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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.

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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 $f$. $\Box$

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  • $\begingroup$ This is unclear to me, Did you take $u_{1},\cdots u_{j} \in Ker \hspace{0.1cm} f?$ That's why they don't satisfy $(1)?$ Why since $v \not\in Ker \hspace{0.1cm} f$ we know all Jordan bases of $\mathbb{C}^{n}$ for $f$ ''must'' contain the same $v$ satisfying property $(1)$ $\endgroup$ – jacopoburelli Aug 31 at 8:20
  • $\begingroup$ @jacopoburelli I realized there was one important part left out by accident. I also added a lot more details. Let me know if this still isn't clear. $\endgroup$ – dylan7 Aug 31 at 17:38
  • $\begingroup$ So the vector i'm looking for is $v \not\in Kerf^{m}$. One more question and I will accept the answer, How this could be seen on a given matrix ? According with the decreasing ordering of Jordan-blocks ? $\endgroup$ – jacopoburelli Aug 31 at 18:04
  • $\begingroup$ The response I posted above is a full solution. The question wanted a subspace, not a particular vector. The subspace it wanted $W$, constructed in my post, consists of the set $\{ f^m z : z \notin Ker f^m\} = span(f^m v), v \notin Ker f^m$. Every element of $W$ is actually in $Ker f$. For some $v \notin ker f^m$, the Jordan Block for this $v$ consists of $m$ ones in the super-diagonal, and 0 elsewhere. This is the block over $f^m v, \dots, fv, v$ I'm not sure what you mean by "decreasing order of Jordan-Blocks". $\endgroup$ – dylan7 Aug 31 at 18:37
  • $\begingroup$ I meant arranging blocks ($= dim Kerf$) by decreasing size $\endgroup$ – jacopoburelli Aug 31 at 18:40

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