In this note, we show that the number of digraphs with n vertices and with cycles of length k, 0 ≤ k ≤ n, is equal to the number of n × n (0,1)-matrices whose eigenvalues are the collection of copies of the ent...In this note, we show that the number of digraphs with n vertices and with cycles of length k, 0 ≤ k ≤ n, is equal to the number of n × n (0,1)-matrices whose eigenvalues are the collection of copies of the entire kth unit roots plus, possibly, 0’s. In particular, 1) when k = 0, since the digraphs reduce to be acyclic, our result reduces to the main theorem obtained recently in [1] stating that, for each n = 1, 2, 3, …, the number of acyclic digraphs is equal to the number of n × n (0,1)-matrices whose eigenvalues are positive real numbers;and 2) when k = n, the digraphs are the Hamiltonian directed cycles and it, therefore, generates another well-known (and trivial) result: the eigenvalues of a Hamiltonian directed cycle with n vertices are the nth unit roots [2].展开更多
文摘In this note, we show that the number of digraphs with n vertices and with cycles of length k, 0 ≤ k ≤ n, is equal to the number of n × n (0,1)-matrices whose eigenvalues are the collection of copies of the entire kth unit roots plus, possibly, 0’s. In particular, 1) when k = 0, since the digraphs reduce to be acyclic, our result reduces to the main theorem obtained recently in [1] stating that, for each n = 1, 2, 3, …, the number of acyclic digraphs is equal to the number of n × n (0,1)-matrices whose eigenvalues are positive real numbers;and 2) when k = n, the digraphs are the Hamiltonian directed cycles and it, therefore, generates another well-known (and trivial) result: the eigenvalues of a Hamiltonian directed cycle with n vertices are the nth unit roots [2].
