How do I find the determinant of a permutation matrix that interchanges rows of an $n \times n$ matrix?

Question

Let $T_n$ be the permutation matrix that interchanges rows of an $n \times n$ matrix in the following way: row $j$ is moved to row $j + 1$ for $j \in \{1, 2, \dots, n − 1\}$ and the last row is moved to the first. Find $\det(T_3)$.

I really don't know where to start with this problem, I tried to think of a $3 \times 3$ matrix and just follow the interchanges, but I'm not sure if that's the right of way of solving this problem. Any help is appreciated.

Answer 1

You’re on the right track. The permutation matrix that effects this rearrangement is itself the corresponding permutation of the identity matrix.

Start by swapping the last two rows. This moves row $n$ up one space and row $n-1$ down one space into its final position. Proceeding up the rows, continue to swap adjacent rows in turn. Eventually, the last row will have migrated to the top and all the other rows moved down by one, as required. This took $n-1$ swaps, therefore $\det(T_n)=\dots$?

Answer 2

It is a circulant permutation matrix. (The eigenvalues of a known permutation are well determined. Here $1$, $i$ and $-i$ are the three eigenvalues. ) You could figure the signature of a permutation as it is the determinant.