I read this question which I have no idea how to start. Could anyone provide me with some detailed answer, please? Thanks.
Suppose that a linear operator $F$ from a Hilbert space $\mathcal H$ to itself has finite rank, i.e., $\dim F(\mathcal H) < \infty$. Show that there are $n$ vectors $f_1, \dots, f_n$ such that their span $S=\mathrm{span}\{f_1, \dots, f_n\}$ satifies $$\mathcal H = \ker F \oplus S.$$ And hence show that for $k \in \mathcal H$, $F$ has the form $$F(k)=\sum_{i=1}^n g_i\langle f_i, k \rangle.$$
First step: let $S$ be the image of the operator $F^*$. One can show that the rank of $$\operatorname{rank} F^*=\operatorname{rank} F.$$ See, for example, this question.
Second step: prove that $\ker F=S^\bot$.
Indeed, let $x\in \ker F$, $s\in\operatorname{im } F^*$, then $$(x,s) = (x,F^*y) = (Fx,y)=0,$$ therefore $\ker F\subset S^\bot$.
On the other hand, if $z\in S^\bot$, then $$\forall s\in S\quad (z,s)=0,$$hence $$\forall y\quad 0=(z,F^*y) = (Fz,y),$$hence $Fz=0$. This implies that $S^\bot \subset \ker F$, which allows us to say that $S^\bot = \ker F$.
Third step: now we need to prove that $H=S\oplus S^\bot$: it is sufficient to show that $S$ is closed (and this is true because it has finite dimension).