Consider a vector field $\underline{{f}}:\mathbb{R}^3\rightarrow \mathbb{R}^3$.
We know that $\underline{\nabla}\cdot\underline{f} = tr(D\underline{f})$,
$D\underline{f} =
\begin{pmatrix}
\frac{\partial f_{1}}{\partial x} & \frac{\partial f_{1}}{\partial y} & \frac{\partial f_{1}}{\partial z} \\
\frac{\partial f_{2}}{\partial x} & \frac{\partial f_{2}}{\partial y} & \frac{\partial f_{2}}{\partial z} \\
\frac{\partial f_{3}}{\partial x} & \frac{\partial f_{3}}{\partial y} & \frac{\partial f_{3}}{\partial z} \\
\end{pmatrix}
$
and $tr : M_{3\times 3}(\mathbb{R})\rightarrow\mathbb{R},\ tr(A) = \sum_{i=1}^3a_{i,i}$
Can $\underline{\nabla}\times\underline{f}$ be defined as a matrix operation on the matrix $D\underline{f}$?
I know that $\underline{a}\times\underline{b} = $
$\begin{vmatrix}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
a_x&a_y&a_z \\
b_x&b_y&b_z
\end{vmatrix}$
Can a function be defined $T:M_{3\times 3}(\mathbb{R}) \rightarrow \mathbb{R}^3$ such that
$T(D\underline{f}) = \underline{\nabla}\times \underline{f}$ explicitly?
Best Answer
Notice that $Df - (Df)^T$ produces an antisymmetric matrix with the correct components--this is similar to how cross products can be computed by turning one of the vectors into an antisymmetric matrix and performing a matrix multiply.
This is a general result---any matrix $T$ can be decomposed as follows:
$$T = \frac{\alpha}{n} I + T_- + S$$
where $\alpha$ is the trace, $T_- = (T-T^T)/2$ is the antisymmetric part, and $S$ is a symmetric, tracefree matrix. The trace corresponds to divergence, and the antisymmetric part corresponds to curl in 3d. In fact, this definition can be used to define a "curl-like" operation in higher dimensions, for while cross products do not generalize, the antisymmetric part of $Df$ does.