Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $0<r<m$

share|improve this question
    
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
1  
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

4 Answers 4

up vote 2 down vote accepted

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 $r<m$.

share|improve this answer

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 :-)

share|improve this answer
    
(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.

share|improve this answer

The image of $A^m$ is contained in the kernel of $A$, because their composition is $0$. Now rank nullity for $A$ gives your inequality.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.