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Let $T$ be an $F$-linear operator on a finite dimensional $F$-vector space $V$ and let $W$ be a subspace of $V$. Define $ann(W)=\{f \in V^* : \forall w \in W, f(w)=0_F\}$. We say that $W$ is $T$-invariant if $T(W) \subseteq W$. First prove that $ann(W)$ is a subspace of $V^*$. Also prove that $W$ is $T$-invariant iff $ann(W)$ is $T^t$-invariant.

My proof so far: For part a) Let $f_1,f_2 \in ann(W)$. Then $f_1+f_2=f_1(w)+f_2(w)=0+0=0 \in ann(W)$. Therefore $0 \in ann(W)$ and the sum is also in $ann(W)$. Let $\alpha \in F$. So $\alpha f=\alpha f(w) = \alpha \cdot 0 = 0 \in ann(W)$ Thus $ann(W)$ is a subspace of $V^*$.

For the second part, I'm a little more unclear: Assume that $W$ is $T$-invariant. So $T(W) \subseteq W$. From a previous result, $T^t(f)=f \cdot T(w)$.
Conversely suppose that $ann(W)$ is $T^t$-invariant. So $T^t(ann(W)) \subseteq ann(W)$. Then $f \cdot T(W) \subseteq ann(W)$. Unsure how to conclude the last part, which states that $T(W) \subseteq W$

I'm unsure on where to go from here, and this is happening for both ways. Any help is appreciated. Thanks in advance

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For part (a), you are on the right track, but there are some incorrect details in your proof. For example, you can't say that $f_1+f_2$ and $f_1(w)+f_2(w)$ are equal; one's a function and the other is a vector! – Aleksandar Bahat Oct 17 '12 at 19:25
Ok so I should say that $(f_1+f_2)(w)=f_1(w)+f_2(w)$ – tk2 Oct 17 '12 at 19:32
up vote 1 down vote accepted

$\newcommand{\ann}{\operatorname{ann}}$Your argument for (a) is correct apart from a consistent notational error: the two calculations should begin $(f_1+f_2)(w)=f_1(w)+f_2(w)=\dots$ and $(\alpha f)(w)=\alpha f(w)=\dots$, respectively. Your $f_1+f_2=f_1(w)+f_2(w)$ and $\alpha f=\alpha f(w)$ make no sense: in each case the object on the lefthand side is an element of $V^*$, while the object on the righthand side is an element of $V$, so they aren’t even the same kind of object, let alone equal to each other.

You’ve made a reasonable start on (b) by assuming that $W$ is $T$-invariant and rewriting this as $T[W]\subseteq W$. Now you want to show that $\ann(W)$ is $T^t$-invariant, i.e., that $T^t[\ann(W)]\subseteq\ann(W)$. The most natural thing to try here is to let $f$ be an arbitrary member of $\ann(W)$ and try to show that $T^t(f)\in\ann(W)$; this of course means showing that $T^t(f)(w)=0_F$ for each $w\in W$, so suppose that $w\in W$. Then $T^t(f)\cdot w=f\cdot T(w)$. $W$ is $T$-invariant, so $T(w)\in W$, and $f\in\ann(W)$, so $f\cdot u=0_F$ for every $u\in W$, so $T^t(f)\cdot w=f\cdot T(w)=0_F$, just as we wanted. It follows that $T^t(f)\in\ann(W)$ and hence that $\ann(W)$ is $T^t$-invariant.

You’ve begun the converse direction reasonably as well. Suppose that $\ann(W)$ is $T^t$-invariant, so that $T^t[\ann(W)]\subseteq\ann(W)$ and hence $T^t(f)\in\ann(W)$ for each $f\in\ann(W)$. We want to show that $W$ is $T$-invariant, i.e., that $T(w)\in W$ for each $w\in W$, so let $w\in W$. Then for each $f\in\ann(W)$ we have $f\cdot T(w)=T^t(f)\cdot w=0_F$, since $T^t(f)\in\ann(W)$. In other words, every $f\in\ann(W)$ annihilates $T(w)$. To complete the proof, you must show that if $v\in V$, and $f\cdot v=0_F$ for every $f\in\ann(W)$, then $v\in W$. (Equivalently, the vectors in $W$ are the only ones annihilated by every $f\in\ann(W)$.) In other words, you need to prove this

Lemma. If $v\in V\setminus W$, there is an $f\in\ann(W)$ such that $f\cdot v\ne 0_F$.

Here’s an extended hint: start with a base $B$ for $W$ and extend it to a base $B\,'$ for $V$ such that $v\in B\,'$. Find an $f\in V^*$ such that $f\cdot b=0_F$ for each $b\in B$ and $f\cdot b\ne 0_F$ for each $b\in B\,'\setminus B$.

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