How do I prove that the eigenspace of an $n\times n$ companion matrix $$ C_p=\begin{bmatrix} 0 & 1 & 0 &\cdots & 0\\ 0 & 0 & 1 &\cdots & 0 \\ \vdots&\vdots &\vdots&\ddots&\vdots\\ 0 & 0 & 0 &\cdots &1 \\ -\alpha_0 &-\alpha_1 &-\alpha_2 &\cdots&-\alpha_{n-1} \end{bmatrix} $$ equals $\operatorname{Span}\{v_{\lambda} \} $ where $v_{\lambda}$ is an eigenvector of the companion matrix w.r.t. the eigenvalue $\lambda$: $$ v_{\lambda} = \begin{bmatrix} 1 \\ \lambda\\ \lambda^{2} \\ \vdots\\ \lambda^{n-1} \end{bmatrix}. $$

up vote 5 down vote accepted

If you write $C_p{\bf x}=\lambda {\bf x}$, where ${\bf x}=(x_1,\dots,x_n)^T$, you get that $x_2=\lambda x_1$, $x_3=\lambda x_2$, $\dots$ , $x_n=\lambda x_{n-1}$ which means that $x_2=\lambda x_1$, $x_3=\lambda^2 x_1$, $\dots$ , $x_n=\lambda^{n-1}x_1$ and using your notation ${\bf x}=x_1v_{\lambda}$.

  • Aah! Thanks! :) – L1meta Oct 29 '12 at 18:32

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