$\ker S\subset\ker T\Leftrightarrow\exists R, T=RS$ Let $V,W$ be two vector spaces over $F=\mathbb{R}$ or $\mathbb{C}$, such that $W$ is finite dimensional and $S,T\in L(V,W)$.

Show that $\ker S\subset\ker T$ if, and only if, there exists a linear operator $R:W\to W$ such that $T=RS$.


The converse was easy to verify, since if $Sv = 0$, then $Tv = RSv = R(0) = 0$.
I tried to prove the other assertion by constructing such $R$ considering basis for $\operatorname{range} T$ and $\operatorname{range} S$ and send one to the other, but this is as far as I got. Could anyone give me a hint? 
 A: You are on the right track.
Define a linear map $R_0\colon S(V)\to W$ as follows: For $w\in S(V)$, pick $v\in V$ with $S(v)=W$ (possible by definition of $S(V)$). Now let $R_0(w)=T(v)$. This is well-defined because if $S(v')=w$, too, then $v-v'\in \ker S\subseteq \ker T$, hence $T(v)=T(v')$.
Of course, we can extend $R_0$ to a linear map $R$ defined on all of $V$ by picking a complementary subspace (or: by extending a basis of $S(V)$ to a basis of $W$).
A: Consider a basis $\{v_1,\dots,v_r\}$ of $\ker S$, which we can extend to a basis $\{v_1,\dots,v_m\}$ of $V$.
The set $B=\{S(v_{r+1}),\dots,S(v_n)\}$ is a basis for the image of $S$, as it is readily verified.
The map $R$ must satisfy $RS=T$, so $RS(v_i)=T(v_i)$ is the forced choice when $r+1\le i\le n$. 
Just complete $B$ to a basis of $W$ and define $R$ to be zero (or whatever) on the remaining elements.
For $1\le i\le r$, we have $RS(v_i)=0=T(v_i)$. Thus $RS$ and $T$ coincide on a basis of $V$.

Note. Finite dimensionality is not required, provided one assumes that every vector space has a basis.
