Div, curl and linear algebra I came across this post lying dormant on some online forum. I am putting it here verbatim, it seems to me worth a lot.

By Prof. S. D. Agashe, IIT Bombay
(Source: Vector Calculus, by Durgaprasanna Bhattacharyya, University
Studies Series,Griffith Prize Thesis, 1918, published by the University
of Calcutta, India, 1920, 90 pp)
Chapter IV: The Linear Vector Function, article 15, p.24:
"The most general vector expression linear in $r$ can contain terms only
of three possible types, $r$, $(a\cdot r)b$ and $c\times r$, $a$, $b$, $c$ being constant unit vectors. Since $r$, $(a\cdot r)b$ and $c\times r$ are in general non-coplanar, it follows from the theorem of the parallelepiped of vectors that the most general linear vector expression can be written in the form $\lambda \cdot r + \mu (a\cdot r)b + \nu (c\times r)$, where $\lambda, \mu, \nu$ are scalar constants".
Bhattacharyya does not prove this. Has anyone seen a similar result and its proof?
Bhattacharyya uses this to show that the divergence of the linear
function is ($3 \lambda + a\cdot b$), that the curl is ($a \times b + 2c$). He goes on to define div and curl of a differentiable function as the div and curl of the (linear) derivative function. The div and curl of a linear function are defined in terms of certain surface integrals.
I am excited about this result because it seems to provide an excellent
route to div and curl, as Bhattacharyya himself remarks.
Sorry for a rather long and "technical" communication.
 A: The claim is true.  Any $3\times3$ matrix can be expressed as 
$$
A= \lambda I+  a b^T + B
$$
where $\lambda$ is real, $a$ and $b$ are 3-vectors and $B$ is skew
(so that $Bx=c\times x$ for some vector $c$).
To prove this, choose an orthogonal matrix
$Q$ to diagonalize the symmetric part of $A$.
Then $Q^TAQ=D+K$ where $D$ is diagonal and $K$ is skew.
If the diagonal entries of $D$ are not all distinct then it is
easy to write
$D=\lambda I+\hat a \hat b^T$ and we finish as below.
If the entries are all distinct, 
we can suppose that $Q$ was chosen so that 
the largest eigenvalue of $D$ is first, the smallest 
second and the middle last.  Then for some positive 
$\mu$ and $\nu$, the matrix
$D$ can be written
$$
D = \lambda I + \mu
\begin{pmatrix}
1 & 0 & 0\cr 0 &-\nu^2&0\cr
0&0&0
\end{pmatrix}
=\lambda I + \hat a\hat b^T+\hat K,
$$ with $$
 \hat a= \mu\begin{pmatrix}1\cr \nu\cr0\end{pmatrix}, 
\quad
\hat b=
\begin{pmatrix} 1 \cr -\nu\cr 0 \end{pmatrix},
\quad \hat K=\mu\begin{pmatrix}
0&\nu&0\cr -\nu & 0&0\cr 0&0&0
\end{pmatrix} .
$$
Let $a=Q\hat a$, $b=Q\hat b$, and $B=Q(\hat K+K)Q^T$,
and you're done.
A: I'm not sure if it is such a good approach. The most general vector linear in $r$ is $M r$ where $M$ is a 3x3 matrix. The number of unfixed constants in your formula is 12, while you only need 9 in general. If you set $b=c$ it seems more sensible, assuming $r\times c\neq0$. Then it's essentially just the expansion of a vector in the basis $r$, $c$, $r\times c$.
The matrix $M$ can be divided into 3 parts, the trace part, a traceless symmetric part and a antisymmetric part:
$$ M = \frac13\mathrm{tr}(M)\,I + \frac12(M+M^t-\frac23\mathrm{tr}(M)\,I) + \frac12(M-M^t) $$
where $I$ is the identity matrix and ${}^t$ denotes transpose.
So, the terms in $M\cdot r$ correspond to the terms
$$ x r + y (r\cdot a)c + z (r\times c)\ .$$ You can see exactly how by just taking the gradient of both sides to get
$$ M = x\, I + y\, c a^t + z\, \varepsilon\cdot c$$
where $\varepsilon$ is the totally antisymmetric tensor and $c a^t$ can be chosen to be traceless, ie $a \cdot c=0$. In terms of components, this reads
$$ M_{ij} = x\, \delta_{ij} + y\, c_i a_j + z\, \sum_k\varepsilon_{ijk} c_k \ ,$$
where $\delta$ is the Kronecker delta symbol.
