Let $\newcommand{\CCl}{\mathbb{C}l}\CCl_{p,q}$ be the complex Clifford-Algebra associated to the Minkowski space $\mathbb{R}^{p,q}$ of signature $(p,q)$, where we consider $\mathbb{R}^{p,q}$ as a linear subspace of $\CCl_{p,q}$ in the usual way. Moreover, let $\Phi:\CCl_{p,q}\to\operatorname{End}_\mathbb{C} S$ be a Dirac spinor representation, i.e. a finite-dimensional, irreducible, complex representation of $\CCl_{p,q}$.
Question: Is it possible that $\Phi(v)=1_S$ for some $v\in\mathbb{R}^{p,q}$?
For example, if $n=p+q$ is even, the representation theory for $\CCl_{p,q}$ implies that $\Phi$ is an isomorphism of $\mathbb{C}$-algebras, thus $\Phi(v)=1_S=\Phi(1_{\CCl_{p,q}})$ implies $v=1_{\CCl_{p,q}}$ - a contradiction. So the question is only interesting for the odd-dimensional case.
Okay, the answer is No, and the reason is quite simple: It can be shown that the Dirac matrices $\gamma_i:=\Phi(e_i)$ are traceless, so the same is true for all $\Phi(v)$ with $v\in\mathbb{R}^{p,q}$. Now, if $\Phi(v)=1$ we can take traces and find $0=\dim S$, a contradiction.