Is every invariant subspace equal to some $\text{null}(T-\lambda I)^n$? This is a natural follow-up to this question.
Let $V$ be a finite-dimensional complex vector space and let $U$ be a subspace of $V$ invariant under the linear operator $T$:
$$\forall u\in U: Tu \in U.$$
Assume further that $U$ is not the subspace of an eigenspace of $V$.
Must there exist $\lambda\in\mathbb C,n\in\{1,\dots,\dim V\}$ such that
$$U=\text{null}\,(T-\lambda I)^n \; ?$$

The linked question is similar except for requiring $n=\dim V$, and the answer is "no" with a counter-example given. But the counterexample has $\dim V=3$ and still satisfies the above condition for $n=2$. Moreover, there is some reason to suspect this might be true because of the role played by subspaces of this form in the Jordan decomposition.
 A: No.
For one thing, $V$ is always $T$-invariant, and unless $T$ has only one eigenvalue, will not be equal to a generalized eigenspace.
For another example, consider $T\colon\mathbb{C}^3\to\mathbb{C}^3$ given by $T(x,y,z) = (x,2y,z)$. Then $U=\{(x,y,0)\in\mathbb{C}^3\mid x,y\in\mathbb{C}\}$ is $T$-invariant, but is not a generalized eigenspace, or even generated by a union of generalized eigenspaces, because you would need $\lambda=1$ to get the vectors $(x,0,0)$, but that would force you to include the vectors $(0,0,z)$ which are not contained in $U$.
Moreover, you can have a subspace spanned by eigenvectors corresponding to distinct eigenvalues, or by suitably chosen portions of generalized eigenspaces corresponding to distinct eigenvalues, and have an invariant subspace that cannot be expressed as a single generalized eigenspace; and if you don't include all the the generalized eigenspace, you will not be able to express them in terms of generalized eigenspaces either. 
Now, for a finite dimensional complex vector space, you can show that any invariant subspace has a basis of generalized eigenvectors; but that is simply because you can find a Jordan canonical form for the restriction of $T$; the generalized eigenvectors for $T|_U$, when $U$ is $T$-invariant, are also generalized eigenvectors of $T$. So in a sense you can "come down" to some generalized eigenvectors fo $T$, but you won't get just a generalized eigenspace.
