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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 $p(t)=t^{n}$.

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  • $\begingroup$ 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 $T\equiv 0$ $\endgroup$ – Peter Melech Sep 19 '18 at 11:46
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The degree of the characteristic polynomial gives the dimension of the space the operator acts upon, so that dimension cannot be allowed to be infinite for the characteristic polynomial to be defined.

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.

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