Local fermionic symmetry and GS action I have a trouble understanding an argument which I think has a simple answer but I am not getting it.
The question is that if you don't impose local fermionic symmetry the GS action has only one term and one of the equation of motion looks as $Γ\cdot p\partial_0\psi=0$, where $\psi$ is a Majorana-Weyl spinor in 10 dimensions and $Γ$ are just generalized gamma matrices.
Now $(Γ\cdot p)^2 = p^2= 0$ for the on-shell condition, but now they say that since the square is zero hence half of the eigenvalues of $Γ\cdot p$ is zero and this is where I am not getting it. If the square of a matrix is zero then how can we say that half of it's eigenvalues are zero. Since the eigenvalues of a squared matrix is square of the eigenvalues of the matrix whose square is taken, so how can we say that half of the eigenvalues are zero? May be it can happen when spinor contain complex component but here the spinor is Majorana too (thinking in some different way that how it will impose constraint on half of the component of $\psi$ but this is not working either).
(For reference, see chapter 5 of GSW Vol.1.)
 A: If a linear map $A:V\to V$ in a finite vector space $V$ squares to zero 
$$\tag{1} A^2~=~0,$$ 
then the image of $A$ must lie in the kernel of $A$:
$$\tag{2} {\rm Im}(A)~\subseteq~{\rm Ker}(A),$$ 
and therefore
$$\tag{3} \dim {\rm Im}(A) ~\leq ~\dim {\rm Ker}(A).$$ 
Now we also know that
$$\tag{4}  \dim {\rm Im}(A) ~= ~\dim V - \dim {\rm Ker}(A). $$
Together (3) and (4) yield that the dimension of the kernel of $A$ is at least half the dimension of the full space $V$,
$$\tag{5} \dim {\rm Ker}(A)~\geq ~ \frac{1}{2}\dim V , $$
which implies that at least half the eigenvalues of $A$ are zero.
A: Let us go into a frame where the light-like vector $p$ has only $p^0$ and $p^1$ as non-zero components with equal magnitude. So the question reduces to the properties of $\Gamma^0 + \Gamma^1$ (or perhaps $\Gamma^0 - \Gamma^1$ but the argument below works the same way for that).
If we look at the explicit bootstrapping of gamma matrices from 2d via repeated Kronecker products, we see that the $\Gamma^0 + \Gamma^1$ in even dimensions $d=2k$ is $(\sigma^1 + -\mathrm{i}\sigma^2)(\otimes \sigma^3)^{k-1}$, where by $(\otimes \sigma_3)^{k-1}$ I mean that we take the Kronecker product with $\sigma_3$ $k-1$-times.
Since $\sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$, the rank of $A\otimes \sigma^3$ is twice the rank of $A$. So in particular, if the rank of $A$ is half the dimension, so is the rank of $A\otimes \sigma^3$.
Finally, $\sigma^1 + -\mathrm{i}\sigma^2 = \begin{pmatrix}0 & 0 \\ 2 & 0\end{pmatrix}$, which obviously has half-rank, and therefore so does $\Gamma^0 + \Gamma^1$ in all even dimensions. Note that that the statement "half the eigenvalues are zero" is actually wrong, since $\Gamma^0 + \Gamma^1$ is nilpotent and nilpotent operators only have zero eigenvalues. The correct statement is that the rank of the operator is half the total dimension.
