removing finite rows for a infinite dimension compact operator Consider a compact operator $T$ of a separable, infinite dimensional Banach space $X$(or Hilbert space). Expand $T$ into matrix form, if I remove finite number of rows inside, is the resultant operator compact again?
The idea is since the removal of rows only affect finite number of terms in the image, it should not affect the convergence of any subsequence, but I am not sure.
 A: From my understanding I am not sure if what you're asking makes sense. If $T$ is an infinite dimensional operator and you remove a finite number of rows from its matrix representation (let us call this $\tilde{T}$) then $\tilde{T}$ will also be infinite dimensional but I don't think the two operators will share the same domain. For example if $(e_{n})_{n\geq 1}$ is a basis of $X$ then $T$ is uniquely determined by $(Te_{n})_{n\geq 1}$. However if we remove the rows $n=N$, $N+1$, $N+2$, $N\in\mathbb{N}$, to generate $\tilde{T}$ then $\tilde{T}e_{N}$, $\tilde{T}e_{N+1}$ and $\tilde{T}e_{N+2}$ are not well defined so how can we associate how $\tilde{T}$ acts on $X$ to how $T$ acts on $X$?
Now, if by "removal of finitely many rows" you mean to set them to all zeros then I believe the answer is yes, if we define $\tilde{T}$ as $T$ but with the rows $N$, $N+1$ and $N+2$ set to all zeros. So we can think of $T$ as $\tilde{T}+K$ where $K$ is the operator which is composed of the rows removed from $T$ and zeroes elsewhere. Since $K$ is of finite rank it is compact and so $\tilde{T}=T-K$ is compact as the space of compact operators is a vector space.
EDIT: To add to the first paragraph of my response, now that I have thought a little more about this. I believe removing a finite number of rows in the literal sense is actually possible, as there would exist an isomorphism on the images in both cases.
So to elaborate, let us define $\tilde{T}$ by removing the rows $N$, $N+1$ and $N+2$, $N\in\mathbb{N}$, from $T$ and define $T'$ by setting the rows $N$, $N+1$ and $N+2$ of $T$ to all zeros. Then, as per above, $T'$ is compact. So take $B_{X}$ and consider $x\in T'(B_{X})$ then $x$ must have zeros in the $N$, $N+1$ and $N+2$ position. Now consider the space $\tilde{X}$ defined as all $x\in X$ with their $e_{N}$, $e_{N+1}$ and $e_{N+2}$ positions removed.  Take $\tilde{B}_{\tilde{X}}$ analogous to $B_{X}$ using the same construction as in $\tilde{X}$ then there exists an isomorphism between $\tilde{T}(\tilde{B}_{\tilde{X}})$ and $T'(B_{X})$. So it seems that even $\tilde{T}$ is compact as well.
