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I'm attempting this problem:

Assuming that $A$ is an $n \times n$ matrix with non-negative entries and $A^2 = 0$. What's the maximal number of positive entries in $A$?

The answer uses a graph theory approach. However, I'm not so familiar with graph theory. I have two questions

  1. Is there a way to do this without using graph theory?
  2. The answer implies when $n = 6$, there should be $3^2 = 9$ positive entries. I'm struggling to find an example for an $6 \times 6$ matrix with $9$ positive entries that would satisfy $A^2 = 0$ with all entries being non-negative. I can only find a matrix with $8$ positive entries. Can you give me an example of $A$ when $n = 6$?
user1691278
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1 Answers1

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The first question has been answered, also without using graph theory in this duplicate:

Let $A$ be a binary $n \times n$ matrix, such that $A^2=0$. What is the max num of $1$'s that $A$ could have?

The answer is given by the function $f(n)=\lfloor \frac{n}{2} \rfloor \lceil \frac{n}{2} \rceil$. For $n=2k$ this $k^2$, and for $n=2k+1$ this is $k^2+k$.

So let me answer the second question. Here is an example of a binary matrix $A$ with nine entries equal to $1$ and all others equal to zero with $A^2=0$: $$ A=\begin{pmatrix} 0 & 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$

Dietrich Burde
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