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.

  • $\begingroup$ The eigenvectors do not change: If the roots of $f(x)$ are distinct, then they are $[1,\lambda_i,\cdots,\lambda_i^{n-1}]^t$. Notice that if $a_i\in F$, then $\lambda_i\in \overline F\subseteq \overline{F(\alpha)}$ $\endgroup$ – Exodd Mar 23 '18 at 9:19
  • $\begingroup$ I want to see the eigenvectors in $F(\alpha)$; I was unable to write them in terms of powers of $\alpha$ and $\lambda_i$'s $\endgroup$ – Beginner Mar 23 '18 at 9:20
  • $\begingroup$ $\lambda_i$ are not generally contained in $F(\alpha)$ if $F$ is not algebrically closed. This means that the eigenvectors may not be writable in $F(\alpha)$ and $L_\alpha$ is not diagonalizable $\endgroup$ – Exodd Mar 23 '18 at 9:22
  • $\begingroup$ Btw, if $F(\alpha)$ is a Normal extension of $F$, then it is true that $\lambda_i$ are contained in $F(\alpha)$. For a counterexample, take $F=\mathbb Q$, $\alpha=\sqrt[3]{2}$, $f(x) = x^3-2$. $\endgroup$ – Exodd Mar 23 '18 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.