
#1
Jun1911, 02:39 AM

P: 162

1. The problem statement, all variables and given/known data
What is the permutation matrix associated to the permutation of [itex]n[/itex] indices defined by [itex]p(i) = n  i + 1[/itex]? What is the cycle decomposition of [itex]p[/itex]? What is it's sign? 2. Relevant equations Prop. A permutation matrix [itex]P[/itex] has a single 1 in each row and in each column, the rest of its entries being 0. 3. The attempt at a solution I. So I'm a bit confused on how to find the matrix associated with [itex]p[/itex]. Here's my attempt: Given [itex]p(i) = n  i + 1[/itex] defines a permutation of [itex]n[/itex] indices, then by our proposition, we know the associated matrix with [itex]p[/itex], say [itex]A[/itex], is an [itex]n \times n[/itex] matrix with a single 1 in each row and each column, the rest of its entries being 0. Therefore it is of the form: [tex]A = \sum_i e_{p(i),i} = \sum_i e_{ni+1,i}[/tex] where [itex]e_{i,j}[/itex] denotes an [itex]n \times n[/itex] matrix with a single 1 in the i^{th} row and j^{th} column. From this we find that: [tex]A = e_{n,1} + e_{n1,2} + \cdots + e_{2,n1} + e_{1,n}[/tex] I guess I am a bit confused on whether I can deduce that [itex]A[/itex] is an [itex]n \times n[/itex] matrix from the fact that [itex]p[/itex] defines a permutation of [itex]n[/itex] indices. If so, does that mean I can sum [itex]i[/itex] from 1 to n in the formula above to find [itex]A[/itex]? II. To find the cycle of decomposition of [itex]p[/itex], provided that my answer from I. is correct, would I just write: [tex](n,1)(n1,2) \cdots[/tex] ? III. I'm not sure on how to determine the sign of [itex]A[/itex] seeing as it depends on the oddness or evenness of [itex]n[/itex]. 



#2
Jun1911, 07:06 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

That function just reverses the order of the indices. The corresponding matrix has all 0s except that the diagonal from lower left to upper right is all 1s. It's sign is [itex](1)^n[/itex].
Yes, it can be decomposed into cycles, (1, n)(2, n1), .... 



#3
Jun1911, 11:52 AM

P: 162

I'm confused about the sign being [itex](1)^n[/itex]. Suppose that [itex]n=2[/itex]. Then:
[tex]A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}[/tex] Then [itex]\det(A) = 1[/itex] and therefore the sign is 1. But according to your formula, the determinant of [itex]A[/itex] should be 1 and therefore the sign should be 1. Here's what I think. After some investigation, I noticed that the sign of the determinant changes every [itex]2n[/itex]. We know that our permutation is a product of k transpositions [itex]\tau_1 \tau_2 \cdots \tau_k[/itex], where k equals either [itex]\frac{n1}{2}[/itex], if [itex]n[/itex] is odd, or [itex]\frac{n}{2}[/itex], if [itex]n[/itex] is even. Thus the sign of the permutation is [itex](1)^k[/itex], with [itex]k[/itex] defined as above. 


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