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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.

Rodrigo de Azevedo
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ImBatman
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    your mention of $P^2=I$ is called an involution and that occurs iff $P=P^T$. As for your question -- this is the class of traceless permutation matrices, or the class of matrices representing derangements – user8675309 Jun 19 '20 at 06:34
  • @user8675309 nicely interpreted! But do they follow any rule as direct as $P^{2} = I$ – ImBatman Jun 19 '20 at 06:43
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    So, as you already know traceless permutation matrices are derangements. You can check that composition of permutations is the same as multiplication of their matrices. If you factor a permutation in cycles, the order of the permutation is the lcm of the sizes of the cycles. Take the permutation $(123)(45)$ what is its order? is it $5$? – Phicar Jun 19 '20 at 13:21
  • @Phicar. No it should be 6. Thanks! So this is sufficient to say that the following rule does not hold. Can you think of an algaebraic property of such kind of matrices? – ImBatman Jun 19 '20 at 13:50
  • @Kabir Indeed! Well, you can check that if your derangement is a cycle, for example $(13542)$ then its order is $5.$ So cycles have that property and you can try to force the length of the cycles to have $lcm =n$ but i do not see a way right now. – Phicar Jun 19 '20 at 13:59
  • @Phicar Fine! But permutation matrices of our interest are those whose trace = $0$ Not those with $P^{n} = I$ – ImBatman Jun 19 '20 at 14:02
  • @Kabir Well, $P^{\ell}=I_n$ where $\ell$ is the lcm of the cycles of the derangement(traceless permutation matrix). – Phicar Jun 19 '20 at 14:28
  • Let us continue this discussion in chat. – ImBatman Jun 19 '20 at 14:41
  • It's not clear exactly what kind of condition you want. Are you looking for a characterization of such matrices among all matrices, or just among permutation matrices? – Travis Willse Jun 24 '20 at 06:07
  • @TravisWillse I want algaebraic equation which could characterize traceless permutation matrices. – ImBatman Jun 24 '20 at 06:17
  • That's clear already but do you want algebraic conditions that so that if $A$ is any matrix (of the right size) then it satisfies the conditions iff $A$ is such a permutation matrix? Or do you want conditions such that of $P$ is a permutation matrix then it satisfies the conditions iff $P$ is tracefree? Either way, $\operatorname{tr} A = 0$ itself is already a compact algebraic condition. – Travis Willse Jun 24 '20 at 06:23
  • I wanted algaebraic conditions so that if A is any matrix(of right size) then it satisfies the conditions iff A is a traceless permutation matrix. I wanted such characterization to use in the following problem https://math.stackexchange.com/questions/3725338/an-n-times-n-matrix-that-has-exactly-one-1-and-one-1-in-each-row-and-colu – ImBatman Jun 24 '20 at 06:52
  • As you probably know orthogonal matrices are characterized by $A^\top A = I$, so you've almost produced a condition yourself: Namely, $A^\top A = I$, $\min{a_{ij}} = 0$, $\operatorname{tr} A = 0$. – Travis Willse Jun 24 '20 at 17:22

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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 $A^5\neq I_5$.

  • Thanks for clarification! But that does not helps us characterizing the traceless permutation matrices. – ImBatman Jun 24 '20 at 05:19
  • Yes, this answer was to contradict your previous claim and an insight about traceless matrix. –  Jun 24 '20 at 05:23