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?
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.