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As I understand it, to get an $n$-dimensional cross product, you need $n-1$ vectors of dimension $n$. However my lecture notes are quite miss leading in the fact that they suggest this isn't always the case.

A linear map $\Phi : \mathbb{R}^n \to \mathbb{R}^n$ has the form $ x \mapsto Ax$ for an $n\times n$ matrix $A$ and is orthogonal if $A^TA = AA^T = I$.

Example Exercise: Prove that if $\Phi$ is orthogonal then $\Phi(x \wedge y) = \pm\Phi (x) \wedge \Phi (y) $, where the sign depends on whether it is orientation preserving or reversing.

Is it the case that this example must just be referring to $3$-dimensional vectors? To calculate the cross product of $n-1$, $n$ dimensional vectors do you do the same technique of stacking up the vectors in a matrix and finding the determinent as with $3$ dimensional vectors?

Does anyone have any hints as to how to solve this example exercise?

Thanks

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This is not the cross-product but the wedge-product. And it is defined for any two vectors in $\mathbb{R}^n$ (you do not need $n-1$ of them). If you wedge-multiply $n-1$ vectors together you can interpret it again as a vector via Hodge duality (that is how I understand your statement). –  Fabian Jan 7 '13 at 20:04
    
There is a sense in which the cross-product exists only in 3 dimensions, and in 7 dimensions, which "turns out to be the only other non-trivial bilinear product of two vectors that is vector valued, anticommutative and orthogonal." See the Wikipedia article –  Joseph O'Rourke Jan 7 '13 at 22:05
    
So if you had 3 vectors in $\mathbb{R}^4$ is there a way to find a vector that is orthogonal to all of them? –  user53076 Jan 8 '13 at 13:57
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