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I am studying on nilpotent matrices over finite fields. By definition a square matrix $A$ is $p$-nilpotent if a power of $A$ modulo $p$ is the zero matrix. For example. Let $J_2$ be the $2\times 2$ matrix of ones. Then $J_2$ is 2-nilpotent. Now, if we consider a $p\times p$ matrix $A$ in which entries are from the set $\{0,1\}$ ($A$ is usually called binary or $(0,1)$-matrix) and $A$ is $p$-nilpotent so that the entries all the main diagonal of $A$ are all ones. Does any one has any idea about the form of matrix $A$? I believe that in this case $A$ should be equal to the all ones matrix.

Hieu
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  • Any triangular matrix with zeros on the diagonal is nilpotent. – pancini Aug 17 '18 at 15:22
  • Yes, you are right. I forgot to include here another condition for $A$ that the entries all the main diagonal are all ones. Thanks for your comment. – Hieu Aug 17 '18 at 15:25
  • If $p$ is even, then $A$ need not be the all-ones matrix. For example, $$ A = \pmatrix{1&1&0&0\1&1&0&0\0&0&1&1\0&0&1&1} $$ is nilpotent over $\Bbb F_2$ and has $1$s on the diagonal – Ben Grossmann Aug 17 '18 at 15:29
  • I need that condition, because I am considering the directed graph in which every vertex has a loop. As we know, there is a theorem stating that a directed graph is acyclic iff the adjacency matrix is nilpotent. But I haven't seen any version of that theorem for finite fields. – Hieu Aug 17 '18 at 15:29
  • Yes, you are right. That example is for the case of $4\times 4$ matrix. If the matrix is $p\times p$, it should be the all ones matrix. – Hieu Aug 17 '18 at 15:30
  • @Hieu is $p$ supposed to be prime? – Ben Grossmann Aug 17 '18 at 15:31
  • Yes, $p$ is prime. – Hieu Aug 17 '18 at 15:31
  • It would probably be helpful to use the Perron-Frobenius theorem. This states our ${0,1}$ is interpreted as a real matrix, it must have a non-negative eigenvector associated with its spectral radius – Ben Grossmann Aug 17 '18 at 16:18
  • Also, if $x$ is such an eigenvector, and if $\hat x$ is $x$ with $0$s replaced with $1$s, then $$ \operatorname{diag}^{-1}(\hat x)A\operatorname{diag}(\hat x) $$ is a non-negative matrix with constant row sums equal to $\rho(A)$ – Ben Grossmann Aug 17 '18 at 16:39
  • We have a quite powerful theorem which states that Given a graph, then the associated matrix A is primitive if and only if the graph is strongly connected and has two cycles of relatively prime lengths.1 Although with this theorem you can not prove that a matrix is primitive over finite field, you can detriment that a matrix is non-primitive or not. In the other words, this theorem for finite field is necessary but not sufficient. – user0410 Aug 19 '18 at 22:45
  • Thanks for your comments. it is a question if we may adapt the theorem for $F_p$, likes if the graph has at least one cycle of length 1 then the number of cycles of length $p$ in the graphs is a multiple of $p$ iff the adjacency matrix is $p$-nilpotent. – Hieu Aug 20 '18 at 15:38
  • If we consider $A$ as a matrix over $\mathbb{R}$, we have that the matrix $A$ is non-negative. If $A$ is also irreducible then $A$ is primitive by the theorem. – Hieu Aug 20 '18 at 15:44

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