Can I always write a bounded operator $T$ as $T=R^{*}S$ If $K$ and $H$ are Hilbert spaces and $T\in B(K)$, can I always express $T$ as a linear combination of products $R^{*}S$ for $R,S \in B(K,H)$ ?
I think I already showed this is true when $K$ and $H$ are finite dimensional. Is this true for any Hilbert spaces $K$ and $H$ ?
If it is not true, are linear combinations of such products at least dense in $B(K)$ ?
 A: Construction
Suppose one has:
$$\dim\mathcal{H}\geq\dim\mathcal{K}$$
Remember the unitaries:
$$\#S=\dim\mathcal{H}:\quad U_\mathcal{H}(\varphi):=(\langle \sigma_s,\varphi\rangle)_{s\in S}\in\ell^2(S)$$
$$\#T=\dim\mathcal{H}:\quad U_\mathcal{K}(\psi):=(\langle \tau_t,\varphi\rangle)_{t\in T}\in\ell^2(T)$$
Regard an embedding:
$$\iota:T\hookrightarrow S:\quad (Ex)_{s=\iota(t)}:=x_t\quad(Ex)_{s\neq\iota(t)}:=0$$
Then one obtains:
$$J:=U_\mathcal{H}^{-1}\circ E\circ U_\mathcal{K}:\quad J^*J=1_\mathcal{K}$$
Construct operators:
$$R^*L:=(JT^*)^*J=TJ^*J=T$$
Concluding construction:
Counterexample
Given the Hilbert spaces $\mathbb{C}$ and $\mathbb{C}^2$.
Consider the operator:
$$T\in\mathcal{B}(\mathbb{C}^2):\quad T:=1_{\mathbb{C}^2}$$
Then this fails since:
$$\dim\mathcal{R}R^*L\leq\dim\mathcal{R}R^*\leq1\\
<2=\dim\mathcal{R}1_{\mathbb{C}^2}=\dim\mathcal{R}T$$
Concluding counterexample.
A: Assuming that for finite dimensional Hilbert spaces $K$ and $H$, I can write any $T$ as a linear combination of these products, then by Goldstine's Theorem, these linear combinations are $w^{*}$ dense in $B(K)$.
