Minimal Polynomial of null operator

I think that is a simple question, but there is some details confusing me. I need to calculate the minimal polynomial and the characteristic polynomial of the null operator in a $\mathbb{F}$-space. My problem is that I cannot assume $\dim(V)<\infty.$

The minimal polynomial is $m(t)=t$. Can I talk about characteristic polynomial in a vector space of infinite dimension? Cause the charactheristic polymial of the null operator in a space $V$ with $n=\dim(V)$ is $p(t)=t^{n}$.

• I think You have to assume $\dim(V)<\infty$ to define a characteristic polynomial in the sense of a determinant of a characteristic matrix (c.f. mathoverflow.net/questions/126464/…) However if for a linear operator $T:V\rightarrow V$ You want a (normed) polynomial $p$ of minimal (positive) degree so that $p(T)=0$, then clearly $p(t)=t$ is that polynomial for the zero operator $T\equiv 0$ – Peter Melech Sep 19 '18 at 11:46

As for the minimal polynomial, it does not have to exist when the dimension is infinite (what would be the minimal polynomial of multiplication by $X$ in the space of polynomials?), but it does exist for the zero operator. By definition, $X$ always annihilates a zero operator, and its unique monic strict divisor $1$ does not, unless the space has dimension $0$. So the minimal polynomial of the zero operator on$~V$ is $1$ if $\dim(V)=0$, and $X$ otherwise.