Upper bound for the rank of a nilpotent matrix , if $A^2 \ne 0$

The inequality generalises easily to $$ \sum_{j=1}^k\mbox{rank}(A_j)\leq \mbox{rank}\left(\prod_{j=1}^k A_j\right)+n(k-1). $$ In particular, if $A^k=0$, $\mbox{rank}(A)\leq \frac{n(k-1)}{k}$.

Suppose $A$ is an element of $M_n(\mathbb{C})$. By Jordan form, if $A^k=0$, we have: (i) all eigenvalue are $0$; (ii) each Jordan block of $A$ has at most $k-1$ entries of $1$. To obtain the maximal value of $\operatorname{rank}(A)$, we consider the case of minimal number of Jordan blocks. The minimal number of Jordan blocks is equal to $\frac{n}{k}$ if $k$ divides $n$ and $\left\lfloor\frac{n}{k}\right\rfloor+1$ if $k$ doesn't divide $n$ (where $\lfloor x\rfloor$ is the unique integer such that $x\leqslant\lfloor x\rfloor<x+1$). Hence the maximal rank is equal to $n-\frac{n}{k}$ if $k$ divides $n$, and $n-\left\lfloor\frac{n}{k}\right\rfloor-1$ if $k$ doesn't divide $n$ . This can also be written as$$\operatorname{rank}(A)\leqslant\left\lfloor\frac{n(k-1)}{k}\right\rfloor.$$

This proves Jonas's inequality, plus that the equality is obtainable (by choosing $A=\begin{pmatrix}D_1&\dots&\dots&\dots&0\\0&\dots&\dots&\dots&0\\\dots&\dots&\dots&\dots&\dots\\0&\dots&\dots&D_k&0\\0&\dots&\dots&\dots&E\end{pmatrix}$ where $D_i=\begin{pmatrix}0&1&\dots&0\\0&0&\dots&0\\0&0&\dots&1\\0&0&\dots&0\end{pmatrix}$ of size $k$ and $E=\begin{pmatrix}0&1&\dots&0\\0&0&\dots&0\\0&0&\dots&1\\0&0&\dots&0\end{pmatrix}$ of size $n(\mathrm{mod}k)$).

If the existence of Jordan form is not assured, the above construction shows that the equality is obtainable, assuming the inequality.

Well, the matrix $$\begin{bmatrix}0&I\\0&0\end{bmatrix}$$ is nilpotent (zeros are $k\times k$, zero matrices, $I$ is a $k\times k$ identity matrix) it is of size $2k$ and has a rank of $k$.