In some proofs about compactness of operators and Fredholm theory, I've seen they are using some fact about the dimension of the image of a composition I am not familiar with. My professor said that $AK_1$ has finite rank ,where $K_1$ is a compact operator of finite rank between Hilbert spaces and $A$ another operator between Hilbert spaces of finite or infinte dimension. Then:
$dimIm(AK_1)\leq dimIm(K_1)<\infty$
Why is this true? Is this a general fact ?
We were proving that $H$ is a Hilbert space and $T\in B(H)$,$T=I+K$ a compact perturbation of the identity, then $dim KerT= dimKerT^*$
We analyse the case of K with finite rank and then that of K compact.
In the second case the proof starts like this:
Since $K$ is compact, $\exists K_1$ of finite rank such that $\|K-K_1\|<1$. Then we write $T=I+K=I+(K-K_1)+K_1$. Since $I+(K-K_1)=A$ is invertible by Neumann's criterion, we can write $T=A(I+A^{-1}K_1)$ where $A^{-1}K_1$ has finite rank because $K_1$ does….
This is where they are using the fact I descrived first to justify that $A^{-1}K_1$ has finite rank, Can someone explain why?
Best Answer
Let $M$ be te image of $K_1$. If $\{x_1,x_2,..,x_n\}$ is a basis for $M$ then $A^{-1}x_1,A^{-1}x_2,..,A^{-1}x_n$ spans the image of $A^{-1}K_1$ so the image of $A^{-1}K_1$ is finite dimensional. Its dimension is at most $n$.