# How does one represent the gradient $\nabla \mathbf A$ of a vector field $\mathbf A$ as a matrix?

I have been searching google about the gradient of a vector. I have found two distinct types of matrix notation for the gradient of a vector. I can't understand which one is correct, or if both are correct. The two are:

$$\nabla A = \partial_iA_je_i\otimes e_j=\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix}$$

or the transpose of the first, that is

$$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix}.$$

• 'the Gradient' Gradient is a subset of the Jacobian, the diagonal of the Jacobian Matrix Jacobian – f5r5e5d Jul 22 '17 at 23:06
• 1. What do you mean by "correct" here, i.e. why can't these just be two differing conventions? 2. How is this a physics rather than a Mathematics question? – ACuriousMind Jul 22 '17 at 23:13
• The the second one seems to be the transpose of the first one and the second one also can be described by Jacobian of (A1,A2,A3/X1,X2,X3).Where the first one can not be written as Jacobian transformation as like the second one.So,what's the right way to write the matrix for gradient of vector -the first one or the second(as a jacobian trans?And of course this is a physics-question.The first guy I have found in-eng.uc.edu/~beaucag/Classes/Processing/Chapter2html/… , the second one is in -umich.edu/~bme456/ch1mathprelim/bme456mathprelim.htm#vectordyad . – Abu sayed Jul 23 '17 at 4:06
• To second ACuriousMind, these are two different notation conventions. The difference can actually be important because equations look different, so just make sure what notation the author is using ... – Sanya Jul 23 '17 at 10:46

The matrix notation for the gradient of a vector field is generally not-that-well-defined, because as you note there are two reasonable matrix encodings (transposes of each other), which are about equally reasonable. Because of that, whenever the notation $\nabla\mathbf A$ as a matrix is used in the literature, the responsible thing to do is to specify explicitly which of the two encodings is meant.

Both of the two representations have features that make them reasonable:

• The second one you mention, $$\nabla A = \begin{bmatrix}\frac{\partial A_1}{\partial X_1} & \frac{\partial A_1}{\partial X_2} & \frac{\partial A_1}{\partial X_3}\\\frac{\partial A_2}{\partial X_1} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_2}{\partial X_3}\\\frac{\partial A_3}{\partial X_1}& \frac{\partial A_3}{\partial X_2}& \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the contravariant index as a row index and the covariant index of the gradient as a column index, which is reflective of both of those natures.

• On the other hand, the first representation in your post, $$\nabla A =\begin{bmatrix}\frac{\partial A_1}{\partial X_1} &\frac{\partial A_2}{\partial X_1} & \frac{\partial A_3}{\partial X_1}\\\frac{\partial A_1}{\partial X_2} & \frac{\partial A_2}{\partial X_2} & \frac{\partial A_3}{\partial X_2}\\\frac{\partial A_1}{\partial X_3} & \frac{\partial A_2}{\partial X_3} & \frac{\partial A_3}{\partial X_3}\end{bmatrix},$$ has the nice property that if $\mathbf r$ and $\mathbf v$ are column vectors, then $$\mathbf r^T \cdot \nabla \mathbf A\cdot \mathbf v = x_i \frac{\partial A_j}{\partial x_i} v_j$$ matches what you would expect as a matrix product, i.e. matching the action of the operator $\mathbf r \cdot \nabla = x_i \frac{\partial}{\partial x_i}$.

For an example of this in action in the literature, see this paper of mine: to avoid confusion, it is important to specify both how the matrix representation is chosen, and how it acts via components. It takes an extra line and it adds a whole lot of clarity to the text.

The gradient of a function is well defined in the literature. The gradient of a vector field $A = A^i\partial_i$ seems to be the gradient of its components (which are functions). I think both matrix representations ($\partial_i A^j$ and $\partial_j A^i$) are "good" since they are simply representations of the same thing : $$\nabla A = (\partial_i A^j)dx^i\otimes \partial_j = (\partial_j A^i)dx^j \otimes \partial_i$$ Choose one of them and stay with it (at least until the end of the proof where you are using gradients of vector field).

Geometrically speaking, the gradient of a vector field you are talking about can be written either as the Lie derivative $\mathcal{L} A$ (along $\partial_i$ or $\partial_j$) or the covariant derivative $\nabla A$ (with vanishing Christoffel symbols $\Gamma$'s) (along $\partial_i$ or $\partial_j$). Both $\mathcal{L} A$ and $\nabla A$ are a little bit overkill for your setting (probably Euclidean flat space) but are good to keep in mind if you need to generalize from $\mathbb{R}^n$ to some manifold with global properties.

• What you have mentioned "same thing",but the second one is actually the jacobian of (A1,A2,A3/X1,X2,X3) whereas the first one don't seems to be the jacobian transformation like J(,,/,_,).So,what is the fact? – Abu sayed Jul 23 '17 at 4:15
• Yes, same geometrical thing. A matrix representation is only a table of numbers, or functions and doesn't give any insight about the geometry behind it (i.e. in terms of sections of bundles). – Noé AC Jul 23 '17 at 7:05
• Both matrix representations are obviously different, so ignoring that is really unhelpful. Mentioning "the" Lie derivative here is also pretty pointless. – Sanya Jul 23 '17 at 10:43
• @Sanya I said it "seems to be related" to the Lie derivative. Looking at Deep's answer probably it's a better idea to look for covariant derivative with null Christoffel $\Gamma$'s. Though here we don't know the context of the original question so both Lie derivative and covariant derivative are somewhat overkill. – Noé AC Jul 23 '17 at 16:11
• @Sanya Also, in local coordinates, $\nabla A = (\partial_i A_j)dx_i \otimes \partial_j = (\partial_j A_i) dx_j \otimes \partial_i$, so using the matrix $(\partial_i A_j)$ or it's transpose $(\partial_j A_i)$ is ok, as long as you know what $i$ and what $j$ are linked to ($e_k$ or $e_k^*$) – Noé AC Jul 23 '17 at 16:22

$\nabla A$ is a tensor and it acts on a vector, say $\mathbf{u}$, to produce another vector: $\mathbf{v}=\nabla A(\mathbf{u})$. In matrix representation, you multiply the matrix of $\nabla A$ with the specified column/row vector representing $\mathbf{u}$. If you write $\mathbf{u}$ as a column vector (usual convention) then the second matrix representation is the correct one, while if you write it as a row vector the first matrix representation is the correct one.

However matrix representation is not necessary. You have $\nabla A=\partial_iA_j~\mathbf{e}_i\otimes\mathbf{e}_j$, and the action of $\nabla A$ on $\mathbf{u}$ is defined to be the vector $\mathbf{v}=\partial_iA_j~\mathbf{e}_i(\mathbf{u})~\mathbf{e}_j=(\partial_iA_j)u_i~\mathbf{e}_j$. No more worries about how the matrix should be written.