# Transformation rule for a wedge product

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).