Prove there are no other invariant subspaces Let $f \in End(V)$ has $n\times n$ matrix at basis $v_1, … ,v_n$ which is jordan block($n \times n$)
$A=\left[\begin{array}{ccc}a&1&0&0&…&0&0\\0&a&1&0&…&0&0\\0&0&a&1&…&0&0\\…&…&…&...&…&...\\0&0&0&0&…&a&1\\0&0&0&0&…&0&a\end{array}\right]$
Let $V_i=span\{v_1,…v_i\}$ Prove $V_i$ are the only $f$-invariant subspaces of $V$
To be honest I don't have any good idea to prove this, the only way it think can work is via contradiction
 A: Suppose $W$ is a non-zero invariant subspace, show that $v_1\in W$, and let $$k=\max\{k\in\{1,\dots,n\}:\text{$v_i\in W$ if $1\leq i\leq k$}\}.$$ I want to show now that $W$ is in fact spanned by the set $S=\{v_1,\dots,v_k\}$, and I'll do it by contradiction. Suppose then that $W$ is not spanned by $S$. This means that not all elements of $W$ are linear combinations of elements of $S$ and then there must exist a vector $x=c_1v_1+\cdots+c_rv_r$ in $W$ with $r>k$ and $c_r\neq0$. 
Computing, we see that $(f-a)(x)=c_2v_1+c_3v_2+\cdots+c_rv_{r-1}$ and since $W$ is $f$-invariant, this is an element of $W$. Computing again, we find that $(f-a)^2(x)=c_3v_1+c_2v_2+\cdots+c_rv_{r-2}$ and by induction we find that $$(f-a)^{r-k-1}(x)=c_rv_{k+1}+c_{r-1}v_{k}+\cdots\in W$$  Since $v_1,\dots,v_k$ are also in $W$ and $c_r\neq0$, this tells us that in fact $v_{k+1}\in W$, and this is absurd in view of the way that we chose the number $k$.
A: This answer is abstract in nature and assumes familiarity with basic module theory.

Endow $V$ with the structure of a $F[X]$-module by letting $X$ act on $V$ via $X·v = f(v)$. Then $f$-invariant subspaces of $V$ are exactly $F[X]$-submodules of $V$.
From this viewpoint, $V$ is cyclic as $V = F[X]e_n$, since $e_{k-1} = (a - X)e_k$ for $k = 1, …, n$. And on the other side, if $μ ∈ F[X]$ is the minimal polynomial of $f$, then $(μ)$ is the annulator of $V$ as $F[X]$-module and so $V \cong F[X]/(μ)$.
The minimal polynomial of $f$ is $(X-a)^n$.
Therefore, $V \cong F[X]/(X-a)^n$ as $F[X]$-modules and the submodules/ideals of the latter correspond exactly to the submodules/ideals of $F[X]$ containing $(X-a)^n$ which hence have to be principal ideals ($F[X]$ is euclidean), each generated by a divisor of $(X-a)^n$. Therefore they are $(X-a)^n, (X-a)^{n-1}, …, (1)$.
Either by counting (there are only $n$ non-trivial submodules and you already know that many) or by retracing them through the isomorphism, you will identify the non-trivial submodules of $V$ as generated by $e_k$ for $k=1, …, n$.
A: Hint:


*

*Show first that every invariant subspace contains $v_1$. This is easy, as it spans the one-dimensional subspace of eigenvectors.

*The map $f$ induces a map $\bar f:V/\langle v_1\rangle\to V/\langle v_1\rangle$ which is also a Jordan block, but one dimension smaller. You can therefore use induction to describe the $\bar f$-invariant subspaces of $V/\langle v_1\rangle$.

*Glue everything, and conclude what you want,noticing that the $f$-invariant subspaces of $V$ are either contained in the kernel of the projection $p:V\to V/\langle v_1\rangle$ or are preimages under $p$ of $\bar f$-invariant subspaces of $V/\langle v_1\rangle$.
