Can we find a characterising property of all the permutation matrices with trace = $0$?
I know that any traceless matrix is a commutator.
i.e. any traceless matrix $A$ can be written as:-
$$A = PQ-QP$$
for some P and Q.
Also, A being a permutation matrix, is a non-negative orthogonal matrix.
I just wanted to see if we could club all these to give a simple characterization of the traceless permutation matrices in form of an equation.
Permutation matrices can be regarded as "matrix representation" of permutations. Like the permutation matrix $A$ given by,
$A=\pmatrix{ 0&1&0&0&0\\ 0&0&1&0&0\\ 1&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&1&0}$
is a matrix representation of the permutation $\sigma=(123)(45)\in S_5$. Because this permutation is the map $(1,2,3,4,5)^t\rightarrow A(1,2,3,4,5)^t=(\sigma1,\sigma2,\sigma3,\sigma4,\sigma5)$.
If we denote the matrix of the permutation $\sigma$ by $P_{\sigma}$, then we can get $M_{\sigma}M_{\rho}=M_{\sigma \rho}$.
Now if a permutation matrix $P$ of order $n$ has trace zero, then its corresponding permutation $\sigma\in S_n$ has no fixed point. Hence if $\sigma=c_1c_2\dots c_k$ is the cycle decomposition of $\sigma$, then $\sum_{i=1}^{k}l(c_i)=n$.
Now as mentioned in the above comment, the permutation matrix $A=M_{\sigma}$ of $\sigma=(123)(45)$ has trace zero and $A^5=M_{\sigma^5}$. But $M_{\sigma^5}=I_5$ implies $ \sigma^5$ is identity permutation, which is not true, since $o(\sigma)=6$. Thus $A^5\neq I_5$.