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.


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 $f$. $\Box$

  • $\begingroup$ 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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