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I am doing a project for a graph theory course and would like to prove the Matrix Tree Theorem. This proof uses the Cauchy-Binet formula which I need to prove first. I have found many different proofs of the formula but I am confused about one step. My basic understanding of linear algebra is holding me back.

I am confused about how

$\sum\limits_{1\leq k_1 < ,...,<k_m \leq n} A\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} \sum\limits_{\sigma \in S_m} \text{sgn}(\sigma)b_{{k}_{\sigma(1)}1}b_{{k}_{\sigma(2)}2}...b_{{k}_{\sigma(m)}m}$

is equal to

$\sum\limits_{1\leq k_1 < ,...,<k_m \leq n} A\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} B\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} $

This is the last line step from this proof https://planetmath.org/cauchybinetformula

I understand this is similar to Cauchy-Binet Formula (Matrix Proof)

Thank you in advance.

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  • $\begingroup$ Basically this is the definition of the determinant. $\endgroup$
    – Mick
    Apr 11, 2020 at 5:26
  • $\begingroup$ Thank you @Mick. Where could I find this definition? $\endgroup$
    – user771314
    Apr 11, 2020 at 5:28
  • $\begingroup$ E.g. at en.m.wikipedia.org/wiki/Determinant $\endgroup$
    – Mick
    Apr 11, 2020 at 5:30
  • $\begingroup$ Thank you @Mick. The answer I was looking for is that this is the Leibniz formula for the determinant $\endgroup$
    – user771314
    Apr 11, 2020 at 5:35

1 Answer 1

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The answer is that the determinant is expressed using the Leibniz formula for determinants.

https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

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