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Suppose two sets of covectors on a vector space $V, \beta^1,...\beta^k $ and $\gamma^1,...,\gamma^k,$ are related by

$$\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$$

where $i=1,...,k$, for a $k\times k$ matrix $[a^i_j]$.

I want to show

$$\beta^1 \wedge...\wedge\beta^k=(det A) \gamma^1 \wedge...\wedge\gamma^k$$

but my efforts fail. Can anyone help me?

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1 Answer 1

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Suppose $\beta^i=\sum_{j=1}^k a^i_j \gamma ^j$ then consider, \begin{align} \beta^1 \wedge \cdots \wedge \beta^k &= \biggl[ \sum_{j_1=1}^k a^1_{j_1} \biggr] \gamma ^{j_1} \wedge \cdots \wedge \biggl[\sum_{j_k=1}^k a^k_{j_k} \gamma ^{j_k} \biggr]\\ &= \sum_{j_1=1}^k \sum_{j_k=1}^k a^1_{j_1} \cdots a^k_{j_k} \gamma ^{j_1} \wedge \cdots \wedge \gamma ^{j_k} \\ &= \sum_{j_1=1}^k \sum_{j_k=1}^k a^1_{j_1} \cdots a^k_{j_k} \epsilon^{j_1 \cdots j_k}\gamma ^1 \wedge \cdots \wedge \gamma ^k \\ &= det[A] \gamma^1 \wedge \cdots \wedge \gamma^k. \end{align} Here I use the anti-symmetric symbol to define the determinant (as is my custom).

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  • $\begingroup$ Thank you James, this is excellent! $\endgroup$ Commented Aug 7, 2013 at 3:06
  • $\begingroup$ @Jellyfish you might look at page 183 of supermath.info/math332.pdf (these notes are not optimal, but that example is what you want to see) $\endgroup$ Commented Aug 7, 2013 at 3:27
  • $\begingroup$ :( I wish I am in your class... $\endgroup$ Commented Aug 7, 2013 at 4:11
  • $\begingroup$ Oh sorry I forgot to say, those notes are amazing! Thanks James. You definitely should get it published! It will help so many students!! They are detailed and comprehensable, I feel your notes talk to me! $\endgroup$ Commented Aug 7, 2013 at 4:15
  • $\begingroup$ @Jellyfish I'll send you my more recent notes once I've fixed them... I haven't posted them in a very accessible way due to the lack of quality control... many errors. Thanks for your comment. $\endgroup$ Commented Aug 7, 2013 at 12:58

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