What does it mean to take the gradient of a vector field? $\nabla \vec{v}(x,y,z)$? I only understand what it means to take the grad of a scalar field.

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    $\begingroup$ The short answer is: the gradient of the vector field $\sum v_i(x, y, z)e_i$, where $e_i$ is an orthonormal basis of $\mathbb{R}^3$, is the matrix $(\partial_i v_j)_{i, j=1, 2, 3}$. $\endgroup$ Jun 11, 2012 at 8:48
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    $\begingroup$ The long answer involves tensor analysis and you can read about it on books such as Itskov, Tensor algebra and tensor analysis for engineers. $\endgroup$ Jun 11, 2012 at 8:49
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    $\begingroup$ Another possible explanation is that the dot is missing between $\nabla$ and $\vec v$, and the "gradient" is actually divergence. $\endgroup$
    – user31373
    Jun 11, 2012 at 14:27
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    $\begingroup$ Guiseppe Negro's short answer is off, switch his i's and j's and its fixed. $\endgroup$
    – user78853
    May 21, 2013 at 21:10
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    $\begingroup$ @Thomas I've converted your answer into a comment. In the future please only use answers to give answers to the question posed in the original post. I realize that you do not yet have the reputation to comment on other people's posts, but that will come before long if you contribute to the site. $\endgroup$ May 21, 2013 at 22:12

6 Answers 6


The gradient of a vector is a tensor which tells us how the vector field changes in any direction. We can represent the gradient of a vector by a matrix of its components with respect to a basis. The $(\nabla V)_{\text{ij}}$ component tells us the change of the $V_j$ component in the $\pmb{e}_i$ direction (maybe I have that backwards). You can check out the Wikipedia article for the details of calculating the components.

To get a physical picture of its meaning we can decompose it into 1) the trace (the divergence) 2) an anti-symmetric tensor (the curl) 3) a traceless symmetric tensor (the shear)

If the vector field represents the flow of material, then we can examine a small cube of material about a point. The divergence describes how the cube changes volume. The curl describes the shape and volume preserving rotation of the fluid. The shear describes the volume-preserving deformation.


Assume the vector $\vec{\bf F} = (F_1, F_2, F_3)$ exists in a 3D space with basis $x_1, x_2, x_3$, then its gradient is the 3 × 3 matrix: $\partial_i$$F_j$

$\nabla\vec{\bf F}=\left( {\begin{array}{c} \frac{\partial F_1}{\partial x_1}&\frac{\partial F_1}{\partial x_2}&\frac{\partial F_1}{\partial x_3}\\ \frac{\partial F_2}{\partial x_1}&\frac{\partial F_2}{\partial x_2}&\frac{\partial F_2}{\partial x_3}\\ \frac{\partial F_3}{\partial x_1}&\frac{\partial F_3}{\partial x_2}&\frac{\partial F_3}{\partial x_3}\\ \end{array}} \right).$

That is, each column is a "usual" gradient of the corresponding scalar component function.

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    $\begingroup$ I think that's wrong because the correct result is the transpose of what you wrote. take a look at the page 7 of this document: $\endgroup$
    – Freshman42
    Apr 20, 2018 at 20:57
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    $\begingroup$ According to Riley, Hobson and Bence (3rd), we have $\nabla\vec{\bf F}=\left( {\begin{array}{c} \frac{\partial F_1}{\partial x_1}&\frac{\partial F_1}{\partial x_2}&\frac{\partial F_1}{\partial x_3}\\ \frac{\partial F_2}{\partial x_1}&\frac{\partial F_2}{\partial x_3}&\frac{\partial F_2}{\partial x_3}\\ \frac{\partial F_3}{\partial x_1}&\frac{\partial F_3}{\partial x_2}&\frac{\partial F_3}{\partial x_3}\\ \end{array}} \right).$ It states on page 937: $\endgroup$ Apr 9, 2021 at 17:37
  • $\begingroup$ The gradient of a vector. Suppose $v_i$ represents the components of a vector; let us consider the quantities generated by forming the derivatives of each $v_i, i =1,2,3$, with respect to each $x_j, j =1,2,3$ i.e. $$T_{ij} = \frac{\partial v_i}{\partial x_j}\ .$$ $\endgroup$ Apr 9, 2021 at 17:41
  • $\begingroup$ @ Freshman42 Actually, that PDF is incorrect. If you for example consider a vector field of 2-vectors in 3-space, multiplying the resulting gradient matrix with the 3-vector along which we want to take the directional derivative in order to get the derivative, which is a 2-vector, only works if the matrix is what Mussé Redi describes. $\endgroup$
    – Smiley1000
    Dec 8, 2021 at 19:12
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    $\begingroup$ I believe the discussion about whether this or the transposed is correct comes down to whether numerator or denominator layout is used. The one used in the answer is numerator layout. $\endgroup$
    – lucidbrot
    Mar 21, 2022 at 14:29

