Show each eigenvalue of a companion matrix has geometric multiplicity $=1$.

I'm not sure this is the answer Teschl has in mind, but it is clear that $A - \lambda I$ has at least $n-1$ linearly independent columns for any $\lambda \in \mathbb{C},$ so $|\mathcal{N}(A - \lambda I)|\leq 1$.

Claim. If $A$ has a cyclic vector $v$, then $\dim \ker (A-\lambda I)\le 1$ for every $\lambda$.

Proof. Let $m$ be the smallest integer such that the span of $v,Av,\dots, A^m v$ intersects $\ker (A-\lambda I)$. Then we have a vector $\sum_{k=0}^{m} c_k A^k v$ such that $c_m\ne 0$ and $$A\sum_{k=0}^{m} c_k A^k v = \lambda \sum_{k=0}^{m} c_k A^k v $$ It follows that $A^{m+1}v$ is a linear combination of $v,\dots, A^m v$. Since $v$ is cyclic, $m$ must be $n-1$. Thus, the span of $v,\dots, A^{n-2}v$ is disjoint from $\ker (A-\lambda I)$, which by dimension count makes the latter at most one-dimensional. $\quad \Box$