[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.5.1

Show that the inner product $$\bex \sef{x_1\wedge \cdots \wedge x_k,y_1\wedge \cdots\wedge y_k} \eex$$ is equal to the determinant of the $k\times k$ matrix $\sex{\sef{x_i,y_j}}$.

Solution. $$\beex \bea &\quad \sef{x_1\wedge\cdots \wedge x_k,y_1\wedge \cdots \wedge y_k}\\ &=\frac{1}{k!} \sum_{\sigma,\tau} \ve_\sigma \ve_\tau \sef{x_{\sigma(1)},y_{\tau(1)}} \cdots \sef{x_{\sigma(k)},y_{\tau(k)}}\\ &=\frac{1}{k!} \sum_{\sigma,\tau} \ve_{\sigma^{-1}} \ve_\tau \sef{x_1,y_{\tau(\sigma^{-1}(1))}} \cdots \sef{x_k,y_{\tau(\sigma^{-1}(k))}} \\ &=\frac{1}{k!} \sum_{\sigma}\sez{ \sum_{\tau}\ve_{\tau\sigma^{-1}} \sef{x_1,y_{\tau(\sigma^{-1}(1))}} \cdots \sef{x_k,y_{\tau(\sigma^{-1}(k))}}} \\ &=\frac{1}{k!} \sum_{\sigma}\det \sex{\sef{x_i,y_j}}\\ &=\det \sex{\sef{x_i,y_j}}. \eea \eeex$$