Let $f \in \mathbb{F}[x]$ and $A_f=c(f) $ be the companion matrix of $f$. Given $ g \in \mathbb{F}[x] $ show that $ \dim \ker g(A_f) \leq \deg g $.

Well, if you try to calculate the power of the matrix $A_f$, you will get it at once. First, let's do it for $A_f^2$, its first n-2 columns are exactly the 2~ n-1 columns of $A_f$ : $$\begin{pmatrix} 0 & & & &*&*\\ 0 &0& & &*&*\\ 1 &0&0&& *&*\\ \vdots&\vdots& \vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&* &* \end{pmatrix}$$ it is obvious that the first $n-2$ columns are linearly independent, and we can ignore how the last two columns looks like exactly. Similarly, for $A_f^3$, its first $n-3$ columns are the 3~n-1 columns of $A_f$, and so on.

Then, the matrix $g(A_f)$ looks like $$\begin{pmatrix} b_0& & & &*&*\\ b_1&\ddots& & &*&*\\ b_2 &\ddots&b_0&& *&*\\ \vdots&\ddots&b_1&\ddots&\vdots&\vdots\\ b_m&\vdots &b_2 & \cdots& *&*\\ \vdots&\ddots &\vdots& &\vdots&\vdots\\ 0&\cdots&b_m&\cdots&* &* \end{pmatrix}$$ Since $b_m\neq 0$, we derive what we want.