direct sum of image and kernel in a infinitedimensional space Is it true that in an infinitdimensional Hilbert space the formula $$\text{im} S\oplus \ker S =H$$holds, where $S:H\rightarrow H$ ?
I know it is true for finitely many dimensions but I'm not so sure about infinitely many. Would it be true under some additional assumption, like assuming that the rank of $S$ is finite ?
 A: Assuming you intended to ask what I proposed in a comment, the answer is yes, this also holds for infinite-dimensional vector spaces, assuming the axiom of choice. Take a basis of $\ker S$ and extend it to all of $S$. The additional basis vectors induce a basis of $S/\ker S$, which by the first isomorphism theorem is isomorphic to $\operatorname{im} S$. This established a linear bijection between $S$ and $\operatorname{im} S\oplus\ker S$. Note that there is no canonical choice for the basis of $S/\ker S$, and you need the axiom of choice to get one.
A: Consider $l_2$, and the translation operator $T:~e_n\mapsto e_{n+1}$, it's injective but not surjective. So that $ker~T=0,im~T\neq l_2,ker~T\oplus im~T\neq l_2$.
If $rank~im T$ is finite, i remember i have learned that the equality holds in some book. (but it's vague to me now, so take care)
You can try to consider the coordinate function on the finite dimensional space $im~T$ in this case.
Other kinds of additional assumptions may concern with 


*

*Projection operator

*Compact operator

*Fredholm operator
Try your best
