7

I am trying to prove that an $ n \times n $ permutation matrix $ P $ that is formed by switching two rows of an $ n \times n $ identity matrix will always be symmetric.

This is what I am trying to use thus far but I can't quite figure out how to piece it all together:

  • A matrix is symmetric if it is its own transpose.

  • The transpose of the identity matrix is still the identity matrix.

  • Any permutation $ P$ of the identity matrix satisfies $ P(P^T)=I$
    (where $ P^T $ is the transpose of $ P$ ).

  • A permutation matrix is always nonsingular and has a determinant of $ \pm 1$ .

  • Basic transpose property: For matrices $ A $ and $ B$ , $ (AB)^T=(B^T)(A^T)$

Any help/advice would be greatly appreciated!

Rodrigo de Azevedo
  • 23,223
Carly
  • 311
  • 1
    Have you tried this with the $2\times 2$ identity matrix? I know it sounds like an over-simplification, but because you are swapping only two rows, that would actually be a representative case. – hardmath Jun 05 '15 at 03:01

1 Answers1

9

If $P$ switches two rows, then $P^2 = I$. Combine this with $P P^T = I$ to get $P=P^T$.

copper.hat
  • 178,207
  • Is that something that I can just infer? I mean I know that it is true because I did the matrix multiplication but I don't know if that is something I would have to prove, and I'm not sure how I would prove it. Any thoughts? – Carly Jun 05 '15 at 03:04
  • 5
    A matrix is defined by its action on a basis. Take the basis to be the standard unit vectors. If $P$ switches rows $i,j$, then $Pe_i = e_j, Pe_j= e_i$ and so we see $P^2 e_i = e_i$ and $p^2 e_j = e_j$. All other rows are unchanged, hence $P^2 = I$. – copper.hat Jun 05 '15 at 03:07