Let \(f\) be a bilinear form on a vector space \(V\). Let \(L_f\) and \(R_f\) be the mappings of \(V\) into \(V^*\) associated with \(f\) in Section 10.1. Prove that \(f\) is skew-symmetric if and only if \(L_f = -R_f\).
Solution
Recall the two associated maps \[ L_f,R_f:V\longrightarrow V^* \] defined by \[ (L_f\alpha)(\beta)=f(\alpha,\beta), \qquad (R_f\alpha)(\beta)=f(\beta,\alpha). \]
1. If \(f\) is skew-symmetric, then \(L_f=-R_f\).
For every \(\alpha,\beta\in V\), \[ (L_f\alpha)(\beta) =f(\alpha,\beta) =-f(\beta,\alpha) =-(R_f\alpha)(\beta). \] Thus the functionals \(L_f\alpha\) and \(-R_f\alpha\) agree on every \(\beta\), so \[ L_f\alpha=-R_f\alpha \] for every \(\alpha\). Hence \[ L_f=-R_f. \]
2. If \(L_f=-R_f\), then \(f\) is skew-symmetric.
Assume \(L_f=-R_f\). For arbitrary \(\alpha,\beta\in V\), \[ \begin{aligned} f(\alpha,\beta) &=(L_f\alpha)(\beta)\\ &=-(R_f\alpha)(\beta)\\ &=-f(\beta,\alpha). \end{aligned} \] Therefore \(f\) is skew-symmetric.
Hence \[ \boxed{f\text{ is skew-symmetric}\iff L_f=-R_f.} \]