Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 2 / c / iii

Let $P={\chi }_M$, and suppose $P(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$. The reciprocal identity gives, for $x\ne 0$,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. HenceThis proves equality of the algebraic multiplicities, not merely that the inverse is another eigenvalue. Invertibility excludes zero eigenvalues throughout.