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Given a permutation matrix that is not full rank, is there an algebraic criterion to tell if matrix contains more than one disjoint non-trivial cycle or exactly one non-trivial cycle?

Example: Consider $$\begin{pmatrix} 0& 0& 0& 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0 \end{pmatrix}$$

This matrix embeds 1st row->2nd row->6th row->8th row->5th row->1st row cycle (essentially a 5-cycle in $8\times 8$ matrix).

When previous example acts on left on \begin{pmatrix} a1& a2& a3& a4& a5& a6& a7& a8\\ b1& b2& b3& b4& b5& b6& b7& b8\\ c1& c2& c3& c4& c5& c6& c7& c8\\ d1& d2& d3& d4& d5& d6& d7& d8\\ e1& e2& e3& e4& e5& e6& e7& e8\\ f1& f2& f3& f4& f5& f6& f7& f8\\ g1& g2& g3& g4& g5& g6& g7& g8\\ h1& h2& h3& h4& h5& h6& h7& h8 \end{pmatrix}

we get \begin{pmatrix} e1& e2& e3& e4& e5& e6& e7& e8\\ a1& a2& a3& a4& a5& a6& a7& a8\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ h1& h2& h3& h4& h5& h6& h7& h8\\ b1& b2& b3& b4& b5& b6& b7& b8\\ 0& 0& 0& 0& 0& 0& 0& 0\\ f1& f2& f3& f4& f5& f6& f7& f8 \end{pmatrix}

Acting on left $5$ times gives\begin{pmatrix} a1& a2& a3& a4& a5& a6& a7& a8\\ b1& b2& b3& b4& b5& b6& b7& b8\\ 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0\\ e1& e2& e3& e4& e5& e6& e7& e8\\ f1& f2& f3& f4& f5& f6& f7& f8\\ 0& 0& 0& 0& 0& 0& 0& 0\\ h1& h2& h3& h4& h5& h6& h7& h8 \end{pmatrix}

Turbo
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1 Answers1

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Note: The below answers an older version of the question, which has since been changed.

If $P$ is a permutation matrix, it will be exactly one cycle if and only if its characteristic polynomial is $x^n - 1$.

Here's a way to verify the characteristic polynomial without actually computing the associated determinant: first, verify that $P^n = I$. Then, verify that $q(P) \neq 0$ for any polynomial $q$ that divides $(x^n - 1)/(x-1)$. If this is the case, then $P$ consists of exactly one cycle.

In any example like that you've presented, we can compute $P^TP$ to get a diagonal matrix consisting of $1$s and $0$s. If we consider the submatrix obtained by deleting all rows and columns corresponding to a diagonal zero, we obtain the smaller corresponding permutation of full rank.

In your example, we compute $$ P^TP = \pmatrix{ 1\\ &1\\ &&0\\ &&&0\\ &&&&1\\ &&&&&1\\ &&&&&&0\\ &&&&&&&1 } $$ And get the submatrix $$ \pmatrix{ 0& 0& 1& 0& 0\\ 1& 0& 0& 0& 0\\ % % 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0\\ % 0& 0& 0& 1& 0 } $$

Ben Grossmann
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  • Why is $q(P)\neq0$ if $q|\frac{x^n-1}{x-1}$? In above matrix, if (3,3),(4,4),(7,7) entry is 1, we get 1 cycle. Mathematica does not give form you suggest. I am also interested in situation where $P$ is not full rank as in example above. – Turbo Feb 15 '15 at 23:27
  • I cannot delete rows. I understand what you are saying. But char poly on mathematica does not look way you suggest if cycle is not over $n$ letters. cycle could be over $m<n$ letters. Still we could have just 1 non trivial cycle (rest of letters be fixed). – Turbo Feb 15 '15 at 23:29
  • What do you mean you can't delete rows? Would it be better if we expressed row-deletion as a matrix operation? – Ben Grossmann Feb 15 '15 at 23:30
  • @Turbo you never said "one non-trivial cycle" earlier. That changes things drastically. – Ben Grossmann Feb 15 '15 at 23:32
  • ahhhh ok I will update. – Turbo Feb 15 '15 at 23:33