# Vector Calculus Notation for "Gradient of a Vector"

Given (differentiable) functions $\,n_{1,2}:\mathbb{R}\to\mathbb{R}\,$ we write vector $\renewcommand{\arraystretch}{2}$

\begin{align} \vec{\boldsymbol{n}} = \begin{bmatrix} n_{1} \\ n_{2} \end{bmatrix} \end{align}

and define "gradient of a vector" operation $\,:\mathbb{R}^2\to\mathbb{R}^{2\times2}\,$ as

\begin{align}\label{1}\tag{$\boldsymbol{\ast}$} \begin{bmatrix} \,^{\partial}/_{\partial x}\, \\ \,^{\partial}/_{\partial y}\, \end{bmatrix} \begin{bmatrix}\,n_{1} & n_{2} \end{bmatrix} = \begin{bmatrix}\, \big(n_{1}\big)_{x} & \big(n_{2}\big)_{x} \\\, \big(n_{1}\big)_{y} & \big(n_{2}\big)_{y} \end{bmatrix}. \end{align}

Is it possible to write the expression above in terms of nabla operator and vector $\,\vec{\boldsymbol{n}}$? For example, I was thinking about something like

\begin{align} \begin{bmatrix}\, \big(n_{1}\big)_{x} & \big(n_{2}\big)_{x} \\\, \big(n_{1}\big)_{y} & \big(n_{2}\big)_{y} \end{bmatrix} = \nabla \circ \vec{\boldsymbol{n}}^{T} \end{align}

where $\,\circ\,$ stands for some (vector calculus?) operation.

What would be an appropriate symbol to write instead of $\,\circ\,$? Is there standard notation for such operation?

I know that the matrix on the right hand side of equation $\eqref{1}$ can be viewed as Jacobian matrix of vector-valued function $\,\vec{\boldsymbol{n}}\,(x,y):\mathbb{R}^{2}\to\mathbb{R}^{2},\,$ but I would like to be able to express in using just the vector $\,\vec{\boldsymbol{n}}\,$ and differential operations of vector calculus.

Yes, there exist actually a standard notation for that. If you look up into a tensor calculus book and read a little about the Tensor Product or Dyadic Product or sometimes called the outer product of vectors, $\otimes$, you will find what you want. Usually, it is written in this way

$\nabla \otimes {\bf{n}} \equiv \nabla {\bf{n}}$

where the second one is mostly used for abbreviation! This link may help you a little. Also, see this one. This book by Itskov is a good one if you want to see the theory and application at the same time. Also, if you are some rigorous mathematician you can take this book by Marsden.

• Did you mean to write $\,\nabla\otimes{\bf{n}} \equiv \nabla {\bf{n}}^{T}$?
• @Vlad: Your answer is ${\bf{H}} = \nabla \otimes \nabla \otimes {\bf{n}} \equiv \nabla \nabla {\bf{n}}$. :) Nov 1, 2015 at 17:24