Let \(n\) be a positive integer, and let \(V\) be the space of polynomials over \(\RR\) which have degree at most \(n\) (throw in the 0-polynomial). Let \(D\) be the differentiation operator on \(V\). What is the minimal polynomial for \(D\)?
Solution
On \(V=\mathbb R[x]_{\le n}\), repeated differentiation gives \[ D^{n+1}=0, \] so the minimal polynomial divides \(x^{n+1}\). On the other hand, \[ D^n(x^n)=n!\ne0, \] so \(D^n\ne0\). Therefore no smaller power of \(x\) annihilates \(D\), and \[ m_D(x)=x^{n+1}. \]