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

I came across the fact that the rank of a $$n \times n$$-matrix A with $$A^2=0$$ is at most $$\frac{n}{2}$$. The easiest way to proof this is using the inequality $$\operatorname{rank}(A)+\operatorname{rank}(B)-n\leqslant\operatorname{rank}(AB)$$. With $$A=B$$ and $$A^2=0$$, this immediately yields $$\operatorname{rank}(A)\leqslant\frac{n}{2}$$

Can this result be generalized? What is the upper bound for the rank of a $$n\times n$$-matrix $$A$$ with the property $$A^k=0$$?

Additional question : Is the above bound for $$k = 2$$ sharp? In other words, is there a $$n \times n$$ matrix $$A$$ with $$\operatorname{rank}\left\lfloor\frac{n}{2}\right\rfloor$$ and $$A^2=0\;\forall n\in\mathbb N$$?

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 $$x\leqslant\lfloor x\rfloor). Hence the maximal rank is equal to $$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.

• So, the generalized bound is also sharp. Jul 10 '14 at 11:26
• Yes, sharp indeed Jul 10 '14 at 11:41

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 $k$.

• I don't know, does it? Calculate $A^2$, it's not hard.
– 5xum
Jul 10 '14 at 10:02
• $A^2=0$ does indeed hold and to cover the case of odd n, we can add a zero-column and a zero-row. Jul 10 '14 at 10:54

I'm not sure that the given inequality is the easiest proof. Indeed, $A^2=0\iff \operatorname{im}A\subset \ker A$ and with the rank-nullity theorem we find the desired result. This bound is sharp: let $n=2p$ and let $(e_1,\ldots,e_n)$ a basis and define $A$ by $$A e_k=0,\quad 1\le k\le p$$ and $$A e_k=e_{n+1-k},\quad p+1\le k\le n$$ then we see that $A^2=0$ and that $\operatorname{rank}A=p=\frac n2$.