= Solution
If $Av=\lambda v$ for $A=\alpha_i$ or $\beta$, then $A^2=I$ implies $\lambda^2=1$. The polynomial $x^2-1$ has distinct roots, so such a matrix is diagonalizable and
$$
\boxed{\lambda=+1\text{ or }-1.}
$$
Anticommutation with another invertible Dirac matrix maps a $+1$ eigenspace to a $-1$ eigenspace, so both signs occur with equal multiplicities in a nontrivial representation. A Hermitian operator is appropriate for the physical <Hamiltonian> and additionally guarantees an <orthonormal> <eigenbasis>.
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