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) ${\mathrm{End}}_{D}(V)$ of $V$. Then $R$ is said to be a dense ring of linear transformations (over $D$) if we are given

1.
any positive integer $n$,

2.
any set $\{{v}_{1},\mathrm{\dots},{v}_{n}\}$ of linearly independent^{} vectors in $V$, and

3.
any set $\{{w}_{i},\mathrm{\dots},{w}_{n}\}$ of arbitrary vectors in $V$,
then there exists an element $f\in R$ such that
$$f({v}_{i})={w}_{i}\mathit{\hspace{1em}}\text{for}i=1,\mathrm{\dots},n.$$ 
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 ${d}_{1}\ne 0$. 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 ${\mathrm{End}}_{D}(V)$ the compactopen topology^{}, then $R$ is dense in ${\mathrm{End}}_{D}(V)$ 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 ${\mathrm{End}}_{D}(V)$ 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 $g\in {\mathrm{End}}_{D}(V)$ 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 $i=1,\mathrm{\dots},n(\alpha )$. Since each ${K}_{i\alpha}$ is finite, so is $K:=\bigcup {K}_{i\alpha}$. After some reindexing, let $\{{v}_{1},\mathrm{\dots},{v}_{n}\}$ be a maximal linearly independent subset of $K$. Set ${w}_{j}=g({v}_{j})$, $j=1,\mathrm{\dots},n$. 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 ${\mathrm{End}}_{D}(V)$. Let ${v}_{1},\mathrm{\dots},{v}_{n}$ be linearly independent, and ${w}_{1},\mathrm{\dots},{w}_{n}$ 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 $i=1,\mathrm{\dots},n$. So $g$ lies in the neighborhood $B=\cap B({K}_{i},{U}_{i})\subseteq {\mathrm{End}}_{D}(V)$. Since $R$ is dense in ${\mathrm{End}}_{D}(V)$, 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 $R={\mathrm{End}}_{D}(V)$. This can be easily observed by using the second half of the proof above. Take a basis ${v}_{1},\mathrm{\dots},{v}_{n}$ of $V$ and any set of $n$ vectors ${w}_{1},\mathrm{\dots},{w}_{n}$ 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 . 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.
Title  dense ring of linear transformations 

Canonical name  DenseRingOfLinearTransformations 
Date of creation  20130322 15:38:46 
Last modified on  20130322 15:38:46 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 16K40 
Synonym  dense ring 
Related topic  SchursLemma 
Defines  Jacobson Density Theorem 