I have always been scared by exterior algebra, which means I don't really have any background. So here's a very basic question I would like to clear out:

Consider $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, where $k$ = dim$V$ and $\phi_i$s are the basis for the dual space $V^*$. For $[\phi_i(v_j)]$, should I think of $\phi \in V^*$ as a linear map, where $\phi_i$ is the $i$th column vector of the matrix $\phi$, and $v_j$ picks the $j$th row of the column vector $\phi_i$?

This question arise from wedge product and determinant.


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    $\begingroup$ I don't understand the question. $\endgroup$ – Qiaochu Yuan Jun 21 '13 at 0:00
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    $\begingroup$ More specifically, what is $\phi$? $\endgroup$ – Matt Jun 21 '13 at 0:07
  • $\begingroup$ $\phi_i$s are the basis for the dual space $V^*$. $\endgroup$ – 1LiterTears Jun 21 '13 at 0:21
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    $\begingroup$ Yes, but what is $\phi$? $\endgroup$ – Qiaochu Yuan Jun 21 '13 at 1:53
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    $\begingroup$ @Jellyfish: yes, that's what linear functionals do. $\endgroup$ – Qiaochu Yuan Jun 21 '13 at 3:01

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