Let \(F\) be a subfield of \(C\), \(V\) the space of \(n \times 1\) matrices over \(F\), \(A\) an invertible \(n \times n\) matrix over \(F\), and \(f\) the bilinear form on \(V\) given by \(f(X, Y) = X^tAY\). If \(M\) is an \(n \times n\) matrix over \(F\), show that \(M\) preserves \(f\) if and only if \(A^{-1}M^tA = M^{-1}\).
Solution
1. If \(M\) preserves \(f\), then \(A^{-1}M^tA=M^{-1}\).
Preservation means \[ f(MX,MY)=f(X,Y) \] for all \(X,Y\). Thus \[ X^tM^tAMY=X^tAY \] for all \(X,Y\), and hence \[ M^tAM=A. \] Taking determinants and using \(\det A\ne0\) gives \[ (\det M)^2=1, \] so \(M\) is invertible. Now multiply \[ M^tAM=A \] on the left by \(A^{-1}\) and on the right by \(M^{-1}\) to obtain \[ \boxed{A^{-1}M^tA=M^{-1}.} \]
2. If \(A^{-1}M^tA=M^{-1}\), then \(M\) preserves \(f\).
The displayed equation already implies that \(M\) is invertible. Multiply it on the left by \(A\) and on the right by \(M\): \[ M^tAM=A. \] Therefore, for all \(X,Y\), \[ \begin{aligned} f(MX,MY) &=X^tM^tAMY\\ &=X^tAY\\ &=f(X,Y). \end{aligned} \] Thus \(M\) preserves \(f\).
Hence \[ \boxed{M\text{ preserves }f\iff A^{-1}M^tA=M^{-1}.} \]