From what i have seen different places on the web $\nabla$ and $\vec \nabla$ is being used on many of the same things. Is there a difference?

I thought that $\nabla f$ was an operator on a function. And $ \vec\nabla f = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} f$ was a vector on a function / scalar. In this example i would think i should get the same result. But if i look at $\vec u = (u,v,w)= \begin{pmatrix}u\\ v\\ w\end{pmatrix}$ i get:

$\nabla \vec u = \begin{pmatrix}\frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial y}\\ \frac{\partial w}{\partial z}\end{pmatrix}$ (My teacher said so). But i dont understand why this would not be $\nabla \vec u = \begin{pmatrix}\frac{\partial \vec u}{\partial x}\\ \frac{\partial \vec u}{\partial y}\\ \frac{\partial \vec u}{\partial z}\end{pmatrix}$. Wich is still a vector.

And if i take $\vec\nabla \vec u = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} \begin{pmatrix}u\\ v\\ w\end{pmatrix}$ i get $ \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix} \begin{pmatrix}u\\ v\\ w\end{pmatrix} = \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}$ Wich is not a vector. And this seems to be $\nabla \cdot \vec u$

So from what i see and think should $\nabla \vec u = \begin{pmatrix}\frac{\partial \vec u}{\partial x}\\ \frac{\partial \vec u}{\partial y}\\ \frac{\partial \vec u}{\partial z}\end{pmatrix} = \begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}&\frac{\partial w}{\partial x}\\ \frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}&\frac{\partial w}{\partial y}\\ \frac{\partial u}{\partial z}&\frac{\partial v}{\partial z}&\frac{\partial w}{\partial z}\end{pmatrix} = \vec \nabla \vec u^{T} = \begin{pmatrix}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\\ \frac{\partial }{\partial z}\end{pmatrix}\begin{pmatrix}u&v&w\end{pmatrix}$. What is wrong with this or why is it correct?

  • $\begingroup$ Arrows over operators (or ordinary vectors) are an author's preference, based on whether they think the reader needs them to understand. Omitting the arrow doesn't change the meaning; it just expects more of readers. $\endgroup$
    – J.G.
    Feb 9, 2022 at 14:40
  • 1
    $\begingroup$ I'm very skeptical about your “my teacher said so” statement. I've never seen $\nabla$ used in a way where element-wise application makes sense. Except if you do the dot product, where you sum the results as well, as you did later in your post. $\endgroup$
    – MvG
    Feb 9, 2022 at 14:48

2 Answers 2


Your teacher is wrong! You are indeed correct that, at least in the standard basis, $$(\nabla \vec u)^i{}_j=\frac{\partial u^i}{\partial x^j}$$ E.g in your example with $\vec{u}=(u~~v~~w)^\intercal~~$ and coordinates $x,y,z$, $$\nabla \vec{u} =\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}\\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix}$$

And the divergence is simply $$\operatorname{div} \vec u=\operatorname{tr}\nabla \vec u$$

(Normally I would use slightly different notation but I am sticking to the standard of your question.)


It's the same. The arrow is just used to clarify the fact that $\nabla$ is a vector.


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