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.
A Closed subset of $M_n(\mathbb{R})$
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matrices
metric-spaces
2 Answers
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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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0Dear Davide: and +1 for your answer, by the way! – 2012-04-01
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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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0@ Georges Yes !!, thnx for the remark. – 2012-04-01