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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?

Giuliano Cantina
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  • Polynomials by definition have only finitely many nonzero coefficients. – Marc van Leeuwen Jun 06 '19 at 12:48
  • N.B. actually, the characteristic polynomial is just $f$, without the alternating sign. The reason they should be defined this way is related to the fact that people factor (split) polynomials using factors $(X-\lambda)$ (for roots $\lambda$), not using factors $(\lambda-X)$. The only valid reason one could advance to prefer $A-XI_n$ rather than (the correct) $XI_n-A$ for defining characteristic polynomials is laziness: flipping the signs of all coefficients of $A$ might be "more work" then flipping the signs of the $X$'s. – Marc van Leeuwen Jun 06 '19 at 13:01

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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.

peek-a-boo
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