Let \(\displaystyle V\) be a vector space, \(\displaystyle \dim (V)=n\) and \(\displaystyle \omega \in \mathrm{Alt}^n V,\ \omega \neq 0 \) an alternating form. Let \(\displaystyle M\) be the change of basis matrix from the basis \(\displaystyle a=(a_1,...,a_n)\) for \(\displaystyle V\) to the basis\(\displaystyle b=(b_1,...,b_n) \) for \(\displaystyle V\).

Proof that \(\displaystyle \det M=\frac{\omega(a_1,...,a_n)}{\omega(b_1,...,b_n)}

\)

I would like to use the Leibniz formula

\(\displaystyle \det M = \sum_{\sigma \in S_n} \mathrm{sign} (\sigma) \cdot M_{1 \sigma(1)}\cdot ... \cdot M_{n \sigma(n)} \)

Since \(\displaystyle a=(a_1,...,a_n)\) and \(\displaystyle b=(b_1,...,b_n)\) are bases for \(\displaystyle V\), I can write

\(\displaystyle a_i = \sum_{j=1}^{n}M_{ij}b_j\)

and therefore

\(\displaystyle \frac{\omega(\sum_{j=1}^{n}M_{1j}b_j,..., \sum_{j=1}^{n}M_{nj}b_j)}{\omega(b_1,...,b_n)} \)

\(\displaystyle =\frac{\omega( M_{11}b_1+...+M_{1n}b_n ,..., M_{n1}b_1+...+M_{nn}b_n )}{\omega(b_1,...,b_n)} \)

How should I continue? Can I write the last as \(\displaystyle \frac{\sum_{j=1}^n M_{1j} \cdot ... \cdot M_{nj} \cdot \omega(b_1,...,b_n)}{\omega(b_1,...,b_n)} \) because \(\displaystyle \omega\) is alternating?

Thanks in advance!

Bye,

Lisa