Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$T^3(T^@(w))=T^2(w) + [w +T(w)+T^2(w)] + T^2(T^2(w))?$$

share|cite|improve this question
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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.