the extension of a quasi-nilpotent operator is also quasi-nilpotent Let $T$ be a bounded linear operator  acting on a complex Banach space $X.$ Let $M$ be a subspace of $X$ invariant by $T$ such that the restriction of $T$ on $M;$ $T_{M}:M\longrightarrow M$ is quasi-nilpotent. What we can say about the operator $T_{\overline{M}}: \overline{M}\longrightarrow \overline{M};$ where $\overline{M}$ is  the closure of $M$)?  Under what conditions on $M,$ the extension $T_{\overline{M}}$ becomes quasi-nilpotent?
 A: Since $M$ is dense in $\overline{M}$ and $T$ is continuous we have $\|(T_M)^n\|=\|(T^n)_M\|=\|(T^n)_\overline{M}\|=\|(T_\overline{M})^n\|$ $(n \in \mathbb{N}).$ Thus
$$
\lim_{n \to \infty}\|(T_\overline{M})^n\|^{1/n}=\lim_{n \to \infty}\|(T_M)^n\|^{1/n}=0
$$
which means that $T_\overline{M}$ is quasi-nilpotent.
Edit: The conclusion is also true under the spectral definition of "quasi-nilpotent" if the resolvent is defined as the set of complex numbers $\lambda$ such that $T_M-\lambda I_M:M \to M$ has a continuous inverse $(T_M-\lambda I_M)^{-1}:M \to M$. Again since $M$ is dense in $\overline{M}$ the inverse can be extended to an inverse of  $T_\overline{M}-\lambda I_\overline{M}$. To see this note that $(T_M-\lambda I_M)^{-1}:M \to M$ can be extended to a continuous linear operator $B:\overline{M} \to \overline{M}$, say, without changing the norm, by the BLT-theorem (see https://en.wikipedia.org/wiki/Continuous_linear_extension, for example). Since $B(T_\overline{M}-\lambda I_\overline{M})x=(T_\overline{M}-\lambda I_\overline{M})Bx=x$ for all $x \in M$
we also have $B(T_\overline{M}-\lambda I_\overline{M})x=(T_\overline{M}-\lambda I_\overline{M})Bx=x$ for all $x \in \overline{M}$. Thus $B=(T_\overline{M}-\lambda I_\overline{M})^{-1}$.
Thus $\sigma(T_M) \subseteq \{0\}$ implies $\sigma(T_\overline{M}) = \{0\}$.
