TITLE: dual map and transpose of matrix
QUESTION [2 upvotes]: Let $V, W$ be vector spaces over a field $F$.
And let $f:V\to W$ and $h:W\to F$ so that $h\in W^*$ and $h\circ f\in V^*$.
Now we have a dual map $f^*:W^*\to V^*$ s.t. $f^*(h)=h\circ f$.
Then, $f^*(h)(v)=h(f(v))$ for all $v\in V$ and $h\in W^*$.
But for example, set $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, B=\begin{bmatrix}e&f\\g&h\end{bmatrix},v=\begin{bmatrix}i\\j\end{bmatrix}$ so that $A$ represents $f$ and $B$ represents $h$, we have $A^TBv\not=BAv$.
What am I missing here?

REPLY [1 votes]: All the elements in the dual space $W^*$ are functionals, this means that it's a function from the vector space to the field: $f: W \to F$.
You said "let $B$ represent $h$". We have that $h \in W^*$, so $g(v)= Bv$ needs to be an element of the field. But $B \cdot v$ is a $2 \times 1$ matrix, and the set of all $2 \times 1$ matrices is not a field because the "multiplication property" is not defined. In fact, the function $g(v)= B\cdot v$ is actually a linear transformation from $W$ to $W$ .
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