minimal polynomial of multiplication endomorphism in polynomial vector space

Question

Let $k[x]$ be the vector space of polynomials with only finitely many non-zero coefficients. Define the subspace $U_{f}=k[x]\cdot f$ where $f$ is a monic polynomial of degree n, say $f=a_{0}+\ldots+a_{n-1}x^{n-1}+x^{n}$. I succeeded to show that $\left(1+U_{f},\ldots,x^{n-1}+U_{f}\right)$ is a Basis of $k[x]/U_{f}$ and found out that multiplication by $x$ induces an endomorphism from $k[x]/U_{f}$ to $k[x]/U_{f}$. Furthermore, I found out that the corresponding matrix looks like this: $A=\begin{bmatrix} 0&\cdots&0&0&-a_{0}\\1&0&\cdots&0&-a_{1}\\ 0&1&\ddots&\vdots&-a_{2}\\ \vdots & \ddots& \ddots&0&\vdots \\ 0& \cdots&0&1&-a_{n-1}\end{bmatrix}$. The characteristic polynomial turns out to be just $\chi(\lambda)=(-1)^{n}f(\lambda)$. My task is to find the minimal polynomial. My intuition tells me that it hast to equal $f$, but somehow I cannot prove it. Can anyone help me out?

Answer 1

The matrix $A$ is what's known as a companion matrix of the polynomial $f(x)$, which has the property that the minimal polynomial is also $f(x)$ (read up about them; they have other interesting properties). Notice that \begin{equation} \{ e_1, A(e_1), \dots, A^{n-1}(e_1)\} = \{ e_1, e_2, \dots, e_n\} \end{equation} So the set on the left is in fact a basis for $\mathbb{K}^n$ ($\mathbb{K}$ being the field we're working over). Now, the result follows directly from my answer to this question.