(adsbygoogle = window.adsbygoogle || []).push({}); 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_{n-i+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_{n-1,2} + \cdots + e_{2,n-1} + 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)(n-1,2) \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].

**Physics Forums - The Fusion of Science and Community**

# Need help finding permutation matrix

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Need help finding permutation matrix

Loading...

**Physics Forums - The Fusion of Science and Community**