# An quasi-nilpotent operator restricted to a subspace is a nilpotent?

I am reading a paper about operator theory, there is a proposition I could not understand.

Let $T\in L(X)$ be a quasi-nilpotent operator and $X_{1}$ be a non-zero finite-dimensional subspace of X, then the restriction of T to $X_{1}$ is nilpotent (there exist a positive integer n such that $T^{n}=0$)?

Is this correct?

• Yes, it is correct. – scineram Sep 16 '13 at 16:03
• @scineram why is it true? – Norbert Sep 16 '13 at 16:32
• Is $X_1$ an invariant space of $T$? – Jonathan Y. Sep 16 '13 at 19:11
• Do you have a link to the paper? – Davide Giraudo Sep 16 '13 at 19:19
• I think there should be other assumptions, otherwise Volterra's operator en.wikipedia.org/wiki/Nilpotent_operator seems to give a counter-example. – Davide Giraudo Sep 16 '13 at 20:33

I'll assume $X_1$ is an invariant subspace of $T$. In that case, we know for all $\lambda\neq 0$ that $(T|_{X_1}-\lambda I_{X_1})$ is injective, and, since $X_1$ is finite-dimensional, invertible. Therefore $\sigma(T|_{X_1})\subset\{0\}$. It's also non-empty, so $T|_{X_1}$ is also quasinilpotent. Finally, as is apparent from the Jordan form of matrices, every quasinilpotent matrix is nilpotent.