1
$\begingroup$

Below is proof of the Cauchy-Binet Formula found on Wikipedia.

Why are the last two equations equal?

Let $\textbf{A}$ be a $m\times n$ matrix and $\textbf{B}$ be a $n\times m$ matrix.

$\begin{align*} \displaystyle \det \left({\mathbf A \mathbf B}\right)&=\sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) \left({\sum_{k \mathop = 1}^n a_{1 k} b_{k l_1} }\right) \cdots \left({\sum_{k \mathop = 1}^n a_{m k} b_{k l_m} }\right)\\ &=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) b_{k_1 l_1} \cdots b_{k_m l_m}\\ &=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \det \left({\mathbf B_{k_1 \cdots k_m} }\right)\\ &=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} \epsilon \left({k_1, \ldots, k_m}\right)a_{1 k_1} \cdots a_{m k_m} \det \left({\mathbf B_{j_1 \cdots j_m} }\right) \tag{1}\\ &=\sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \det \left({\mathbf A_{j_1 \cdots j_m} }\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right) \tag{2}\\ \end{align*}$

This is how far I got starting with equation $(2)$... $\begin{equation} \sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \det \left({\mathbf A_{j_1 \cdots j_m} }\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right) = \sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} (\sum_{1\leq k_1,...,\ k_m\leq n}\epsilon(k_1, k_2, ...,k_m)a_{j_1,k_1}a_{j_2,k_2}...a_{j_m,k_m}) \det \left({\mathbf B_{j_1 \cdots j_m} }\right) \end{equation}$

$\endgroup$
1
$\begingroup$

The last $=$ just factors out $\det A_{j_1\cdots j_m}$, defined in terms of the Levi-Civita symbol.

$\endgroup$
  • $\begingroup$ Why does the index chain on the lower limit of the sum then? The index for equation $(1)$ is $1\leq k_1, k_2, ..., k_m\leq n$ and for the second equation is $1\leq j_1 \leq j_2 \leq ... \leq j_m \leq n$. How does it go from there to there? $\endgroup$ – W. G. May 29 '19 at 14:02
  • 1
    $\begingroup$ I added some more to the problem rewriting the determinant with the Levi-Civita symbol. $\endgroup$ – W. G. May 29 '19 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.