Tell me more ×
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.
up vote 2 down vote favorite
1

I can guess that set of Nilpotent Matrices are closed in $M_n(\mathbb{R})$, But I am not able to make it rigorous; I have thought the map $A\mapsto A^k$ is continuous. But then? Please help.

share|improve this question
3  
0% accept rate? – user2468 Apr 1 '12 at 20:16

2 Answers

up vote 8 down vote accepted

If $A$ is nilpotent, then $A^n=0$. Since the map $f(A):=A^n$ is continuous, $\operatorname{Nil}(M_n(\mathbb R)):=\{A\in M_n(\mathbb R): \exists p\in\mathbb N_{>0}, A^p=0\}=f^{-1}(\{0_{M_n}\})$ is closed.

share|improve this answer
thank you Dear Sir. – Taxi Driver Apr 1 '12 at 20:15
1  
Caro Davide, $\operatorname{Nil}(M_n(\mathbb R))=F_n$. – Georges Elencwajg Apr 1 '12 at 20:22
@GeorgesElencwajg In fact, yes, I realize that the $F_p$ are nested. And we have less characters to type. Thanks. – Davide Giraudo Apr 1 '12 at 20:33
Dear Davide: and +1 for your answer, by the way! – Georges Elencwajg Apr 1 '12 at 20:43
up vote 1 down vote

Since the map $m_k: A\rightarrow A^k$ is continuous and $A$ is nilpotent of order $\leq k$

iff $A\in m_k^{-1}(0)$ is closed (because $\{0\}$ is closed)

we have $\operatorname{Nil}(M_n(\mathbb R))=\bigcup_{k=1}^n m_k^{-1}(0)$ also is closed.

share|improve this answer
3  
Dear Matrix, you should definitely not write $Ker(m_k)$, because $m_k$ is not a linear mapping. – Georges Elencwajg Apr 1 '12 at 20:48
@ Georges Yes !!, thnx for the remark. – Abdelmajid Khadari Apr 1 '12 at 20:58

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.