Symbols of elliptic operators First let me state the problem, then I'll  explain its origin and finally, I'll ask the main question..
Problem S.  Fix a positive integer $n$. Find all the pairs $(V, S)$, whith the following properties.
1. $V$ is a finite dimensional  complex  vector space  equipped with a Hermitian metric. $\DeclareMathOperator{\Sym}{Sym}$ We denote by $\Sym(V)$ the space of symmetric complex linear operators $V\to V$.
2. $\newcommand{\bR}{\mathbb{R}}$  $S$ is a linear map $S:\bR^n\to\Sym(V)$ such  that for any $\xi\in\bR^n\setminus 0$ the symmetric operator $S(\xi)$ is invertible.
Readers familiar with the basics of p.d.e.-s will surely recognize $S(\xi)$  as the principal symbol of an elliptic, first order  partial differential operator with constant coefficients that acts on $C^\infty(\bR^n, V)$. That explains the letter $S$ in the name of the problem.
We  denote by $\newcommand{\eS}{\mathscr{S}}$ $\eS_n$ the space of  solutions of Problem S for a given positive integer $n$.
Observe that $\eS_n$  is equipped with  two basic algebraic operations $\oplus,\otimes$ $\newcommand{\one}{\boldsymbol{1}}$
$$(V_1, S_1)\oplus (V_2, S_2):= ( V_1\oplus V_2, S_1\oplus S_2), $$
$$ (V_1, S_1)\otimes (V_2, S_2):=  ( V_1\otimes V_2, S_1\otimes\one_{V_2}+ \one_{V_1}\otimes  S_2). $$
The group $\DeclareMathOperator{\GL}{GL}$ $\GL(n,\bR)$ acts in an obvious way on $\eS_n$.  More precisely if $S:\bR^n\to\Sym(V)$  is a solution $S\in\eS_n$, and $T\in \GL(n,\bR)$, then $S\circ T\in \eS_n$.
Let us  also observe that  for each $n$, the set $\eS_n$ is not empty.    We can obtain maps $S: \bR^n\to\Sym(V)$ with the desired properties by using complex representations of the Clifford algebra generated by an    Euclidean  inner product on the   space $\bR^n$.   I will  refer to such examples as Clifford examples and I will denote by $\newcommand{\eC}{\mathscr{C}}$ $\eC_n$ the subset of  $\eS_n$ constructed  as above using representations of    Clifford algebras. Observe that $\eC_n$ is also closed  under the operations $\oplus,\otimes$ and invariant under the above action of $\GL(n,\bR)$
Main Question.  Fix $n$ Are there non Clifford solutions to Problem S? In other words, is the set $\eS_n\setminus \eC_n$ non-empty?
Addendum.   Apparently this question is related to a  classical question  discussed by Porteous  in his  book Topological Geometry.          For a given real vector space $V$ find  the largest $n$  find the maximal subspaces  $\DeclareMathOperator{\Endo}{End}$  $S\subset \Endo(V)$ such that $S\setminus 0 \subset \GL(V)$.   The answer  has to do with Radon-Hurwitz numbers, and  it basically  says that if $S$ is   such a  subspace, maximal or not,  then $V$  s a module over   the Clifford algebra generated by  an inner product  on $S$. 
 A: Maybe I'm misunderstanding something, but it seems that the answer is probably 'no', at least if $d = \dim_\mathbb{C} V$ is large enough.  
What really matters is the $n$-dimensional real subspace $\mathsf{S}=\mathrm{Im}(S)\subset\mathrm{Sym}(V)$.  What you need is that this space not meet the cone of singular matrices in $\mathrm{Sym}(V)$ anywhere but at the origin.  This is an open condition on the element $\mathsf{S}\in\mathrm{Gr}_n\bigl(\mathrm{Sym}(V),\mathbb{R}\bigr)$, and this latter space has dimension $n(d^2{-}n)$.  Thus, the set of such subspaces with your property has this latter dimension, but the group of unitary transformations on $V$ (which, I gather, is the space of symmetries of the problem) is only of dimension $d^2$, so there must be at least an $(n{-}1)d^2-n^2$ parameter family of 'inequivalent' subspaces that meet your criteria.  
However, there are only a finite number of inequivalent $d$-dimensional complex representations of the Clifford algebra on $\mathbb{R}^n$ endowed with a definite inner product.
