Transitivity of dual operators Near the end of this proof of Burnside's theorem: https://core.ac.uk/download/pdf/82680953.pdf, it says that it should be clear that since $\mathcal{A}$ acts transitively on the vector space, the operators dual to things in $\mathcal{A}$ should act transitively on the dual space.
In other words, given any two non-zero functionals $f,f'$ on $V$, there should be some $A \in \mathcal{A}$ such that $f(Av)=f'(v)$ for all $v$. And all we know is that for any non-zero $w,w'$ there is an $A\in \mathcal{A}$ such that $Aw=w'$.
I can't think of any easy explanation for this (that doesn't rely on already having proved the theorem). Can anyone help?
 A: Hint: Because we’re working over finite dimensional spaces (so that $V$ is isomorphic to its double dual), it suffices to prove the inverse statement, which is that if $\mathcal A$ does not act transitively over $V$, then it also fails to act transitively over the dual space.
To that end, let $U$ be a non-trivial, proper, $\mathcal A$-invariant subspace. Take $f$ to be a functional whose kernel contains $U$, and take $f’$ to be a functional whose kernel does not contain $U$.
A: Fix a basis $v_1,\ldots,v_n$ of $V$ and let $f_1,\ldots,f_n$ be its dual basis. If $f=\sum_ja_jf_j$ and $g=\sum_jb_jf_j$, define $x=\sum_ja_jv_j$, $y=\sum_jb_jv_j$. By hypothesis there exists $A$ with $y=Ax$. That is, $\sum_ja_jAv_j=\sum_jb_jv_j$. Thus, representing $A$ as a matrix over the basis $\{v_1,\ldots,v_n\}$,
$$
\sum_jb_jv_j=\sum_ja_j\sum_kA_{kj}v_k=\sum_k(\sum_ja_jA_{kj})v_k.
$$
Comparing coefficients we obtain 
$$\tag1
b_k=\sum_ja_jA_{kj}.
$$
and then
\begin{align}
f(Av_k)&=\sum_ja_jf_j(Av_k)=\sum_ja_jf_j(\sum_hA_{kh}v_h)\\[0.3cm]&=\sum_j\sum_ha_jA_{kh}f_j(v_h)=\sum_ja_jA_{kj}=b_k=g(v_k).
\end{align}
As we can do this for all elements in the basis,
$$f(Av)=g(v)$$for all $v$.
