If $V$ is a f.d. vector space, why is every right ideal of $\mathrm{End}_F(V)$ of form $I=\{T:\mathrm{im}(T)\subseteq W\}$ for $W$? This is a qual problem from Spring 2011 at my university. 

Let $V$ be a finite dimensional vector space. Then every right ideal of $\mathrm{End}_F(V)$ is of form $I=\{T:\mathrm{im}(T)\subseteq W\}$ for some fixed subspace $W$.

A solution is found here, in problem 8, but I'm just curious if there are perhaps alternative, more elegant solutions. The proof there itself admits in an ending remark that it is a naive approach. Thanks.
 A: I would do it as follows.
1. First note that if $A,B\in {\rm End}(V)$ satisfy ${\rm im}(A)\subset {\rm im}(B)$, then one can find $C\in{\rm End}(V)$ such that $A=BC$. This is the key point, and a well-known "exercise".
1'. If $S\in\mathcal I$, then any $T\in{\rm End}(V)$ such that ${\rm im}(T)\subset {\rm im}(S)$ belongs to $\mathcal I$. This follows immediately from 1.
2. If $T_1,T_2\in\mathcal I$, then one can find $T\in\mathcal I$ such that ${\rm im}(T)$ contains ${\rm im}(T_1)$ and ${\rm im}(T_2)$. To see this write ${\rm im}(T_1)+{\rm im}(T_2)=V_1\oplus V_2\oplus V_3$, where $V_i\subset {\rm im}(T_i)$ for $i=1,2$ and $V_3={\rm im}(T_1)\cap {\rm im}(T_2)$. Then choose $E$ such that $V=E\oplus V_1\oplus V_2\oplus V_3$. For $i=1,2,3$, the associated projection $p_i$ onto $V_i$ belongs to $\mathcal I$ by 1'. So $T=p_1+p_2+p_3\in \mathcal I$, and ${\rm im}(T)= {\rm im}(T_1)+{\rm im}(T_2)$.
3. It follows immediately from 2 that there exists $S\in\mathcal I$ such that ${\rm im}(T)\subset {\rm im}(S)$ for every $T\in\mathcal I$. Just take $S\in\mathcal I$ such that ${\rm im}(S)$ has maximal dimension. Setting $W:={\rm im}(S)$, it follows from 1' that $\mathcal I=\{ T\in{\rm End}(V);\; {\rm im}(T)\subset W\}$.
