sign pops up when discussing the del operator in the correct context of an operator, in which case its adjoint is -$div$ as you say. Above, we're discussing del in the heuristic context of an almost vectorial object with some "transpose" akin to that of a proper finite-dimensional vector. $\endgroup$
    – aghostinthefigures
    Jun 12, 2018 at 6:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .