I'm looking for a proof that if $v_1, ..., v_{n-1} \in \mathbb{R}^n$, then $\wedge (v_1, ..., v_{n-1})$ is orthogonal to each $v_1, ..., v_{n-1}$ for all $n \in \mathbb{N}$.
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You don't need the hypothesis that the $v_i$ are orthogonal among each other. – Marc van Leeuwen Mar 05 '17 at 08:40
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What you wrote down is just a mnemonic for the formula giving the coordinates of the generalised cross product with respect to a positively oriented orthonormal basis $\def\e#1{\mathbf e_{#1}}\e1.\ldots,\e n$; it is not a definition, worse, the formula is not defined in any formal sense (since a determinant has to have all its entries in some commutative ring to be defined, which is not the case here).
For a proper definition, one could start with an Euclidean space$~V$ of dimension$~n>0$. The space of $n$-linear alternating forms on$~V$ has dimension$~1$, and each nonzero such form when applied to ordered orthonormal bases of$~V$ takes exactly two (opposite) values, partitioning the set of ordered orthonormal bases into two parts. Choosing one part as the "positively oriented" among these bases makes $V$ into an oriented Euclidean space, and singles out among the $n$-linear alternating forms the one, to be called the volume form$\def\vol{\operatorname{vol}{}}~\vol$ of$~V$, that takes the value$~1$ on the positively oriented orthonormal bases.
Now fixing $n-1$ vectors $v_1,\ldots,v_{n-1}\in V$, the map $V\to\Bbb R$ given by $w\mapsto\vol(v_1,\ldots,v_{n-1},w)$ is a linear form on$~V$. Then there is a unique vector$~u$ such that the map coincides with $w\mapsto u\cdot w$, and by definition $\bigwedge(v_2,\ldots,v_{n-1})=u$; in other words one has the defining relation $$ \vol(v_1,\ldots,v_{n-1},w)=\bigwedge(v_2,\ldots,v_{n-1})\cdot w \qquad\text{for all $w\in V$.} $$
The fact that to volume form vanishes when its arguments are linearly dependent (as does any alternating multilinear form) here means that $\bigwedge(v_2,\ldots,v_{n-1})$ is orthogonal to any vector in the span of $v_2,\ldots,v_{n-1}$, and in particular to each of the $v_i$ themselves.
Marc van Leeuwen
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Thank you. I thought it was a weird formula too and wondered where it came from. Can you recommend a material on this subject? I have no idea what alternating- or linear forms are. – saldukoo Mar 06 '17 at 18:45
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Also, how would one go about constructing an explicit formula for $\wedge$ using the definition above? – saldukoo Mar 07 '17 at 03:38
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The dot product of that vector with a vector $w$ is obtained by replacing the last row of that formal determinant with the coordinates of $w$.
Ted Shifrin
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