{(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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Multi-linear algebra Dual basis

**Physics Forums | Science Articles, Homework Help, Discussion**