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Homedense ring of linear transformations
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dense ring of linear transformations
Let $D$ be a division ring and $V$ a vector space over $D$. Let $R$ be a subring of the ring of endomorphisms (linear transformations) $\operatorname{End}_{D}(V)$ of $V$. Then $R$ is said to be a dense ring of linear transformations (over $D$) if we are given
1. 2. any set $\{v_{1},\ldots,v_{n}\}$ of linearly independent vectors in $V$, and
3. any set $\{w_{i},\ldots,w_{n}\}$ of arbitrary vectors in $V$,
then there exists an element $f\in R$ such that
$f(v_{i})=w_{i}\quad\mbox{ for }i=1,\ldots,n.$ 
Note that the linear independence of the $v_{i}$’s is essential in insuring the existence of a linear transformation $f$. Otherwise, suppose $0=\sum d_{i}v_{i}$ where $d_{1}\neq 0$. Pick $w_{i}$’s so that they are linearly independent. Then $0=f(\sum d_{i}v_{i})=\sum d_{i}f(v_{i})=\sum d_{i}w_{i}$, contradicting the linear independence of the $w_{i}$’s.
The notion of “dense” comes from topology: if $V$ is given the discrete topology and $\operatorname{End}_{D}(V)$ the compactopen topology, then $R$ is dense in $\operatorname{End}_{D}(V)$ iff $R$ is a dense ring of linear transformations of $V$.
Proof.
First, assume that $R$ is a dense ring of linear transformations of $V$. Recall that the compactopen topology on $\operatorname{End}_{D}(V)$ has subbasis of the form $B(K,U):=\{f\mid f(K)\subseteq U\}$, where $U$ is open and $K$ is compact in $V$. Since $V$ is discrete, $K$ is finite. Now, pick a point $g\in\operatorname{End}_{D}(V)$ and let
$B=\bigcup_{{\alpha\in I}}\bigcap_{{i=1}}^{{n(\alpha)}}B(K_{{i\alpha}},U_{{i% \alpha}})$ 
be a neighborhood of $g$, $I$ some index set. Then for some $\alpha\in I$, $g\in\bigcap B(K_{{i\alpha}},U_{{i\alpha}})$. This means that $g(K_{{i\alpha}})\subseteq U_{{i\alpha}}$ for all $i=1,\ldots,n(\alpha)$. Since each $K_{{i\alpha}}$ is finite, so is $K:=\bigcup K_{{i\alpha}}$. After some reindexing, let $\{v_{1},\ldots,v_{n}\}$ be a maximal linearly independent subset of $K$. Set $w_{j}=g(v_{j})$, $j=1,\ldots,n$. By assumption, there is an $f\in R$ such that $f(v_{j})=w_{j}$, for all $j$. For any $v\in K$, $v$ is a linear combination of the $v_{j}$’s: $v=\sum d_{j}v_{j}$, $d_{j}\in D$. Then $f(v)=\sum d_{j}f(v_{j})=\sum d_{j}g(_{j})=g(v)\in U_{{i\alpha}}$ for some $i$. This shows that $f(K_{{i\alpha}})\subseteq U_{{i\alpha}}$ and we have $f\in\bigcap B(K_{{i\alpha}},U_{{i\alpha}})\subseteq B$.
Conversely, assume that the ring $R$ is a dense subset of the space $\operatorname{End}_{D}(V)$. Let $v_{1},\ldots,v_{n}$ be linearly independent, and $w_{1},\ldots,w_{n}$ be arbitrary vectors in $V$. Let $W$ be the subspace spanned by the $v_{i}$’s. Because the $v_{i}$’s are linearly independent, there exists a linear transformation $g$ such that $g(v_{i})=w_{i}$ and $g(v)=0$ for $v\notin W$. Let $K_{i}=\{v_{i}\}$ and $U_{i}=\{w_{i}\}$. Then the $K_{i}$’s are compact and the $U_{i}$’s are open in the discrete space $V$. Clearly $g\in\{h\mid h(v_{i})=w_{i}\}=B(K_{i},U_{i})$ for each $i=1,\ldots,n$. So $g$ lies in the neighborhood $B=\cap B(K_{i},U_{i})\subseteq\operatorname{End}_{D}(V)$. Since $R$ is dense in $\operatorname{End}_{D}(V)$, there is an $f\in R\cap B$. This implies that $f(v_{i})=w_{i}$ for all $i$. ∎
Remarks.

If $V$ is finite dimensional over $D$, then any dense ring of linear transformations $R=\operatorname{End}_{D}(V)$. This can be easily observed by using the second half of the proof above. Take a basis $v_{1},\ldots,v_{n}$ of $V$ and any set of $n$ vectors $w_{1},\ldots,w_{n}$ in $V$. Let $g$ be the linear transformation that maps $v_{i}$ to $w_{i}$. The above proof shows that there is an $f\in R$ such that $f$ agrees with $g$ on the basis elements. But then they must agree on all of $V$ as a result, which is precisely the statement that $g=f\in R$.

It can be shown that a ring $R$ is a primitive ring iff it is isomorphic to a dense ring of linear transformations of a vector space over a division ring. This is known as the Jacobson Density Theorem. It is a generalization of the special case of the WedderburnArtin Theorem when the ring in question is a simple Artinian ring. In the general case, the finite chain condition is dropped.
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