Determinant, Change of basis

May 2009
12
0
Hi!

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
 

NonCommAlg

MHF Hall of Honor
May 2008
2,295
1,663
Hi!

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
the point here is that since \(\displaystyle \omega\) is alternating we have \(\displaystyle \omega(x_1, \cdots , x_n)=0\) if \(\displaystyle x_i=x_j\) for some \(\displaystyle i \neq j.\) so using multilinearity of \(\displaystyle \omega\) to expand \(\displaystyle \omega(\sum_{j=1}^{n}M_{1j}b_j,..., \sum_{j=1}^{n}M_{nj}b_j),\) all terms in which one of \(\displaystyle b_j\) appears more than once will be zero. thus we'll be left with terms in the form \(\displaystyle c_{\sigma} \omega(b_{\sigma(1)}, \cdots , b_{\sigma(n)}),\) where \(\displaystyle \sigma \in S_n\) and \(\displaystyle c_{\sigma}\) is in terms of \(\displaystyle M_{ij}.\) again, using the fact that \(\displaystyle \omega\) is alternating, we have

\(\displaystyle \omega(b_{\sigma(1)}, \cdots , b_{\sigma(n)})=\text{sgn}(\sigma) \cdot \omega(b_1, \cdots , b_n).\) the only thing you need to show now is that \(\displaystyle c_{\sigma}=M_{1\sigma(1)} \cdot \cdots M_{n \sigma(n)}.\) (left for you!)
 
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