Transformation Rule for a Wedge Product of Covectors Suppose two sets of covectors on a vector space $V$, $\beta^1,\ldots,\beta^k$ and $\gamma^1,\ldots,\gamma^k$, are related by $$\beta^i=\sum_{i=1}^ka^i_j\gamma^i,\quad i=1,…,k,$$
for a $k\times k$ matrix $A=[a^i_j]$. Show that $$\beta^1 \wedge\cdots\wedge\beta^k=(\det A)\gamma^1\wedge\cdots\wedge\gamma^k.$$ This is a problem on Tu's textbook "introduction to manifolds" (problem 3.7). I've been working on this, and I just don't seem to understand what to do. When I tried to write out the definitions for the wedge product, everything just seemed to get worse.
If someone would please offer some helpful hints, I would appreciate it. I'm not trying to cheat, I'd really prefer to understand all of this material.
 A: This is a well-known result.  It becomes a bit clearer in index-free notation for linear operators.  Let's start with two covectors.
$$\beta^i \wedge \beta^j = \underline a(\gamma^i) \wedge \underline a(\gamma^j)$$
This can be taken as a definition of $\underline a(\gamma^i \wedge \gamma^j)$, and it can be extended to larger wedge products.
When the manifold has dimension $k$, then $\gamma^1 \wedge \ldots \wedge \gamma^k$ is the pseudoscalar of the manifold.  The highest-ranked wedge product of vectors forms a one-dimensional vector space---all other wedge products of $k$ (co)vectors form only scalar multiples of the pseudoscalar.
Hence, it becomes clear that, for some scalar $\alpha$, 
$$\beta^1 \wedge \beta^2 \wedge \ldots \wedge \beta^k = \underline a(\gamma^1 \wedge \gamma^2 \wedge \ldots \wedge \gamma^k) = \alpha \gamma^1 \wedge \gamma^2 \wedge \ldots \wedge \gamma^k$$
This construction fulfills the needs Micah describes.  The conclusion is that $\alpha = \det \underline a$.  It's certainly a lot simpler than trying to prove things with indices.
A: I realize this is an old question, but I was looking at it for homework and thought I would share with you how I solved it. I felt as if the purpose of the exercise was for the reader to explicitly compute the value as the previous exercises in the text were proving the wedge is alternating and linear.
\begin{align*}
\beta^1\wedge\cdots\wedge\beta^k&=\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)\\
&=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}\\
&=\sum_{\sigma\in S_k}\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{\sigma(1)}\wedge\cdots\wedge \gamma^{\sigma(k)}\\
&=\sum_{\sigma\in S_k}\text{sgn}\sigma\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{1}\wedge\cdots\wedge \gamma^{k}\\
&=(\det A)\gamma^{1}\wedge\cdots\wedge \gamma^{k}
\end{align*}
Here the first equality holds from the fact that the wedge product is linear and antisymmetric, so you just distribute and zero out terms and you'll be left with all permutations of $1,\ldots,k$ of gamma. Then you pull out the constants and rearrange your gammas. Rearranging brings out the sign of the permutation, then you are left with what you desire.
