Extensions of finite-rank operators Let $V$ be a vector space and let $W$ be its subspace of infinite codimension. Let $\mathcal{F}_W$ be the family of all finite-rank operators on $V$ with range contained in $W$. Consider the left-sided ideal in $\mathcal{L}(V)$ generated by $\mathcal{F}_W$. Does this ideal contain all the finite-rank operators on $V$?
 A: If you consider infinite dimensional vector space $V$ and its zero subspace $W=\{0\}\subset V$ you would get the desired counterexample.
From now we consider more interesting case $\operatorname{dim}W=+\infty$. Let $E=\{e_\lambda: \lambda\in\Lambda\}$ be some Hamel basis of $V$. Then for each $v\in V$ there exist a finite sum representation
$$
v=\sum\limits_{\lambda\in\Lambda} v_\lambda e_\lambda
$$
Now we endow $V$ with the structure of inner product space. By definition we take
$$
\langle v_1,v_2\rangle=\sum\limits_{\lambda\in\Lambda} v_{1,\lambda}\overline{v_{2,\lambda}}
$$
For each $v_1,v_2\in V$ by $v_1\bigcirc v_2$ we define rank-one operator on $V$ by equality
$$
v_1\bigcirc v_2 : V\to V: v\mapsto\langle v, v_2\rangle v_1
$$
It is easy to check that for all $a\in \mathcal{L}(V)$, $v_1,v_2\in V$ we have 
$$
a\circ(v_1\bigcirc v_2)=a(v_1)\bigcirc v_2,\qquad (v_1\bigcirc v_2)\circ a=v_1\bigcirc a^*(v_2)
$$
Moreover,  for all $s_1,s_2\in\mathbb{C}$ and $v_1,v_2,v\in V$ we have
$$
(s_1 v_1+s_2 v_2)\bigcirc v = s_1(v_1\bigcirc v)+s_2(v_2\bigcirc v) 
$$
$$
v\bigcirc(s_1 v_1+s_2 v_2)=\overline{s_1}  (v\bigcirc v_1)+\overline{s_2}(v\bigcirc v_2)
$$
Take $a\in\mathcal{F}(V)=\mathcal{F}(V,V)$, then there exist 
$\{v_{1,k}:k=\overline{1,n}\}\subset V$, $\{v_{2,k}:k=\overline{1,n}\}\subset V$ and 
$\{s_k:k=\overline{1,n}\}\subset\mathbb{C}$ such that
$$
a=\sum\limits_{k=1}^n s_k (v_{1,k}\bigcirc v_{2,k})
$$
From the linearity of $\bigcirc$ operation in the first argument it follows that we can assume that vectors $\{v_{1,k}:k=\overline{1,n}\}\subset V$ are linearly independent. 
Since $\operatorname{dim}W=+\infty$ there exist a linearly independent family $\{w_k:k=\overline{1,n}\}\subset W$. We define a linear operator $b\in\mathcal{F}(W,V)$ by equalities $b(w_k)=v_k$ for $k=\overline{1,n}$ and $b(v)=0$ for $v\in\operatorname{span}\{w_k:k=\overline{1,n}\}^{\perp}$. Then there exist $c=\sum\limits_{k=1}^n s_k (w_{k}\bigcirc v_{2,k})\in\mathcal{F}(V,W)$ such that
$$
b\circ c=
b\circ \left(\sum\limits_{k=1}^n s_k (w_{k}\bigcirc v_{2,k})\right)=
\sum\limits_{k=1}^n s_k b\circ (w_{k}\bigcirc v_{2,k})=
$$
$$
\sum\limits_{k=1}^n s_k (b(w_{k})\bigcirc v_{2,k})=
\sum\limits_{k=1}^n s_k (v_{1,k}\bigcirc v_{2,k})=a
$$
Thus for all $a\in\mathcal{F}(V)$ there exist $b\in\mathcal{F}(W,V)\subset\mathcal{L}(V)$ and $c\in\mathcal{F}(V,W)$ such that $a=b\circ c$. Therefore $\mathcal{F}(V)\subset\mathcal{L}(V)\cdot \mathcal{F}(V,W)$
