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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 $A^2=0\;\forall n\in\mathbb N$?

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4 Answers 4

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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 $A^k=0$, $\mbox{rank}(A)\leq \frac{n(k-1)}{k}$.

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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.

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  • $\begingroup$ So, the generalized bound is also sharp. $\endgroup$
    – Peter
    Commented Jul 10, 2014 at 11:26
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    $\begingroup$ Yes, sharp indeed $\endgroup$ Commented Jul 10, 2014 at 11:41
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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 $k$.

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    $\begingroup$ I don't know, does it? Calculate $A^2$, it's not hard. $\endgroup$
    – 5xum
    Commented Jul 10, 2014 at 10:02
  • $\begingroup$ $A^2=0$ does indeed hold and to cover the case of odd n, we can add a zero-column and a zero-row. $\endgroup$
    – Peter
    Commented Jul 10, 2014 at 10:54
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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 $\operatorname{rank}A=p=\frac n2$.

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