The set of all nilpotent matrices is a closed subset of $M(n,\mathbb R).$

Question

Let $M(n,\mathbb R)$ be endowed with $\|.\|_2.$ Then show that the set of all nilpotent matrices is a closed subset of $M(n,\mathbb R).$

I tried using the continuous map $A\mapsto A^n$ on $M(n,\mathbb R).$ But arbitrary union of closed sets is not necessarily closed.

So how should I think? I'm not asking for the complete solution but some hint to start.

This is $u^{-1}({0})$ with $u:A\mapsto A^n$. Thus... – Did Aug 15 '13 at 11:04 — Did, Aug 15 '13 at 11:04

score 10 · Accepted Answer · answered Aug 15 '13 at 11:14

10

If $A$ is nilpotent, then $A^k = 0$ for some $k \in \mathbb{N}$, so the minimal polynomial $p(x)$ of $A$ divides $x^k$. Since $\text{deg}(p(x)) \leq n$, it follows that $p(x) = x^m$ for some $m \leq n$, and hence $A^m = 0$ for some $m\leq n$. Hence, $A^n = 0$.

Now by @Did's comment, just take $u^{-1}(\{0\})$, where $u(A) = A^n$.

answered Aug 15 '13 at 11:14

Prahlad Vaidyanathan

31,554

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Brilliant. Simply brilliant. – Sriti Mallick Aug 15 '13 at 11:16

The set of all nilpotent matrices is a closed subset of $M(n,\mathbb R).$

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