TITLE: Does every linear operator have a minimal polynomial?
QUESTION [27 upvotes]: I know that a linear operator $T$ defined on a finite-dimensional vector space has a minimal polynomial since, by Caley-Hamilton, $g(T)=0$, where $g$ is the characteristic polynomial.  Is there a linear operator defined on an infinite-dimensional vector space that has no minimal polynomial?

REPLY [7 votes]: You can consider the linear space $X$ of all complex continuous functions on $[0,1]$ and let $Tf=\int_{0}^{x}f(t)dt$ be the integration operator, which is definitely linear on $X$. The operator $T$ has no minimal polynomial. To see why, first notice that $(T-\lambda I)$ has trivial kernel for $\lambda \ne 0$ because $Tf=\lambda f$ forces $g(x)=\int_{0}^{x}f(t)dt$ to satisfy
$$
                    g'(x)=\frac{1}{\lambda}g(x),\;\; g(0)=0,
$$
which has the unique solution $g(x)\equiv 0$.
Therefore, if $T$ has a minimal polynomial, it could only have the form $T^n=0$ for some $n$ which would force $g=T^nf$ to be identically $0$ for all $f$, and differentiating $n$ times would give $f=0$ for all $f$.