Eigenvalues of $\not{p}$ Let $p_\mu$ be a generic non-null 4-vector, and let $m=\sqrt{p_\mu p^\mu}\neq 0$ (let us choose $\operatorname{Im}m\ge0$). Let also $\gamma^\mu$ be the Dirac gamma matrices.
Prove that the eigenvalues of $\not p=\gamma^\mu p_\mu$ are $\pm m$, and they are both 2-degenerate.
I know that must be true because of the explicit solutions of the equations $(\not p\pm m)u=0$ when I use a given representation of the gamma matrices. But is there a way to prove it independently of the representation? (I assume anyway that $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$.)
I also proved the first part of the statement, by using $\not p \not p = m^2$, but I can't think of any way to determine the multiplicity of the two eigenvalues. I know the only possibilities for the multiplicities of $m$ and $-m$ can be $(1,3)$, $(2,2)$, $(3,1)$, but I can't rule out the $1,3$ possibilities.

EDIT: My answer below relies on the assumption that the multiplicity of the eigenvalues $m$ and $-m$ sum to $4$, which is equivalent to say that $\not p$ is diagonalizable. But since $\not p^\dagger=\gamma^0\not p \gamma ^0\neq \not p$, this seems to me anything but trivial! How can I prove that $\not p$ is diagonalizable?
I have found that the whole statement is equivalent to $$\exists S\:\colon\:\not p=S^\dagger \gamma^0DS,S^\dagger\gamma^0=S^{-1},D=\operatorname {diag}(m,m,-m,-m)\,,$$ but I can't manage to prove it.
 A: After struggling for hours, the answer appeared clearly in my head just after having posted.
Since $\not p \not p = m^2$, then, for each eigenvector $u$, corresponding to the eigenvalue $\lambda$, $$\not p u = \lambda u \qquad \implies \qquad \not p \not p u 
= \lambda ^2 u \qquad \implies \qquad \lambda ^2 = m^2\,,$$ which proves that $\pm m$ are the only possible eigenvalues.

Answer to edit of OP (part 1):
Since $(m^{-1}\not p)^2=1$, then $m^{-1} \not p$ is diagonalizable, and so is $\not p$. As a consequence, the sum of the multiplicities of $m$ and $-m$ is $4$. Note that one of the two multiplicities may be zero at this point.

Let then $n$ be the multiplicity of the eigenvalue $m$. Then $4-n$ is the multiplicity of the eigenvalue $-m$. $$0=\operatorname{tr} \not p=nm+(4-n)(-m)=(2n-4)m\qquad \implies \qquad n=2\,.$$
This proves that $\exists S$ such that $\not p = S^{-1}DS$.

Answer to edit of OP (part 2):
As for the condition $S^{-1}=S^\dagger \gamma^0$ it is not provable as in general it is false; however, it is possible to choose such an $S$, since $$\begin{aligned}\not pu & =mu\\
\not pv & =-mv
\end{aligned}
\quad\implies\quad\bar{v}u=v^{\dagger}\gamma^{0}u=0\,.$$
A: As written, the statement seems to be incorrect: if $p_\mu$ is a space-like vector, $p_\mu p^\mu=-m^2$ (provided the signature is $\{1,-1,-1,-1\}$). It is not assumed in the wording of the statement that the Dirac equation is satisfied. Moreover, the sign of the magnitude || in the statement suggests that $p_\mu$ can indeed be space-like.