It depends on how you define the gradient operator. In geometric calculus, we have the identity $\nabla A = \nabla \cdot A + \nabla \wedge A$, where $A$ is a multivector field. A vector field is a specific type of multivector field, so this same formula works for $\vec v(x,y,z)$ as well.

So we get $\nabla\vec v = \nabla \cdot \vec v + \nabla \wedge \vec v$. The first term should be familiar to you -- it's just the regular old divergence. However the second term is a different type of object entirely (actually, it's a generalization of the familiar $3$D curl $\nabla \times \vec u$ that works in any dimension).

In the same way that a vector field can be though of as associating with every point in your domain an oriented line segment (a vector), $\nabla \wedge \vec v$ associates with every point in your domain an oriented plane segment (which we call bivectors). So $\nabla \wedge \vec v$ is called a bivector field.

So to answer your question, the gradient of a vector field is the sum of a scalar field and a bivector field.

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    $\begingroup$ Interesting! However, how to reconcile this language with the usual one? It seems non trivial: you never add scalars and vectors in "standard" vector analysis. Which is the interpretation? $\endgroup$
    – Quillo
    May 25, 2020 at 1:08

Gradient of a vector field (or a multi-valued function $f: R^m\to R^n$) is jacobian of the multi-valued function $f$, where each row $r_i$ of the $\text{Jacobian}(f)$ represents the gradient of $f_i$ (remember, each component $f_i$ of the multi-valued function $f$ is a scalar).

  • $\begingroup$ Hope this edit is acceptable. $\endgroup$
    – awllower
    Mar 31, 2013 at 8:24

Gradient of a vector field is intuitively the Flux/volume leaving out of the differential volume dV. Visualise in 2D first. Suppose you have a vector field E in 2D. Now if you plot the Field lines of E and take a particular Area (small area..), Divergence of E is the net field lines, that is, (field line coming out of the area minus field lines going into the area). Similarly in 3D, Divergence is a measure of (field lines going out - field lines coming in). If you mathematically implement this you see you get 3 terms of partial derivatives added, which essentially adds the total net field lines.

For a scalar field(say F(x,y,z) ) it represents the rate of change of F along the the 3 perpendicular ( also called orthonormal ) vectors you defined your system with (say x, y, z ).


The gradient of a vector field corresponds to finding a matrix (or a dyadic product) which controls how the vector field changes as we move from point to another in the input plane.

Details: Let $ \vec{F(p)} = F^i e_i = \begin{bmatrix} F^1 \\ F^2 \\ F^3 \end{bmatrix}$ be our vector field dependent on what point of space we take, if step from a point $p$ in the direction $ \epsilon \vec{v}$, we have:

$$ \vec{F(p+ \epsilon \vec{v})} = F^i(p+ \epsilon v) e_i= F^i(p) e_i + \epsilon \left(v \cdot \nabla F^i(p) \right) e_i$$

But, what is $( v \cdot \nabla F^i ) e_i= \begin{bmatrix} \nabla F^1 \\ \nabla F^2 \\ \nabla F^3 \end{bmatrix} v= \begin{bmatrix} \partial_1 F^1 & \partial_2 F^1 & \partial_3 F^1 \\ \partial_1 F^2 & \partial_2 F^2 & \partial_3 F^2 \\ \partial_1 F^3 & \partial_2 F^3 & \partial_3 F^3 \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \\ v^3 \end{bmatrix}$

And that's what the gradient of a vector field is, a big matrix controls how the gradient vector changes when we move in any direction of the input space.

Note that I have assumed the eulicdean basis (i.e: i,j,k)


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