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.
Best Answer
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.