Let $V$ be a vector space, dim$V=n$, $N\in L(V)$ a nilpotent operator of index $n$. Let $W$ be an $N$-invariant subspace of $V$ and $m$ the nilpotency index of $\left.N\right|_W$. Prove that $W=[\{N^{n-m}e,\dots,N^{n-1}e\}]$.

So obviously I'm trying to prove that the set $\{N^{n-m}e,\dots,N^{n-1}e\}$ is linearly independent and spans $W$. The fact that it's linearly independent follows directly from the fact that it is a subset of the cyclic basis for $V$, however, I'm struggling with the fact that it spans $W$. I've tried using the fact that for $w\in W$ we have $N^mw=0$ and $N^{m-1}w\neq 0$ but that didn't get me very far (or maybe I missed something obvious).

Any hints would be most helpful (I'm not looking for a complete proof).

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  • 1
    $\begingroup$ Use $(N\lvert_W)^m = 0$ to conclude $W \subset [\{N^{n-m}e,\dotsc, N^{n-1}e\}]$. Then use $(N\lvert_W)^{m-1}\neq 0$ to conclude equality. $\endgroup$ – Daniel Fischer Jan 11 '16 at 21:45
  • $\begingroup$ Could you give a bit more clarification? And post as an answer so I can accept it. @DanielFischer $\endgroup$ – implicati0n Jan 12 '16 at 6:17

Let's write $e_k = N^ke$ for $0 \leqslant k \leqslant n-1$. Since $\{e_0,\dotsc,e_{n-1}\}$ is a basis of $V$, we can express every $w\in W$ in that basis:

$$w = \sum_{k = 0}^{n-1} c_k\cdot e_k\tag{1}$$

with coefficients $c_k$ from the scalar field. Using the assumption $N^m w = 0$ and the representation $(1)$, we can deduce that $c_0 = \dotsc = c_{n-m-1} = 0$:

$$0 = n^mw = N^m\Biggl(\sum_{k = 0}^{n-1} c_k e_k\Biggr) = \sum_{k = 0}^{n-m-1} c_k e_{k+m} + \sum_{k = n-m}^{n-1}c_kN^me_k = \sum_{k = 0}^{n-m-1} c_k e_{k+m}.$$

Since the $e_k$ are linearly independent, it follows that $c_k = 0$ for $k \leqslant n-m-1$, i.e. $w = \sum\limits_{k = n-m}^{n-1} c_k e_k$. As $w\in W$ was arbitrary, we have

$$W \subset \operatorname{span} \{e_{n-m}, e_{n-m+1},\dotsc, e_{n-1}\}. \tag{2}$$

By the assumption $(N\lvert_W)^{m-1} \neq 0$, there is a $w_0 \in W$ with $N^{m-1}w_0 \neq 0$. Then the set

$$\{ w_0, Nw_0, \dotsc, N^{m-1}w_0\}$$

is a linearly independent subset of $W$: By the $N$-invariance of $W$ we have $N^kw_0 \in W$ for all $k\in \mathbb{N}$. Now suppose we have a linear relation

$$0 = \sum_{k = 0}^{m-1} c_k N^kw_0.\tag{$\ast$}$$

Applying $N^{m-1}$ yields

$$0 = N^{m-1}\Biggl(\sum_{k = 0}^{m-1} c_k N^kw_0\Biggr) = \sum_{k = 0}^{m-1} c_k N^{k+m-1}w_0 = c_0N^{m-1}w_0,$$

since $N^{k+m-1} w_0 = 0$ for $k \geqslant 1$, and then $N^{m-1} w_0 \neq 0$ gives us $c_0 = 0$. If we already know $c_k = 0$ for $0 \leqslant k < r \leqslant m-1$, applying $N^{m-1-r}$ to $(\ast)$ yields $c_r = 0$ in the same way, so indeed we must have $c_k = 0$ for $0 \leqslant k \leqslant m-1$ in $(\ast)$.


$$\dim W \geqslant m = \dim \operatorname{span} \{e_{n-m}, e_{n-m+1},\dotsc, e_{n-1}\}, \tag{3}$$

so in $(2)$ we actually have equality of the spaces.

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