$n$ dimensional vector space over algebraically closed field has at least $n+1$ $T$-invariant subspaces 
Let $V$ be an $n$-dimensional vector space over an algebraically closed field $F$ and $T:V\rightarrow V$ a linear operator. Show that $V$ has at least $n+1$ $T$-invariant subspaces.

Of course, $0$ and $V$ are always $T$-invariant, so I want to collect another $n-1$ $T$-invariant subspaces.
Since we're working over an algebraically closed field, the characteristic polynomial splits, so the minimal polynomial does too, yielding a Jordan canonical form of $T$, where $T$ can be represented as a diagonal block matrix with Jordan blocks $J_1,\ldots ,J_r$ for $1\leqslant r\leqslant n$, with the size of each Jordan block $J_i$ corresponding to the geometric multiplicity of its corresponding generalized eigenspace $E_i \subset V$.
At this point I know that each generalized eigenspace and all possible direct sums of the generalized eigenspaces correspond to nontrivial $T$-invariant subspaces, which gives another $\sum_{k=1}^r \binom{r}{k}$ $T$-invariant subspaces.
But here I'm stuck since this might not collect enough $T$-invariant subspaces in general if $r$ is small and $n$ is large (e.g. if $T=0$ and $r=1$, though this case can be handled trivially anyway).  Since the Jordan blocks cannot be decomposed any further, how do I find more $T$-invariant subspaces?
Intuitively, I would think to look for kernels of $T$, particularly kernels of $T|_{E_i}$, but I can't quite get a grasp on when we would find a nontrivial kernel for each restricted operator. Any pointers on how to continue?
 A: *

*If the minimal polynomial $f$ doesn't have degree $n$ then there is an eigenvalue such that $\ker(T-a)$ has dimension $\ge 2$ which gives infinitely many 1-dimensional eigenspaces.


*If $\deg(f)=n$ then the invariant subspaces $W$ are exactly the $\ker(h(T))$ for each $h$ dividing $f$.
(if $u$ strictly divides $v$ which strictly divides $f$ then $\dim(\ker(u(T)))<\dim(\ker(v(T)))<\dim(\ker(f(T)))$, so $\deg f=n$ implies $\dim(\ker(u(T))= \deg u $,
then let $u$ be the minimal polynomial of $T|_W$)
Factorize $f=\prod_j (x-a_j)^{e_j}$ you get $\prod_j (e_j+1)$ invariant subspaces, and it is minimal when $f=(x-a_1)^n$.
A: Consider the Jordan block
$$T = \begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix}.$$
Then the ambient vector space $F^3$ has four $T$-invariant subspaces: the zero vector space, the 1-dimensional space spanned by $(1, 0, 0)$, the two-dimensional space spanned by $(1, 0, 0)$ and $(1, 1, 0)$, and the whole space. (In fact it is not important that $T$ is a Jordan block for this to work, it just has to be upper-triangular).
In the general case, decompose an operator $T$ on $F^n$ into Jordan blocks which will give you $n$ nonzero distinct $T$-invariant subspaces by a similar argument, then together with the zero subspace you have $n + 1$ of them.
