# Determinant, Change of basis

#### lisa

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?

Bye,
Lisa

#### NonCommAlg

MHF Hall of Honor
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?

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!)