Let $V$ be a $n$ dimensional vector space, $V^*$ is its dual, $\mathscr{A}$ is a linear transform from $V$ to $V$. Given $g\in V^*$, define $$\mathscr{B}(g)=g\circ\mathscr{A}$$ Then it is easy to prove that $\mathscr{B}$ is linear trasnform from $V^*$ to $V^*$.
Let $f\in V^*$. Suppose that any nonzero invariant subspace of $\mathscr{A}$ is not a subspace of $\ker (f)$. Show that $$f,\mathscr{B}f,\cdots,\mathscr{B}^{n-1}f$$ is a base of $V^*$.
My attempt: Let $$k_0f+k_1\mathscr{B}f+\cdots+k_{n-1}\mathscr{B}^{n-1}f=0$$ then $$k_0f(v)+k_1\mathscr{B}f(v)+\cdots+k_{n-1}\mathscr{B}^{n-1}f(v)=0, \,\forall\ v\in V$$ that is, $$k_0f(v)+k_1 f(\mathscr{A}v)+\cdots+k_{n-1}f(\mathscr{A}^{n-1}v)=0, \,\forall\ v\in V$$ or equivalently, $$k_0v+k_1\mathscr{A}v+\cdots+k_{n-1}\mathscr{A}^{n-1}v\in \ker f, \,\forall\ v\in V$$ How to proceed then?