# (very simple) Vector identity proof

I'm completely new with the sum notation and Leva-Civita-Symbols. Can somebody tell me if what I did was correct?

Let $a,b \in \mathbb{R}^3$.

I want to prove $$(b \wedge a) = -(a \wedge b)$$ so I did $$(b \wedge a)_i = \epsilon_{ijk}b_ja_k = - \epsilon_{ikj}b_ja_k = -\epsilon_{ikj}a_kb_j = - (a \wedge b)_i$$

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whats your definition of wedge product (the definitions ive used have anticommutivity built in)? also, in $\mathbb{R}^3$ it's the same as the familiar cross product. – yoyo Feb 22 '11 at 21:18
I know this is not actually a proof, since the anti-symmetry is a deep property of wedge-product itself, but I needed this index permutation in the Levi-Civita-Symbol somewhere else too. And I would like to see how these things hold together. – a1337q Feb 22 '11 at 21:19
This proof looks fine. – mjqxxxx Feb 22 '11 at 22:11
Assuming the first equality is your definition of wedge product, or is a property you have proved already, this is fine. – Ross Millikan Feb 23 '11 at 4:39
I think it's totally wrong because rotating indices in the epsilon tensor must not change sign, because the permutation stays either even or odd as before. The sign only changes when transposing (exchanging two) indices. So again what I mean: $\epsilon_{ijk} = \epsilon_{kij}$ and $\epsilon_{ijk} = -\epsilon_{kji}$ – a1337q Feb 24 '11 at 13:50