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?


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).

  • $\begingroup$ Thank you James, this is excellent! $\endgroup$ – 1LiterTears Aug 7 '13 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) $\endgroup$ – James S. Cook Aug 7 '13 at 3:27
  • $\begingroup$ :( I wish I am in your class... $\endgroup$ – 1LiterTears Aug 7 '13 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! $\endgroup$ – 1LiterTears Aug 7 '13 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. $\endgroup$ – James S. Cook Aug 7 '13 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.