Given an $n\times n$ matrix $A$ with $A^k=0$ for some $k\geq 2$, at most how many nonzero elements could $A$ have if its element is either $0$ or $1$?
My attempt: Sylvester rank inequality gives $rank(A)\leq n(k-1)/k$. Does that mean the answer is $n(k-1)/k$? If $k=2$ and $n=2$, the bound is $1$. It seems ok. But rigorously how to improve?
Thanks!
The proposed bound is not true. Let $n=2m$ and $$a_{ij}=\begin{cases} 0 &j\le m\\ 0 & j>m,\ i>m\\ 1 & j>m,\ i\le m \end{cases}$$ Then $A^2=0.$ The matrix $A$ has $m^2=n^2/4$ positive entries.
For $n=2m+1$ let $$a_{ij}=\begin{cases} 0 &j\le m\\ 0 & j> m,\ i>m\\ 1 & j\ge m,\ i\le m \end{cases}$$ Then $A^2=0$ and the matrix $A$ has $(m+1)m=(n^2-1)/4$ positive entries.