I would like to get an answer to the following question: Let $\Delta_n$ be the vector space of complex $n$spinors. A vector $X \in \mathbb{R}^n$ acts on $\Delta_n$ by Clifford multiplication. We can thus for every $X \in \mathbb{R}^n$ define a map $F_{X}: \Delta_n\rightarrow \Delta_n$ given by $F_{X}(\phi)=X \cdot \phi$ for every $\phi \in \Delta_n$, where $\cdot$ denotes the Clifford multiplication. My question now is: Can one determine the eigenvalues and eigenvectors of $F_{X}$. If so, what are they? Every help will be appreciated!

2$\begingroup$ What you need is a realization of the spin representation in terms of matrices. In the book ''Twistor and Killing spinors or Riemannian manifolds'' of Baum, Friedrich, Grunewald and Kath, one can find an explicit description of the spin representation and so also of the Clifford multiplication. For $n=4$ the spin module $\Delta_{4}=\Delta_{4}^{+}\oplus\Delta_{4}^{}$ is described in page 72 and also the Clifford multiplication, for $\Delta_{6}$ see page 125 and for $\Delta_{7}$ see page 97. See also the paper: ''On Nearly Parallel $G_2$Structures'', page 2, for $\Delta_{7}$. $\endgroup$ – 314159. Jul 3 '16 at 9:30
We can make the following easy computation. Let us assume for simplicity that we are in the realm of complex Clifford algebras and complex Clifford representations. Using OP's conventions, we want to compute the eigenvalues of the linear map defined by $F_{X}(\phi) = X\cdot \phi$ for all spinors $\phi$. Let us assume that $\lambda\in\mathbb{C}$ is such an eigenvalue with eigenvector $\phi$. Then:
$X\cdot\phi = \lambda \phi \Rightarrow X\cdot X\cdot \phi = \lambda X\cdot \phi$
which implies
$X^{2}\cdot\phi = \lambda^{2} \phi \Rightarrow g(X,X) = \lambda^{2}$
So if $F_{X}$ has any eigenvalues at all, they can only be given by $\pm \sqrt{g(X,X)}$. Hence, either both eigenvalues are real, or both are purely imaginary. Notice that in many situations $F_{X}$ will not have any eigenvalues when restricted to chiral representations. For example, in even dimensions $\Delta_{n}$ splits in chiral representations as $\Delta_{n} = \Delta^{+}_{n} \oplus \Delta^{}_{n}$ and in that case Clifford multiplication interchanges chirality and hence it cannot have chiral eigenvectors .

$\begingroup$ Last sentence, $v_+ + i w_i$ is allowed. Complexified already. $\endgroup$ – AHusain Jun 21 '17 at 23:37