Compact operators satisfying a certain relation must be finite-rank Let $H$ be an infinite-dimensional Hilbert space, equipped with a given Hilbert basis $(e_i)_{i \in \mathbb{N}}$. 
Consider the following introductory problem : can we find a compact operator $A$ in $H$ that satisfies the relation
$$ \sum \limits_{k=0}^n c_kA^k=0$$
for some given $(c_k)_k$, $c_k \in \mathbb{R}^{n+1}$ for all $k=0,...,n$ ?
Two cases arise :


*

*$c_0 \neq 0$ : Suppose that $A$ is a compact operator satisfying the given relation. We can rewrite it as 
$$c_nA^n+...+c_1A=-c_0Id$$
Factoring by $A$ and using the fact that $c_0 \neq 0$, we get that


$$ \underbrace{A}_\text{compact} \circ(\underbrace{\frac{-c_n}{c_0}A^{n-1}+...+\frac{c_1}{c_0}Id}_\text{bounded})=Id$$
Set $B=\frac{-c_n}{c_0}A^{n-1}+...+\frac{c_1}{c_0}Id$. The composition $A \circ B$ will compact, because $A$ is compact and $B$ is a bounded operator. Therefore the $Id$ operator is compact, which is absurd because $H$ is infinite-dimensional. Therefore such an A cannot exist.


*

*$c_0=0$ : In this case we can find a compact operator with relative ease. Consider 
$$ V= vect\left\{e_n : n \in \left\{0,...,N\right\}\right\}$$ where the $(e_n)$ are elements of the basis of $H$. It is finite-dimensional and thus closed, so let $A$ be the well-defined projection operator on $V$. We then have that 
$$ A \circ A = A \iff A^2-A=0$$
and can thus create a relation of the given form.
Since $dim(V) < +\infty$, $A$ is finite-rank, and thus compact. 


Now consider again the case where $c_0=0$. My question is the following :
If $A$ is a compact operator satisfying this type of relation for given $(c_k)_k$, then must $A$ be finite-rank ?
I suspect that the answer is yes (maybe an analogy can be made to the case of matrices that have $0$ as an eigenvalue), but don't really have a good idea of where to start. Decomposing a finite-rank operator in $(e_i)$ might shed some light on this but appart from that I am not sure of how to proceed.  

Denote $i$ the smallest index such that $c_i\ne0$. Then we can factorize the polynomial
$$
\sum_{k=i}^n a_k t^k = c_i t^i \prod_{k=i+1}^n (t-\lambda_k)
$$ 
This implies
$$
c_i \left( \prod_{k=i+1}^n (A-\lambda_k I) \right) A^i=0.
$$
Since $A$ is compact, the null spaces of all operators $A-\lambda_k I$ 
are finite-dimensional, hence the range of $A^i$ is finite-dimensional.
In case $i=1$, this shows that $A$ is finite-rank. If $i>1$ then $A^i$ is finite-rank, which does not imply that $A$ is finite-rank:
Define $A:l^2\to l^2$ by
$$
(Ax)_n =\begin{cases} 0 & \text{if $n$ odd}\\
2^{-n}x_{n-1} & \text{if $n$ even}\\
\end{cases}.
$$
So $A$ is a compact multiplication operator times right shift, hence compact. By construction, $A^2=0$ but $A$ has no finite-rank.
