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Let $T:W \to W$ be a linear operator vector space $W$ over $\mathbb{F}$. such that $w \in W$ where $$\{w, T(w) ,T^2(w)\}$$ is linearly independent and $T^3(w)= w +T(w)+T^2(w)$.

Show that $$V := \operatorname{span}{\{w ,T(w),T^2(w)\}}$$ is $T$-invariant.

My attempt

First w implies that $T^3(w)= w +T(w)+T^2(w)$ and is a element $W$ Then $$T^3(T(w)) = T(w) + T^2(w) + T^3(w) = T(w) + T^2(w) + [w +T(w)+T^2(w)]$$ and is a element W I don't know what to do with $T^2(w)$ because then $$T^3(T^@(w))=T^2(w) + [w +T(w)+T^2(w)] + T^2(T^2(w))?$$

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A general element $x$ of $V$ is of the form $$x=aw+bT(w)+cT^2(w).$$ Can you show that $T(x)$ is of the same form (with some other constants in place of $a,b,c$)? –  Jyrki Lahtonen Aug 7 '13 at 10:02
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2 Answers

Let $v \in V$, $v$ can be expressed as $aw + bT(w) + cT^2(w)$. Now it follows:

$\begin{align*} T(v) &= T(aw + bT(w) + cT^2(w)) \\ &= aT(w) + bT^2(w) + cT^3(w) \;\mbox{by linearity}\\ &= aT(w) + bT^2(w) + c\left[w + T(w) + T^2(w) \right] \; \mbox{by the hypothesis} \\ &= cw + (a+c)T(w) + (b+c)T^2(w) \in V \end{align*}$

so $V$ is $T$-invariant.

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Let $v \in V$. Show $T(v) \in V$:
$$T(v) = T(\alpha w + \beta T(w) + \gamma T^2(w) = \alpha T(w) + \beta T^2(w) + \gamma T^3(w)$$ by requirements we have $$T(v) = \gamma w + (\alpha + \gamma) T(w) + (\beta + \gamma) T^2(w) \in V$$ $$\Rightarrow T(V) \subset V$$ q.e.d.

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