Let \(V\) be a finite-dimensional vector space over a field \(F\), and let \(\omega: V \times V \to F\) be a nondegenerate alternating bilinear form on \(V\). Suppose \(T: V \to V\) is a linear transformation that preserves \(\omega\), i.e. \[\omega(T(u), T(v)) = \omega(u, v) \quad \text{for all } u, v \in V.\] Prove that if \(\lambda \in F\) (or in an algebraic closure \(\bar{F}\)) is an eigenvalue of \(T\), then \(\lambda \ne 0\) and \(\lambda^{-1} = 1/\lambda\) is also an eigenvalue of \(T\) (with the same algebraic and geometric multiplicity).
Solution
Choose a basis of \(V\) and let \(J\) be the matrix of \(\omega\) and \(M\) the matrix of \(T\). Nondegeneracy of \(\omega\) means that \(J\) is invertible, and preservation of \(\omega\) gives \[ M^TJM=J. \] Hence \(M\) is invertible, so \(0\) is not an eigenvalue.
From the displayed identity, \[ M^{-1}=J^{-1}M^TJ. \] Thus \(M^{-1}\) is similar to \(M^T\), and every matrix is similar to its transpose. Therefore \[ M\sim M^{-1}. \]
The eigenvalues of \(M^{-1}\) are the reciprocals of the eigenvalues of \(M\), with the same Jordan block sizes. Since \(M\) and \(M^{-1}\) are similar, the multiset of Jordan blocks of \(M\) is invariant under \[ \lambda\longmapsto\lambda^{-1}. \] Consequently, if \(\lambda\) is an eigenvalue of \(T\), then \(\lambda\ne0\) and \(\lambda^{-1}\) is also an eigenvalue, with the same algebraic multiplicity and the same geometric multiplicity.