# Thread: Annihilators and dual spaces

1. ## Annihilators and dual spaces

For these problems:
1. Define the annihilator of $S^0$ as
$S^0 = \{f \in V*: f(x) = 0 \forall x \in S\}$, where $S$ is a subset of finite-dimensional vector space $V$.

2. Let $V,W$ be finite-dimensional vector spaces with ordered bases $\beta, \gamma$ respectively. Then for any linear mapping $T: V \rightarrow W$, the mapping $T^t: W* \rightarrow V*$ is defined by $T^t(g)=gT$ for all $g \in W*$ with the property that $[T^t]_{\gamma *}^{\beta *} = ([T]_\beta^gamma)^t$.

Question 1: Suppose that $W$ is a finite-dimensional vector space and $T: V \rightarrow W$ is linear. Prove that $N(T^t) = (R(T))^0$.

Question 2: Let $T$ be a linear operator on $V$ and let $W$ be a subspace of $V$. Prove that $W$ is $T$-invariant if and only if $W^0$ is $T^t$-invariant.

2. Originally Posted by Last_Singularity
For these problems:
1. Define the annihilator of $S^0$ as
$S^0 = \{f \in V*: f(x) = 0 \forall x \in S\}$, where $S$ is a subset of finite-dimensional vector space $V$.

2. Let $V,W$ be finite-dimensional vector spaces with ordered bases $\beta, \gamma$ respectively. Then for any linear mapping $T: V \rightarrow W$, the mapping $T^t: W* \rightarrow V*$ is defined by $T^t(g)=gT$ for all $g \in W*$ with the property that $[T^t]_{\gamma *}^{\beta *} = ([T]_\beta^gamma)^t$.

Question 1: Suppose that $W$ is a finite-dimensional vector space and $T: V \rightarrow W$ is linear. Prove that $N(T^t) = (R(T))^0$.

Question 2: Let $T$ be a linear operator on $V$ and let $W$ be a subspace of $V$. Prove that $W$ is $T$-invariant if and only if $W^0$ is $T^t$-invariant.

All this doesn't make much sense: you're going to define the annihilator of S^o OR you're going to define the annihilator S^o nof some set S?
I'm almost sure it must be the latter. Now, what is V*? the v.s. of all linear functionals on V?

later on you talk of N(T^t)...what is this? The nullity or kernel of T^t?

Be more careful and if you want write again your question.

Tonio

3. I got it solved - thanks!