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Let $A$ be an injective endomorphic operator on some functions space (wlog say $C^{\infty}(\mathbb{R})$). Then it cannot be nilpotent.

Does the above statement hold? I've tried to prove it like that:

Proof by induction. For $n=1$, this is just injectivity. Assume it holds for $n-1$, ie. $A^{n-1}$ in injective. Then $A^n f = 0$ is equivalent to $A(A^{n-1}f))=0$. As $A$ is injective, then $A^{n-1}f=0$, but as $A^{n-1}$ is injective aswell, we ge that $f=0$.

As every power of the operator $A$, ie. every $A^n$ is injective, A cannot be nilpotent.

numbertheorist
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    Thank you. And the question is...? And $;A;$ is an operator...on what algebraic structure or what? – DonAntonio Jun 25 '18 at 16:45
  • Thanks, I have added the details – numbertheorist Jun 25 '18 at 17:02
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    It's a bit cleaner to prove the contrapositive: that if $A$ is nilpotent then it can't be injective. This is straightforward: if $A^n = 0$ then pick any nonzero vector $v$, so that $A^n v = 0$. Then the sequence $v, Av, A^2 v, \dots A^n v$ must eventually be zero, and the first point at which that happens, $A$ has sent a nonzero vector $A^i v$ to zero. – Qiaochu Yuan Jun 25 '18 at 19:35

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