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$\phi = \mathbb{V} \rightarrow \mathbb{V}$ is an operator satisfying $\phi^n = 0$ for some $n$ and $\phi^{n-1} \ne 0$

Let $v \in \mathbb{V}$ be a vector s.t. $\phi^{n-1} \ne 0$. Is the set {v, $\phi(v)$, $\dots$, $\phi^{n-1}(v)$} linearly independent?

I intuitively understand that it is. Also I've proven that the only eigenvalue of this linear transformation is 0.

As for proving linear independence, I'm trying to prove it by contradiction (is this the correct path?).

Assume the set is not linearly independent. Then there are values $c_1, \dots, c_n$ such that:

$c_1v+c_2\phi(v) +\dots +c_{n}\phi^{n-1}(v) = 0 = \phi^n(v)$ and $c_i \ne 0$ for some $i \in [n]$.

Where do I move from here?

Avgustine
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1 Answers1

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We have $$0 = c_{1}v + c_{2}\phi(v) + \cdots + c_{n}\phi^{n-1}(v),$$ where not all the $c_{i}$ are zero. Apply $\phi^{n-1}$ to both sides. By linearity, we have \begin{align*} 0 = \phi^{n-1}(0) &= \phi^{n-1}\left(c_{1}v + c_{2}\phi(v) + \cdots + c_{n}\phi^{n-1}(v)\right)\\ &= c_{1}\phi^{n-1}(v) + c_{2}\phi^{n}(v) + \cdots + c_{n}\phi^{2n-2}(v)\\ &= c_{1}\phi^{n-1}(v), \end{align*} where the last equality follows because $\phi^{n} = 0$ (and so $\phi^{m} = 0$ for all $m \geq n$). But by assumption $\phi^{n-1}(v) \neq 0,$ so we must have $c_{1} = 0.$

So we actually have $$0 = c_{2}\phi(v) + \cdots + c_{n}\phi^{n-1}(v).$$ Can you use the same trick to continue from here, and arrive at a contradiction?

ckefa
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