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A question I found and that I could not answer so far. Assuming that $A$ is a $n \times n$ matrix with non-negative entries, that fulfills the equation $A^2= 0$, where $0$ is the zero matrix. What is the maximal number of positive entries for $A$?

So far I have tried going through the specific entries, namely to solve $0 = \sum_{k=1}^n a_{i,k} \cdot a_{k,j}$ for all $i,j$, but I was hoping for a simpler way. An example for $n=5$ has been

$ \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $

which fulfills the equation, but I do not know if there are better possibilities

colt_browning
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probgus
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1 Answers1

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Consider the incidence graph of the matrix obtained by replacing every non-zero entry by 1. A path of length 2 in this graph corresponds to the product of two non-zero coefficients, say $a_{i,j}a_{j,k}$. In your example, the graph would have 5 vertices and edges $(2,1), (3,1), (4,1), (2,5), (3,5), (4,5)$. Thus the problem can be reduced to the following graph-theoretic question:

Question. What is the maximum number of edges of a directed graph containing no paths of length 2?

This question was discussed in [1, Theorem 4.3] or [2, Theorem 1] and the answer is $k^2$ if $n = 2k$ and $k^2 + k$ if $n = 2k + 1$. For your case, $n = 5$, $k = 2$ and $k^2 + k = 6$, in agreement with your bound.

[1] Bermond, J.-C.; Sotteau, D.; Germa, A.; Heydemann, M.-C. Chemins et circuits dans les graphes orientés. (French) Ann. Discrete Math. 8 (1980), 293--309

[2] Sotteau, D.; Wojda, A. P., Digraphs without directed path of length two or three. Discrete Math. 58 (1986), no. 1, 105--108.

J.-E. Pin
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