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See http://en.wikipedia.org/wiki/Laplace_expansion

What does $\tau\,=(n,n-1,\ldots,i)\sigma'(j,j+1,\ldots,n)$ stand for as well as the statements follow? "Since the two cycles can be written respectively as $n - i$ and $n - j$ transpositions..."

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Are you aware that there are expressions before and after "and" in the Wikipedia text? If they don't appear in your browser, it's not surprising that you can't understand the sentence :-) It says $n-i$ before the "and" and $n-j$ after it. –  joriki Oct 10 '12 at 15:35
    
@joriki oops, I just pasted the text right from my webpage, and those expressions are lost. I can see them. Still get confused. –  ymfoi Oct 10 '12 at 15:44
    
There's a preview window under the text area where you entered your question, so you can check whether the question looks as intended before posting. –  joriki Oct 10 '12 at 16:02
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Follow the internal link to “cycles” in that proof. You will learn that $(n,n-1,\ldots,i)$ is the permutation that maps $n$ to $n-1$, …, and $i$ to $n$, and similarly for the parenthesis on the right. The whole expression is the product of three permutations, with $\sigma'$ in the middle.

A cycle of length $k$ is the product of $k-1$ transpositions (cycles of length 2).

Addendum: The comment thread makes it seem that a formal definition of a cycle like $(a_1,\ldots,a_k)$ is called for. Assuming $a_1$, …, $a_k$ are distinct elements of $\{1,2,\ldots,n\}$, you can define the cycle as the following permutation on $\{1,2,\ldots,n\}$: $$(a_1,\ldots,a_k)(q) =\begin{cases} a_{j+1}&\text{if $q=a_j$ with $1\le j<k$},\\ a_1&\text{if $q=a_k$,}\\ q&\text{otherwise.} \end{cases}$$

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I don't care much for the proof in wikipedia. A better proof would rely on the characterization of the determinant in terms of its multilinear and alternating properties. –  Harald Hanche-Olsen Oct 10 '12 at 16:34
    
What's the rule for "cycles" multiplication? –  ymfoi Oct 11 '12 at 4:12
    
Like any multiplication of permutations. I.e., a permutation is just an invertible map $\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$, and the “multiplication” just composes two such maps. In particular, the product of two cycles is in general not a cycle. –  Harald Hanche-Olsen Oct 11 '12 at 8:05
    
You mean that $\tau$ is a composition of permutations$(n,n-1,\ldots,i)$, $\sigma'$ and $(j,j+1,\ldots,n)$? But they do not have equivalent length, how can I compose then? –  ymfoi Oct 11 '12 at 8:15
    
You compose them by applying them one by one. For example, to compute $\tau(j)$ you note that the cycle on the right maps $j$ to $j+1$. Then you apply $\sigma'$ and get $\sigma'(j+1)=:k$, say. If $k$ is outside the cycle on the left, the final answer is $k$. Otherwise, you apply the left cycle to get $k-1$, unless $k=i$, in which case the final answer is $n$. Something similar happens if you start with any element inside the cycle on the right. Otherwise, that cycle leaves the element alone, so you apply $\sigma'$ directly to it, then the other cycle. –  Harald Hanche-Olsen Oct 11 '12 at 11:41
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