Apparently it's a theorem by Weyl.
Let $H$ be an infinite dimensional, separable Hilbert space and let $A,B \in B(H)$.
Let $K(H)$ be the set of compact operators acting on $H$.
Show that if $A-B \in K(H)$ and if $\lambda \in \sigma(A)-\sigma_p(A)$, or if $\lambda$ is an eigenvalue of infinite multiplicity, then $\lambda \in \sigma(B)$.
Here $\sigma_p$ is the point spectrum, and it consists of the eigenvalues.
Hint: assume wlog $\lambda=0$(I'm not sure why we can do that) and write $A=B+(A-B)$.
Some thoughts:
We have two cases:
- Suppose $A-B \in K(H)$ and $\lambda \in \sigma(A)-\sigma_p(A)$.
Then we have $\lambda \in \sigma(B+(A-B))-\sigma_p(B+(A-B))$.
So $\lambda \in \sigma(B+ K(H))$ and $\lambda$ is not an eigenvalue of $(B+ K(H))$.
Now I feel like I have to use the compactness of $A-B$. Not sure to continue.
- Suppose $A-B \in K(H)$ and $\lambda$ is an eigenvalue of infinite multiplicity. Then we have $\lambda \in \sigma_p(A-B)$.
Again.. I'm missing something about compactness.
Any help will be appreciated.
Thank you in advance!
Best Answer
An operator $T\in B(H)$ is said to be a Fredholm operator if the null space of $T$ is finite dimensional and the range of $T$ has finite co-dimension. In this case the index of $T$ is defined by $$ \text{ind}(T) = \text{dim}(\text{Ker}(T)) - \text{codim}(\text{Ran}(T)). $$
Among the standard results of the theory of Fredholm opertors one has:
Theorem. If $T$ is Fredholm and $K$ is compact, then $T+K$ is also Fredholm and $\text{ind}(T) = \text{ind}(T+K)$.
Letting $A'=A-\lambda $, and $B'=B-\lambda $, as suggested by @Berci, we need to prove that, if $A'-B'$ is compact, and $A'$ fails to be invertible either because:
its null space is infinite dimensional, or
$A'$ is injective, but not surjective,
then $B'$ is also not invertible.
In order to prove this, suppose by contradiction that $B'$ is invertible. Then $B'$ is clearly Fredholm, with $\text{ind}(B') = 0$, and hence $A'$ is also Fredholm with $\text{ind}(A') = 0$. However this is incompatible with either (1) or (2), so we have arrived at a contradiction, proving that $B'$ is not invertible, and hence that $\lambda \in \sigma (B)$.