In a paper by Halperin and Rosenthal a simple proof for Burnside's theorem on algebras of matrices is given which can be found here.
I understand the majority of the proof but one step is unclear to me:
How does
so there is no nonzero vector $x$ that is annihilated by all such $\phi$
imply
Thus every $T$ of the form $Tx = \phi(x)y_0$ for some $\phi$ is in $\mathscr{A}$?
One way to understand the fact would be to take the elements of $\mathscr{V}^*$ to $\mathscr{V}$ via an isomorphy (because $\mathscr{V}$ is in this case $K^m$ which is self-dual) and use the transitivity of $\mathscr{A}$. But then, why would the author explicitly give the statement about nonzero vectors?
The set of $\phi$ such that $x \mapsto \phi(x)y_0 \in \mathscr{A}$ forms a subspace of $\mathscr{V}^*$, since $\mathscr{A}$ is closed under addition and scalar multiplication, and this set was previously shown to be non-empty (or you could just take $\phi = 0$). If this were a proper subspace, then it would be contained in some hyperplane in $\mathscr{V}^{**}$ (finite dimensions implies that this subspace is closed). That is, every $\phi$ would be annihilated by some element of the second dual.
Since $\mathscr{V}$ is finite-dimensional, it is reflexive, so there would have to be some $x \in \mathscr{V}$ such that $\hat{x}(\phi) = 0$, i.e. $\phi(x) = 0$, for all $\phi$ in this subspace. But, as shown, this is not the case. So, this subspace must not be proper, and hence $x \mapsto \phi(x)y_0$ lies in $\mathscr{A}$ for all $\phi \in \mathscr{V}^*$.