= Solution
Let $P=\chi_M$, and suppose $P(z)=(z-\lambda)^dR(z)$ with $R(\lambda)\ne0$. The reciprocal identity gives, for $x\ne0$,
$$
P(x)=x^{2m-d}(1-\lambda x)^dR(1/x).
$$
Near $x=\lambda^{-1}$ the factor $x^{2m-d}R(1/x)$ is nonzero. Also $1-\lambda x=-\lambda(x-\lambda^{-1})$ has a simple zero there. Hence
$$
\boxed{\lambda^{-1}\text{ is a root of multiplicity exactly }d.}
$$
This proves equality of the <algebraic multiplicities>, not merely that the inverse is another <eigenvalue>. Invertibility excludes zero eigenvalues throughout.
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