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