# A Closed subset of $M_n(\mathbb{R})$

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.

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0% accept rate? – user2468 Apr 1 '12 at 20:16

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.

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thank you Dear Sir. – Taxi Driver Apr 1 '12 at 20:15
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

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.

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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