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

Question

Given the differential equation $$x^{(n)}(t)+c_{n-1}x^{(n-1)}(t) + \dotsb + c_1x'(t) + c_0=0,$$ we can form a vector $\xi = (x, x', \dotsc, x^{(n-1)})$, and then we have $$\xi'(t) = A\xi,$$ where $A$ is the transpose of the companion matrix for the polynomial $$z^n + c_{n-1}z^{n-1}+\dotsb + c_1z + c_0.$$ A problem in Teschl's ODE book is to show that each eigenvalue of $A$ has geometric multiplicity $1.$ The hint it gives is "can you find a cyclic vector for $A$? How does that help?"

(By a cyclic vector he means $v$ such that $\{A^{k}v\}\,\, (0\leq k <n)$ spans the vector space.)

I see that $e_n$ is a cyclic vector for $A$, but I'm not seeing how that helps. Any ideas?

Answer 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. $\quad \Box$

Answer 2

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 $|\mathcal{N}(A - \lambda I)|\leq 1$.

Answer 3

What I had in mind is the following intuitive picture: If the geometric multiplicity of an eigenvalue $\lambda$ is larger than one, there are two Jordan blocks corresponding to this eigenvalue. Now the components of the cyclic vector corresponding to these blocks are affected in the same way by powers of $A$. Consequently you can only scale them but not change their relative direction.

Explicitly, take the generalized eigenvectors $u_1$ and $u_2$ which generate these Jordan blocks and let $u_3,\dots,u_n$ be the remaining generalized eigenvectors such that we have a basis. Let $\nu = \sum_{k=1}^m \nu_k u_k$ be your cyclic vector and note $A^k \nu = \lambda^k(\nu_1 u_1 + \nu_2 u_2) + \dots$ (where the omitted terms are in the span of $\{u_k\}_{k\geq 3}$). In particular, $\nu_1\ne 0$ as otherwise we cannot generate $u_1$ (and similarly $\nu_2\ne 0$). But then $\nu_2 u_1 - \nu_1 u_2$ is nontrivial and cannot be generated.