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I am trying to understand the relationship between the wedge product and linear subspace. Let $e_1,\cdots, e_4$ be the standard basis of $\mathbb{R}^4$. The wedge product $$(e_1+2e_2)\wedge (3e_1+e_3+e_4) $$ can be thought of as the oriented linear subspace generated by the vectors $e_1+2e_2$ and $3e_1+e_3+e_4$.

Upon expanding we get $$e_{13}+e_{14}-6e_{12}+2e_{23}+2e_{24},$$ where $e_{ij}=e_i\wedge e_j$. Each $e_{ij}$ can be thought of as the linear subspace spanned by $e_i,e_j$. But what role do the coefficients $1,1,-6,2,2$ play?

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    I don't see how it is useful to think of a bivector as a linear subspace of the original space. For example, $\{e_1,e_2\}$ and $\{2e_1,e_2\}$ span the same subspace, but $e_1\wedge e_2$ and $2e_1\wedge e_2$ are different bivectors -- one is twice the size of the other. How would identifying them with subspaces work?2011-11-27
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    The identification is not one to one. In your example, the two wedge products correspond to same linear subspace.2011-11-27
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    $v_1\wedge v_2$ is a basis for a linear subspace of $\wedge^2 \mathbb R^4$, not for a subspace of $\mathbb R^4$ itself.2011-11-27
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    From a book by Frank Morgan: The oriented m-planes through the origin in $\mathbb{R}^n$ are in one-to-one correspondence with the unit, simple m-vectors in $\wedge^m\mathbb{R}^n$.2011-11-27
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    @TCL: Ah, "simple" is the operative word.2011-11-27

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