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Some while ago I encountered a theorem which goes like this:

Let $V$ be a $n$-dimensional vector space, and $N:V\to V$ a (linear) nilpotent operator of index $n$. Suppose we're given a cyclic base $E=\{e,Ne,...,N^{n-1}e\}$, where $e\in V$. Then all nontrivial invariant subspaces of $N$ are given as $W_k=\langle N^ke,...,N^{n-1}e\rangle, k=1,...,n-1$ (where $\langle ...\rangle$ denotes the span).

My attempt to prove this contained two steps: 1) Prove that $W_k$ itself is $N$-invariant, and 2) Prove that any arbitrary $(n-k)$-dimensional $N$-invariant subspace is the same as $W_k$. The first step is fairly easy and straightforward, but I have some doubts concerning the second step, so I'll copy my proof here.

$2^{nd}$ step: A general $(n-k)$-dimensional $N$-invariant subspace may be given as $$V_k=\langle v_1,...,v_{n-k}\rangle,$$ where $v_1,...,v_{n-k}$ is a linearly independent set of vectors. Since $E$ is a basis in $V$, we may write the vectors $v_j$ as their linear combination, i.e.: $$v_j=\sum_{i=0}^{n-1}\alpha_{j,i}N^ie.$$ We know that $dim(V_k)=n-k$, so some of $\alpha_{j,i}$ have to be zero. So we focus on the ones which don't perish.

If we consider them as sequences of numbers, then (because of the invariance of $V_k$ to $N$) there are two conclusions:

a) For a fixed $i$, it can't be true that $\alpha_{j,i-1}\neq 0$ and $\alpha_{j,i}=0 \hspace{5pt} \forall j$,

b) For at least one sum, the last coefficient should be joined with $N^{n-1}$

These two statements can be conjoined into one: Each of the last $(n-k)$ vectors of the base $E$ has a nonvanishing coefficient (i.e. $\alpha_{j,i}\neq 0$) in at least one of the sums.

Now, since $V_k$ is $(n-k)$-dimensional, it makes sense to say that the first $k$ vectors in those sums have coefficients equal to zero (if they didn't then $V_k$ wouldn't be $(n-k)$ dimensional since all $N^ie$ are linearly independent).

Overall, we may conclude that $$V_k=\langle\sum_{i=k}^{n-1}\alpha_{1,i}N^ie,...,\sum_{i=k}^{n-1}\alpha_{n-k,i}N^ie\rangle=\langle N^ke,...,N^{n-1}e\rangle=W_k,$$ i.e. $V_k$ and $W_k$ are equal sets. This concludes the proof.

Now, I'm aware that there is a simpler and more elegant proof of this statement, but I want to know if this is also valid, and whether I overlooked or misinterpreted something.

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  • $\begingroup$ It's not enough to prove that a $k$-dimensional subspace is isomorphic to $W_k$ (which would be true for any $k$-dimensional subspace). You need to prove that the sets are equal. $\endgroup$ Dec 15, 2015 at 18:08
  • $\begingroup$ Fair point, but doesn't the equality of the spans still prove that the sets are equal, so the problem in the proof is just terminology? Or is there a deeper issue? $\endgroup$ Dec 15, 2015 at 18:18
  • $\begingroup$ Just proof terminology. I was just making a passing nitpick I suppose $\endgroup$ Dec 15, 2015 at 20:44
  • $\begingroup$ It's alright, we should encourage precise terminology anyway, so I edited the question accordingly. $\endgroup$ Dec 15, 2015 at 20:51

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