Let \(V\) be a vector space over a field \(F\). Show that the set of all skew-symmetric bilinear forms on \(V\) is a subspace of \(L(V, V, F)\).
Solution
Let \[ \mathcal K=\{f\in L(V,V,F): f(\alpha,\beta)=-f(\beta,\alpha) ext{ for all }\alpha,\beta\in V\}. \]
1. The zero form belongs to \(\mathcal K\).
For all \(\alpha,\beta\in V\), \[ 0(\alpha,\beta)=0=-0(\beta,\alpha). \] Hence the zero bilinear form is skew-symmetric.
2. The set \(\mathcal K\) is closed under addition.
Let \(f,g\in\mathcal K\). Then, for all \(\alpha,\beta\), \[ \begin{aligned} (f+g)(\alpha,\beta) &=f(\alpha,\beta)+g(\alpha,\beta)\\ &=-f(\beta,\alpha)-g(\beta,\alpha)\\ &=-(f+g)(\beta,\alpha). \end{aligned} \] Thus \(f+g\) is skew-symmetric and lies in \(\mathcal K\).
3. The set \(\mathcal K\) is closed under scalar multiplication.
Let \(c\in F\) and \(f\in\mathcal K\). Then \[ (cf)(\alpha,\beta) =c f(\alpha,\beta) =-c f(\beta,\alpha) =-(cf)(\beta,\alpha) \] for all \(\alpha,\beta\). Hence \(cf\in\mathcal K\).
By steps <1>1–<1>3, \(\mathcal K\) is a vector subspace of \(L(V,V,F)\).