Show exists a subspace $W \subseteq \mathbb{C}^{n}$ of dimension $1$ such that every Jordan basis of $ \mathbb{C}^{n}$ contains a generator of $W$ Let $n\geq 2$.
Given $f$ nilpotent endomorphism of $\mathbb{C}^{n}$ such that exists an integer $k \geq 1$ such that $dim \hspace{0.1cm} Kerf^{k+1} = dim \hspace{0.1cm} Kerf^{k}+1$.
$(1) \hspace{0.1cm}$Show exists a subspace $W \subseteq \mathbb{C}^{n}$ of dimension $1$ such that every Jordan basis of $ \mathbb{C}^{n}$ contains a generator of $W$.
$(2) \hspace{0.1cm}$Give an example of nilpotent $g$ $\in End(\mathbb{C}^{n})$ with the property that $dim \hspace{0.1cm} Ker \hspace{0.1cm}g^{2} = dim \hspace{0.1cm} Ker\hspace{0.1cm}g +1$ such that exists no subspace $Z \subseteq \mathbb{C}^{n}$ of dimension $1$ such that every basis of Jordan Basis$_{\mathbb{C}^{n}}$ of $g^{2}$ contains a generator of $Z$. 
I think $(2)$ follows directly from truly understsanding $(1)$ so I'd like to solve $(1)$ first.
My guess is that given a Jordan basis $B=\{v_{1},\cdots,v_{n}\}$ such that $$M_{B}^{B}(f) = \begin{pmatrix}0 & 1 & \cdots & 0 \\ 0 & 0 & 1\cdots & 0 \\ \vdots & \cdots & \cdots & 1 \\ 0 & \cdots & \cdots & 0 \end{pmatrix}$$
(Wlog we can restrict our view only to a Jordan block)
The subspace I'm looking for is $v_{1} \in Ker \hspace{0.1cm}f$ for every Jordan basis,
But I'm unable to deduce or prove it directly from the proof of Jordan basis construction,
Any tip,help or solution would be appreciated.
 A: For part 1.
We know there exists an integer $m \geq 1$ such that $ker f^{m+1} = \mathbb{C}^n$, but $ker f^m \neq \mathbb{C}^n$ and $dim(ker f^m) + 1 = dim(ker f^{m + 1})$.
Thus it can be shown
$ker f \subsetneq ker f^2 \subsetneq \dots \subsetneq ker f^m  \subsetneq ker f^{m+1} = \mathbb{C}^n$ where $\subsetneq$ denotes "strict inclusion" to be precise.
By the above, $\exists v,$
$v \in \mathbb{C}^n$ s.t $ v \notin ker f^m$.  Otherwise, $ker f^m = ker f^{m+1}$
Consider linearly independent, $v, u_1, \dots u_j$, such that $ j + 1 = dim(ker f)$, and all $u_1, \dots, u_j \in ker f^m$.
Such that $f^m v, \dots, fv, v, f^{s_1} u_1, \dots, f u_1, u_1, \dots, f^{s_j}u_j, \dots f u_j, u_j$ is a Jordan basis for $\mathbb{C}^n$.  


*

*The part $j+1 = dim(ker f)$ is important because for any Jordan basis
$f^{k_0} w_0, \dots, fw_0, w_0, f^{k_1} w_1, \dots, f w_1, w_1 \dots f^{k_j}w_j, \dots f w_j, w_j$ ($f^{k_i + 1}w_i = 0$), the list
$f^{k_0}w_0, \dots, f^{k_j}w_j$ is a basis for $ker f$. Thus for any Jordan basis the number of $k_i$'s is fixed to the dimension of $ker f$. Otherwise, it isn't a basis for $\mathbb{C}^n$
Since $v \notin ker f^m$, we know all Jordan bases of $\mathbb{C}^n$ for $f$ must contain at least one $v$, such that $v \notin ker f^m$, ($ker f^m \neq \mathbb{C}^n$). We know at least one, since $dim(ker f^m) + 1 = dim(ker f^{m+1})$.
Fix the $v$, and the $u_i$'s above. Suppose $\exists v_2 \notin ker f^m, v_2 \neq v$ and   $v_2,u_1, \dots, u_j$ is linearly independent.
Such that
$f^m v_2, \dots, fv_2, v_2, f^{s_1} u_1, \dots, f u_1, u_1, \dots f^{s_j}u_j, \dots f u_j, u_j$ 
is a Jordan basis of $\mathbb{C}^n$ for $f$.
Now consider some vector $w \notin ker f^m$ and its linear expansion over both bases. 
$w = \lambda_{0,1} f^m v + \dots + \lambda_{0,(m-1)}fv + \lambda_{0,m}v + \lambda_{1,1} f^{s_1} u_1 + \dots + \lambda_{1,s_1-1}f u_1 + \lambda_{1,(s_1 -1)}u_1 + \dots + \lambda_{j,1}f^{s_j}u_j + \dots + \lambda_{j, (s_j -1)}f u_j + \lambda_{j, s_j}u_j$
$ = $
$\eta_{0,1} f^m v_2 + \dots + \eta_{0,(m-1)}fv_2 + \eta_{0,m}v_2 + \eta_{1,1} f^{s_1} u_1 + \dots + \eta_{1,s_1-1}f u_1 + \eta_{1,(s_1 -1)}u_1 + \dots + \eta_{j,1}f^{s_j}u_j + \dots + \eta_{j, (s_j -1)}f u_j + \eta_{j, s_j}u_j$
The $\lambda_{i,j}, \eta_{i,j} \in \mathbb{C}$
Applying $f^m$ to both sides of both equalities, we get
$f^m w = \lambda_{0,m}f^mv = \eta_{0,m} f^m v_2$
and $f^m w \neq 0 \implies \lambda_{0,m},\eta_{0,m} \neq 0$ 
So $(\lambda_{0,m}/ \eta_{0,m})f^mv =f^m v_2$. Let $W = span(f^mv) $.
Since $v_2$ is arbitrary, a scalar multiple of $f^m v$ is in  every Jordan Basis of $\mathbb{C}^n$ for $f$. $\Box$
