correspondence between subspaces and left ideals in End$(V)$ Let $V$ be a right- n-dimensional vector space over a division ring $D$. Let $R=\text{End}(V_D)$. We know that for every subspace $W$ of $V$ the set $J_W = \{f \in R \mid f(W) = 0\}$ is a left ideal of $R$.
What I have to prove is that the mapping $W_D \mapsto J_W$ is a bijection between the the lattice of subspaces of $V$ and the lattice of all left ideals of $R$.
Injectivity is actually really clear. What about surjectivity? My attempt was defining, for each $J$ left ideal of $R$, the subspace $W_J = \{v \in V \mid f(v) = 0\, \forall f \in J\}$. Then we have to show that $W_J$ is mapped into $J$ by our correspondence. The fact that $J \subset J_{W_J}$ is trivial, but what about the other inclusion?
Any hint\solution is highly appreciated. Thanks!
 A: I am not sure if anything breaks down for division rings which are not fields, but this is how I would do it in a regular linear algebra setting, and then you can tell us if it works for division rings:
Take a basis $\mathcal{B}=\left\{e_1,\ldots,e_n\right\}$ of $V$, and consider the functionals $\phi_i\colon V\to D$, $\phi_i(e_j)=\delta_{ij}$. Also consider the co-functionals $\psi_i\colon D\to V$, $\psi_i(d)=e_id$.
By definition, $J_{W_J}=\left\{g\in \operatorname{End}(V):\bigcap_{f\in J}\ker f\subseteq\ker g\right\}$. Let $g\in J_{W_J}$.
Let us find $f_1,\ldots,f_N\in J$ such that $\bigcap_{j=1}^N\ker(f_j)=\bigcap_{f\in J}\ker(f)(=W_J)$.

*

*Just choose any $f_1\in J$. If $\dim \bigcap_{f\in J}\ker(f)=\dim\ker(f_1)$, we are done;

*If not, choose some $f_2\in J$ such that $\ker(f_2)\cap \ker(f_1)$ is strictly smaller than $\ker(f_1)$. If $\dim \bigcap_{f\in J}\ker(f)=\dim(\ker f_1\cap\ker f_2)$, we are done.

*If not, keep "decreasing the dimension" of $\ker(f_1)\cap\ker(f_2)\cap\cdots$ until you get to the dimension of $\bigcap_{f\in J}\ker(f)$.

Then we have
$$\bigcap_{f\in J}\ker(f)=\bigcap_{j=1}^N\ker f_j=\bigcap_{j=1}^N\bigcap_{i=1}^n\ker(\phi_i f_j).$$
and, for each $k$,
$$\ker(g)\subseteq\ker(\phi_k g)$$
so $\bigcap_{j=1}^N\bigcap_{i=1}^n\ker(\phi_i f_j)\subseteq\ker(\phi_k g)$ for each $k$.
The following is standard (there should be a proof somewhere on this site as well):

Theorem (see Rudin, Functional Analysis, Lemma 3.9): If $p_1,\ldots,p_m,p$ are linear functionals on a vector space $V$, then $p$ is a linear combination of $p_1,\ldots,p_m$ iff $\bigcap_{j=1}^m p_j\subseteq\ker(p)$.

So for each $k$ we can find scalars $\alpha_{i,j,k}$ such that $\phi_k g=\sum_{i,j}\alpha_{i,j,k}(\phi_i  f_j)$ (if I am doing things right, spaces of operators between right $D$-modules should be left $D$-modules... right?)
Now remember the "cofunctionals" $\psi_k$ right from the beggining. We have $\operatorname{id}_V=\sum_k\psi_k\phi_k$. Then
$$g=\sum_k\psi_k\phi_kg=\sum_{i,j,k}\psi_k(\alpha_{i,j,k}(\phi_i f_j)),$$
and all $\psi_k(\alpha_{i,j,k}(\phi_i f_j))$ belong to $J$, as it is a left ideal by hypothesis, so $g$ does as well.
