Prove symmetries of eigenvalues of symplectic matrices

Question

$M \in \mathrm{GL}(2m,\mathbb{R})$ is said to be symplectic if

$$ M^T JM=J,$$

$$J =\begin{bmatrix} 0 & -I_m \\ I_m & 0\end{bmatrix}. $$

Suppose $M \in \mathrm{GL}(2m,\mathbb{R})$ is symplectic and has $\det(M)=1$, let $\chi_M(x)$ be the characteristic polynomial of M, prove that

(a) If $\lambda \in \mathbb{C}, \lambda \neq 0$ is a root of $\chi_M$ with multiplicity $d$, then $\frac{1}{\lambda}$ is a root of $\chi_M$ with the same multiplicity $d$.

(b) $1,-1$ have even multiplicity in $\chi_M$.

It's difficult to argue part (b) rigorously, and I don't think that it is obvious from (a). The given hint is that $\chi_M = \chi_{M^{-1}}$, and therefore for $z \in \mathbb{C}, z \neq 0$ we have $$\chi_M(z) = z^{2m}\chi_M(\frac{1}{z}).$$

Answer 1

Let us first prove the hint. Since $J$ is invertible with $J^{-1} = J^T = -J$, we have that $M$ is invertible with $M^{-1} =J^{-1}M^T J$ (just multiply your first formula by $J^{-1}$). Since eigenvalues are invariant under transposition and conjugation of matrices, so is the characteristic polynomial, hence $\chi_{M^{}} = \chi_{M^{-1}}$. As you can see in this post (in the last part), this implies the given formula, because $$\chi_{M^{}}(z) = \chi_{M^{-1}}(z) = \frac{1}{\det(M)}(-z)^{2m} \chi_{M^{}}\left( \frac{1}{z} \right) = z^{2m} \chi_{M^{}}\left( \frac{1}{z} \right).$$

(a) From the above equation and from the fact that the eigenvalues of $M$ are non-zero, we deduce that $a$ is a zero of $\chi_M$ if and only if $\frac{1}{a}$ is a zero of $\chi_M$. Dividing by $(z-a) = -az\left(\frac{1}{z} - \frac{1}{a}\right)$ and by $\left(z-\frac{1}{a}\right) = -\frac{z}{a} \left(\frac{1}{z}-a\right)$ on both sides, we can go on in the same way and see that also the multiplicities coincide.

(b) You are right in saying that it might not be so clear that this follows from (a). But consider that $1$ and $-1$ are the only complex numbers that are their own multiplicative inverses, and that $-1$ cannot have odd multiplicity, because otherwise the determinant would be negative (being the product of the eigenvalues counted with multiplicity). Since the total number of eigenvalues counted with multiplicity is $2m$, hence even, and since all other eigenvalues counted together take up an even number of slots, if $1$ is present it must also have even multiplicity.

Answer 2

The characteristic polynomial $\chi_M$ satisfies this for $\lambda \ne 0$:

$$\chi_M(\lambda) = \lambda^{2m}\chi_M\left(\frac1\lambda\right)$$

Indeed, note that $\det J = \det M =\det M^T = 1$. We have:

\begin{align} \chi_M(\lambda) &= \det(M - \lambda I)\\ &= \det J \cdot \det(M - \lambda I)\\ &= \det (JM - \lambda J)\\ &= \det M^T \cdot \det (JM - \lambda J)\\ &= \det (M^TJM - \lambda M^TJ)\\ &= \det (J - \lambda M^T J)\\ &= \det J \cdot \det(I - \lambda M^T)\\ &= (-\lambda)^{2m} \det\left(M^T - \frac1\lambda I\right)\\ &= \lambda^{2m} \det\left(M - \frac1\lambda I\right)\\ &= \lambda^{2m} \chi_M\left(\frac1\lambda\right) \end{align}

Now (a) is evident.

Denote $d(\lambda)$ the multiplicity of $\lambda$ in $\chi_M$. From (a) follows $d(\lambda) = d\left(\frac1\lambda\right)$. We have $\lambda \ne \frac1\lambda$ whenever $\lambda \ne \pm 1$ so $$2m = d(1) + d(-1) + 2 \sum_{\lambda \in \sigma(M) \setminus \{-1,1\}, |\lambda| > 1} d(\lambda)$$

Hence $d(1) + d(-1)$ is even. If both $d(1)$ and $d(-1)$ are odd, then we would have $$1 = \det M = \prod_{\lambda \in \sigma(M)} \lambda = (-1)^{d(-1)} = -1$$

Therefore, both $d(1)$ and $d(-1)$ are even.

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Here it is shown that $\lambda^{2m}\chi_M\left(\frac1\lambda\right)$ is precisely the characteristic polynomial of $\chi_{M^{-1}}$. Therefore, we conclude $\chi_M = \chi_{M^{-1}}$, but it was not needed for this solution.