# Show that $\operatorname{rank}A+\operatorname{rank}A^m \leq n$ where $A^{m+1}=0$

Let $A$ be an $n \times n$ nilpotent matrix over a field $F$ with $A^{m+1}=0$. Show that $$\operatorname{rank}A+\operatorname{rank}A^m \leq n.$$

By FTLA, it is equivalent to $$n \leq \operatorname{nullity}A+\operatorname{nullity}A^m$$ By definition, $\operatorname{nullity}A^m>0$. Thus it suffices to show that $$\operatorname{nullity}A^r>\operatorname{nullity}A^{r+1}$$ for $0<r<m$

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What is the FTLA? – Mariano Suárez-Alvarez Dec 16 '13 at 17:43
Fundamental theorem of linear algebra. – Guillermo Dec 16 '13 at 17:45
Yeah, I guessed that. I was trying to ask what that it state... (It is not exactly a standard name, you see) – Mariano Suárez-Alvarez Dec 16 '13 at 17:46

Suppose $\mathrm{nullity}(A^r)=\mathrm{nullity}(A^{r+1})$. Since $\ker(A^r)\subseteq \ker(A^{r+1})$ trivially, this implies that $\ker(A^r)=\ker(A^{r+1})$. But then $$\ker(A^{r+2})=\{x:Ax\in \ker(A^{r+1})\}=\{x:Ax\in \ker(A^r)\}=\ker(A^{r+1})=\ker(A^r)$$ and so by induction we see that $\mathbb R^n=\ker(A^{m+1})=\ker(A^r)$, contradicting $r<m$.

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Using the Jordan canonical form, we can reduce immediately to the case where $A$ is one nilpontent Jordan block, and there the inequality is immediate :-)

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(Either notice that The Jordan canonical form for nilpontent matrices can be computed over any field, or that extending the field to the algebraic closure does not change anything) – Mariano Suárez-Alvarez Dec 16 '13 at 17:54

The dimension of the kernel of $A^m$ plus the rank of $A^m$ is equal to $n$. Since $A^{m+1}$ is zero, the image of $A$ is included in the kernel of $A^m$, and in particular the rank of $A$ is less than the dimension of the kernel of $A^m$. Your formula follows.

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The image of $A^m$ is contained in the kernel of $A$, because their composition is $0$. This implies $\def\rk{\operatorname{rk}}\rk A^m\leq\dim\ker(A)$. Now rank nullity for $A$ gives $\dim\ker(A)=n-\rk A$, and substituting that gives the requested inequality.

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