In Matrix form the determination of eigenvectors of a companion matrix has been appeared in different questions on this site; my question is a little bit different, I was unable to do an interpretation.
Consider the matrix $$ C_p=\begin{bmatrix} 0 & 1 & 0 &\cdots & 0\\ 0 & 0 & 1 &\cdots & 0 \\ \vdots&\vdots &\vdots&\ddots&\vdots\\ 0 & 0 & 0 &\cdots &1 \\ -\alpha_0 &-\alpha_1 &-\alpha_2 &\cdots&-\alpha_{n-1} \end{bmatrix} $$ Suppose that the characteristic polynomial $\chi(x)=x^n+a_{n-1}x^{n-1} +\cdots + a_0$ of this matrix has $n$ distinct roots $\lambda_1,\cdots,\lambda_n$. Then the column vector $[1,\lambda_i,\cdots,\lambda_i^{n-1}]^t$ is an eigenvector of this matrix with eigenvalue $\lambda_i$.
(1) Suppose we have a Galois extension $F(\alpha)$ of $F$ of degree $n$ where minimal polynomial of $\alpha$ is $\chi(x)$. Then multiplication by $\alpha$ is an $F$-linear mapping $L_{\alpha}$ on $F(\alpha)$, whose matrix w.r.t. basis $\{1,\alpha,\cdots,\alpha^{n-1}\}$ is the above companion matrix.
Q. 1 How can we visualize the corresponding eigenvectors of $L_{\alpha}$ in $F(\alpha)$ described above?
(2) Let $V$ be the vector space over $\mathbb{C}$ of polynomials of degree $<n$, and $\{1,x,\cdots,x^{n-1}\}$ a basis. Then $$T(1)=x, \,\, T(x)=x^2,\cdots,\,\, T(x^{n-1})=-a_{n-1}x^{n-1}-a_{n-2}x^{n-2}-\cdots-a_0$$ is a linear map on $V$ whose matrix w.r.t above basis is $C_p$.
Q.2 If $\chi(x)$ has $n$ distinct eigenvalues, what are polynomials which are the eigenvectors of $T$ (analogous to $[1,\lambda_i,\cdots,\lambda_i^{n-1}]$)?
Both questions are essentially same; I want to see the eigenvectors of the companion matrix as elements of some field extension $F(\alpha)$ or as polynomials.