Show that $\ker \hat{T} = \text{ann}(\text{range } T)$ This is an old exam problem: Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $T: V \to W$ be a linear transformation. Define $\hat{T}: W^* \to V^*$ by $(\hat{T}(f))(v)=f(T(v))$. Here, $U^*$ denotes the dual vector space $U$. Since $\hat{T}$ is a linear transformation prove $\ker \hat{T} = \text{ann}(\text{range } T)$ where $\text{ann}(Y) = \{g \in U^* : g(y)=0 \text{ for all } y \in Y \}$ for a subspace $Y$ of vector space $U$.
I'm really not sure how to start this problem. Any advice would be great!
 A: Let $A = \ker \hat T$ and $B = ann(range(T))$.  We want to show that $A \subset B$ and $B \subset A$.
Consider an arbitrary $f \in A$.  By definition, $f \circ T = 0$.  This means that for any $x \in V$, $f(T(x)) = 0$.  Note that for any $y \in range(T)$, there is an $x$ such that $y = T(x)$.  Thus, given such a $y$ and an $x$ with $y = T(x)$, we have
$$
f(y) = f(T(x)) = 0
$$
so that $f \in B$.
Conversely, consider an arbitrary $f \in B$.  For any $x$, we note that $T(x)$ is in the range of $T$, from which it follows that $f (T(x)) = 0$.  However, since $f(T(x)) = 0$ for every $x$, we may conclude that $f \circ T = 0$, so that $f \in A$.
A: Hint: If you write $T$ and $\hat{T}$ as matrices, then $T$ and $\hat{T}$ are transposes.  Now, an element of the kernel of $\hat{T}$ is the same as something that annihilates the columns of $T$ (just take the transpose of the matrix multiplication that equals the zero vector), but the columns of $T$ span the image of $T$.
A: $\widehat{T\mkern-0.5mu}\mkern0.5mu\colon W^*\rightarrow V^*\,$ is the transpose of the linear map $T$, defined by $f\mapsto f\circ T$.
Now $\,f\in\ker\widehat{T\mkern-0.5mu}\mkern0.5mu\iff f\circ T=0$. Observe this simply means the restriction of $f$ to $\operatorname{Im}T$ is $0$, in other words $f\in\operatorname{Ann}_{W^*}(\operatorname{Im}T)$.
