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

Choose a basis and let $A$ be the matrix of $T$. Use the monic conventionUnder a change of basis, $A$ becomes $B=P^{-1}AP$ for an invertible matrix $P$. Thenso multiplicativity of the determinant gives $\det (xI-B)=\det (xI-A)$. Therefore the characteristic polynomial is independent of the chosen basis and is well-defined for the linear transformation itself.

The given $J$ satisfies $J^2=-I_{2m}$ and is invertible. Taking determinants in the defining identity of a symplectic matrix givesSince $\det (M^T)=\det M$ and $\det J\ne 0$, cancellation yields $(\det M{)}^2=1$. HenceIn particular every symplectic matrix is invertible. This argument establishes the requested alternatives; the later parts explicitly supply the determinant-one hypothesis that they use.

The identity is symplectic because $I^{T}JI=J$. For two symplectic matrices $M,N$,Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying $M^{T}JM=J$ on the left by $M^{-T}$ and on the right by $M^{-1}$ givesso it is also closed under inverses. Associativity is inherited from the general linear group. These facts prove that the symplectic group is a subgroup of $\mathrm{GL}(2m,\mathbb{R})$.

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.