TITLE: $\ker S\subset\ker T\Leftrightarrow\exists R, T=RS$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.