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

    Show that the inner product $$ex sef{x_1wedge cdots wedge x_k,y_1wedge cdotswedge y_k} eex$$ is equal to the determinant of the $k imes k$ matrix $sex{sef{x_i,y_j}}$.

     

    Solution. $$eex ea &quad sef{x_1wedgecdots wedge x_k,y_1wedge cdots wedge y_k}\ &=frac{1}{k!} sum_{sigma, au} ve_sigma ve_ au sef{x_{sigma(1)},y_{ au(1)}} cdots sef{x_{sigma(k)},y_{ au(k)}}\ &=frac{1}{k!} sum_{sigma, au} ve_{sigma^{-1}} ve_ au sef{x_1,y_{ au(sigma^{-1}(1))}} cdots sef{x_k,y_{ au(sigma^{-1}(k))}} \ &=frac{1}{k!} sum_{sigma}sez{ sum_{ au}ve_{ ausigma^{-1}} sef{x_1,y_{ au(sigma^{-1}(1))}} cdots sef{x_k,y_{ au(sigma^{-1}(k))}}} \ &=frac{1}{k!} sum_{sigma}det sex{sef{x_i,y_j}}\ &=det sex{sef{x_i,y_j}}. eea eeex$$

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  • 原文地址:https://www.cnblogs.com/zhangzujin/p/4112003.html
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