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

Rearrange $M^{T}JM=J$ asThus $M^{-1}$ is similar to $M^T$. Similar matrices have the same characteristic polynomial by part (a), and transposition also preserves it:Therefore $\boxed{{\chi }_{M^{-1}}={\chi }_M}$. This is the underlying algebraic source of the reciprocal spectrum of a symplectic matrix.

For $\lambda \ne 0$, factor the matrix inside the determinant:The second line uses $\det M=1$; the third uses the even matrix size; the last uses part (i). ConsequentlyIn other words, the characteristic polynomial equals its reciprocal polynomial.

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.

The fixed points of reciprocal root pairing are $1$ and $-1$, so part (iii) alone does not establish their parity. Use the polynomial identity itself.

Write $P(x)=(x-1{)}^{r}A(x)$, where $A(1)\ne 0$ and allow $r=0$ if one is not a root. After substituting in $P(x)=x^{2m}P(1/x)$ and canceling $(x-1{)}^r$ away from $x=1$, obtainLetting $x\to 1$ yields $A(1)=(-1{)}^{r}A(1)$, so $r$ is even.

Similarly write $P(x)=(x+1{)}^{s}B(x)$ with $B(-1)\ne 0$. This time cancellation gives $B(x)=x^{2m-s}B(1/x)$. Letting $x\to -1$ yields $B(-1)=(-1{)}^{2m-s}B(-1)$, so $s$ is even. ThusThe conclusion includes multiplicity zero. It is a consequence of the reciprocal identity in even degree, which is stronger at these fixed points than just pairing the roots.