{(a_i)_j} is the dual basis to the basis {(e_i)_j}(adsbygoogle = window.adsbygoogle || []).push({});

I want to show that

((a_i)_1) \wedge (a_i)_2 \wedge.... \wedge (a_i)_n ((e_i)_1,(e_i)_2,...,(e_i)_n) = 1

this is exercise 4.1(a) from Spivak. So my approach was:

\BigWedge_ L=1^k (a_i)_L ((e_i)_1,...,(e_i)_n) = k! Alt(\BigCross_L=1^k (a_i)_L)((e_i)_1,...,(e_i)_n)= k! Alt(T)((e_i)_1,...,(e_i)_n) = k!(1/k! Sum _ {permutations σ} sgn σ T ((e_i)_σ (1),...,(e_i)_σ (n))

where T = \BigCross_L=1^k (a_i)_L

So there is already a result on what T ((e_i)_1,...,(e_i)_n) is. 1 if all the sub-indices agree, and 0 otherwise. My question is..... is T ((e_i)_σ (1),...,(e_i)_σ (n)) any different?

I'm assuming that in the one dimensional case we would say that T acts on one element in a linear fashion.... but I'm kinda confused by the idea of having several arguments.....

Otherwise,....is there an easier approach to the solution?

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# Multi-linear algebra Dual basis

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