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