Talk:Leibniz formula for determinants

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Is the "physisists formula" usinge the Levi-Civita symbols correct? As I understand it, there is a 1/n! missing. —The preceding unsigned comment was added by 131.130.26.227 (talk • contribs) 14:03, 17 August 2006 (UTC)

So there was. I've now written the formula in a manner which obviates the need for the prefactor and makes the connection with the first formula more transparent. -- Fropuff 05:35, 11 February 2007 (UTC)

[edit] A part of the proof is missing

In the proof, the reason why the n-tuples are reduced to the permutations is missing. The reason why the ordered n-tuples (k_1, \dots, k_n) are reduced to the permutations is because F is alternating, and therefore F(E^{k_1}, \dots, E^{k_n}) is zero for all n-tuples (k_1, \dots, k_n) that repeat indices. Jeff Wu 01:11, 6 December 2006 (UTC)


[edit] A second part of the proof is missing :)

The proof shows that if F is a function that satisfies the conditions, then F is equal to the Leibniz function. This shows the uniqueness of a solution. To prove the existence of a solution, you must show that the Leibniz function actually satisfies the conditions. Ceroklis 11:47, 27 September 2007 (UTC)

[edit] part of proof

This way, it is easier to follow the proof:

\begin{align} F(A)& = F\left(\sum_{k_1 = 1}^n a_{k_1}^1 E^{k_1}, A^2, \dots, A^n\right)\\ & = \sum_{k_1 = 1}^n a_{k_1}^1 F\left(E^{k_1}, A^2, \dots, A^n\right)\\ & = \sum_{k_1 = 1}^n a_{k_1}^1 \sum_{k_2 = 1}^n a_{k_2}^1 F\left(E^{k_1}, E^{k_2}, A^3, \dots, A^n\right)\\ & = \sum_{k_1, k_2 = 1}^n \prod_{i = 1}^2 a_{k_i}^i F\left(E^{k_1}, E^{k_2}, A^3, \dots, A^n\right)\\ & = \cdots\\ & = \sum_{k_1, \dots, k_n = 1}^n \prod_{i = 1}^n a_{k_i}^i F\left(E^{k_1}, \dots, E^{k_n}\right). \end{align}

Ave caesar, 145.97.205.169 23:11, 12 October 2007 (UTC)