Let \(F\) be a subfield of \(C\), \(V\) a finite-dimensional vector space over \(F\), and \(f\) a non-singular bilinear form on \(V\). If \(T\) is a linear operator on \(V\) preserving \(f\), prove that \(\det T = \pm 1\).
Solution
Choose an ordered basis of \(V\). Let \(A\) be the matrix of \(f\) and \(P\) the matrix of \(T\) in this basis. Since \(f\) is non-singular, \[ \det A\ne0. \]
Preservation of \(f\) means \[ f(T\alpha,T\beta)=f(\alpha,\beta) \] for all \(\alpha,\beta\). In matrix form this is \[ P^tAP=A. \] Taking determinants gives \[ \det(P^t)\det(A)\det(P)=\det(A). \] Because \[ \det(P^t)=\det P \] and \(\det A\ne0\), we may cancel \(\det A\) to obtain \[ (\det P)^2=1. \] Over the subfield \(F\subseteq\mathbb C\), the only roots of \(x^2-1\) are \(1\) and \(-1\). Hence \[ \boxed{\det T=\det P=\pm1.} \]