lecture19

lecture19 - Lecture 19 We begin with a review of tensors...

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Unformatted text preview: Lecture 19 We begin with a review of tensors and alternating tensors. We defined L k ( V ) to be the set of k-linear maps T : V k R . We defined → 1 , . . . , e ∗ e 1 , . . . , e n to be a basis of V and e ∗ n to be a basis of V ∗ . We also defined k e ∗ I = e ∗ i 1 ⊗ ··· ⊗ e ∗ i k } to be a basis of L ( V ), where I = ( i 1 , . . . , i k ) , 1 ≤ i r ≤ n is a { k multi-index. This showed that dim L k = n . and T ∈ L k We defined the permutation operation on a tensor. For σ ∈ S n , we defined T σ ∈ L k by T σ ( v 1 , . . . , v k ) = T ( v σ − 1 (1) , . . . , v σ − 1 ( k ) ). Then we defined that T is alternating if T σ = ( − 1) σ T . We defined A k = A k ( V ) to be the space of all alternating k-tensors. We defined the alternating operator Alt : L k → A k by Alt ( T ) = ( − 1) σ T σ , and we defined ψ I = Alt ( e ∗ I ), where I = ( i 1 , . . . , i k ) is a strictly increasing multi-index. We proved the following theorem: Theorem 4.24. The ψ I ’s (where I is strictly increasing) are a basis for A k ( V ) . Corollary 6. If 0 ≤ k ≤ n , then n n ! dim A k = = k !( n − k )! . (4.69) k Corollary 7. If k > n , then A k = . { } We now ask what is the kernel of Alt ? That is, for which T ∈ L k is Alt ( T ) = 0?...
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This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

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lecture19 - Lecture 19 We begin with a review of tensors...

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