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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.