dense ring of linear transformations

Let D be a division ring and V a vector spaceMathworldPlanetmath over D. Let R be a subring of the ring of endomorphisms (linear transformations) EndD(V) of V. Then R is said to be a dense ring of linear transformations (over D) if we are given

  1. 1.

    any positive integer n,

  2. 2.

    any set {v1,,vn} of linearly independentMathworldPlanetmath vectors in V, and

  3. 3.

    any set {wi,,wn} of arbitrary vectors in V,

then there exists an element fR such that

f(vi)=wi for i=1,,n.

Note that the linear independence of the vi’s is essential in insuring the existence of a linear transformation f. Otherwise, suppose 0=divi where d10. Pick wi’s so that they are linearly independent. Then 0=f(divi)=dif(vi)=diwi, contradicting the linear independence of the wi’s.

The notion of “dense” comes from topologyMathworldPlanetmath: if V is given the discrete topology and EndD(V) the compact-open topologyMathworldPlanetmath, then R is dense in EndD(V) iff R is a dense ring of linear transformations of V.


First, assume that R is a dense ring of linear transformations of V. Recall that the compact-open topology on EndD(V) has subbasis of the form B(K,U):={ff(K)U}, where U is open and K is compactPlanetmathPlanetmath in V. Since V is discrete, K is finite. Now, pick a point gEndD(V) and let


be a neighborhoodMathworldPlanetmathPlanetmath of g, I some index setMathworldPlanetmathPlanetmath. Then for some αI, gB(Kiα,Uiα). This means that g(Kiα)Uiα for all i=1,,n(α). Since each Kiα is finite, so is K:=Kiα. After some re-indexing, let {v1,,vn} be a maximal linearly independent subset of K. Set wj=g(vj), j=1,,n. By assumptionPlanetmathPlanetmath, there is an fR such that f(vj)=wj, for all j. For any vK, v is a linear combinationMathworldPlanetmath of the vj’s: v=djvj, djD. Then f(v)=djf(vj)=djg(j)=g(v)Uiα for some i. This shows that f(Kiα)Uiα and we have fB(Kiα,Uiα)B.

Conversely, assume that the ring R is a dense subsetPlanetmathPlanetmath of the space EndD(V). Let v1,,vn be linearly independent, and w1,,wn be arbitrary vectors in V. Let W be the subspaceMathworldPlanetmathPlanetmathPlanetmath spanned by the vi’s. Because the vi’s are linearly independent, there exists a linear transformation g such that g(vi)=wi and g(v)=0 for vW. Let Ki={vi} and Ui={wi}. Then the Ki’s are compact and the Ui’s are open in the discrete space V. Clearly g{hh(vi)=wi}=B(Ki,Ui) for each i=1,,n. So g lies in the neighborhood B=B(Ki,Ui)EndD(V). Since R is dense in EndD(V), there is an fRB. This implies that f(vi)=wi for all i. ∎


  • If V is finite dimensional over D, then any dense ring of linear transformations R=EndD(V). This can be easily observed by using the second half of the proof above. Take a basis v1,,vn of V and any set of n vectors w1,,wn in V. Let g be the linear transformation that maps vi to wi. The above proof shows that there is an fR 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=fR.

  • It can be shown that a ring R is a primitive ring iff it is isomorphicPlanetmathPlanetmathPlanetmathPlanetmath to a dense ring of linear transformations of a vector space over a division ring. This is known as the . It is a generalizationPlanetmathPlanetmath of the special case of the Wedderburn-Artin 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 2013-03-22 15:38:46
Last modified on 2013-03-22 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