Let E be a Banach space of infinite dimension and A a compact operator of E in itself.
I'm looking for this equivalence:
There exists a polynomial P such that P (A) = 0 $\iff$ there exists $m \in \mathbb{N ^ *}$ such that $A ^ m$ be a of finite rank operator.
My attempt to show the sens $\Longrightarrow $
If P exist such that P(A)=0 then necessarily P(0)=0 and we can write $0=P(A)=A^m .(I+a_1A+...a_pA^p)$ forsome integer m>0
Then $Ker (A^m)$ have a supplementary $Ker (I+a_1A+...a_pA^p)$. We know also that $Rank (A^m)$ is a supplementary of $Ker (A^m)$. Then $Rank (A^m)$ and $Ker (I+a_1A+...a_pA^p)$ are isomorphic.
By Riesz The dimension of $Ker (I+a_1A+...a_pA^p)$ is finite, then the dimension of $Rank (A^m)$ is finite
I need help to show the sens $\Longleftarrow$
If $A^m = B$ has finite rank, there is a finite-dimensional subspace $V$ such that $E = V + \text{Ker}(B)$. Let $W = \text{span}(V, BV)$, so $B$ maps the finite-dimensional space $W$ to itself. This corresponds to a matrix $\hat{B}$. If $C(x)$ is the characteristic polynomial of $\hat{B}$, we have $C(B) = 0$. Then we can take $P(x) = x C(x^m)$.