Let \(f\) be a symmetric bilinear form on \(C^n\) and \(g\) a skew-symmetric bilinear form on \(C^n\). Suppose \(f + g = 0\). Show that \(f = g = 0\).
Solution
Because \(f+g=0\), one has \(g=-f\). Hence \(f\) is both symmetric and skew-symmetric.
1. The form \(f\) vanishes identically.
For arbitrary \(\alpha,\beta\in\mathbb C^n\), symmetry gives \[ f(\alpha,\beta)=f(\beta,\alpha), \] while skew-symmetry of \(f=-g\) gives \[ f(\beta,\alpha)=-f(\alpha,\beta). \] Therefore \[ 2f(\alpha,\beta)=0. \] Since the scalar field is \(\mathbb C\), \(2 e0\) and is invertible. Thus \[ f(\alpha,\beta)=0 \] for all \(\alpha,\beta\), so \(f=0\).
2. The form \(g\) also vanishes.
Since \(g=-f\) and \(f=0\) by step 1, one has \(g=0\).
Hence \[ \boxed{f=g=0}. \]