1. ## Dense rings.

Hi:

First, I must make some definitions, following Neal H.
McCoy, The Theory of Rings.

Def.1: A vector space V over a division ring D is a
D-module with the additional requirement that if 1
is the unity of D and x \in V, then x1 = x.

Def.2: Let k be a positive integer. Then the ring R of
linear transformations is said to be k-fold
transitive if given any ordered set x_1, x_2,...,x_k
of k linearly independent vectors and any ordered
set y_1, y_2,...,y_k of k arbitrary vectors, there
exists a \in R such that x_i a = y_i (i=1,2,...,k).
If R is k-fold transitive for every positive
interger k, R is said to be a dense ring of linear
transformations.

Now, I have this theorem:
Theor.1: Let T be the ring of all linear
transformations of a vector space V with a denumerable
basis. Then...

At a certain point in the proof, he says: Since Ve is
now assumed to have infinite dimension, it has a
denumerable basis X = {x_1, x_2,... } [because V has a
denumerable basis; e \in T; that is, Ve is a subspace
of V].

[T being the ring of ALL linear transformations of V, T
is dense.]

Further on he says: Let Z = {z_1, z_2, ...} be a basis
of V. Since there exists an element of T which maps the
basis elements z_i (i = 1, 2,...) onto any specified
elements of V, there exists an elemnt d of T such that
Code:
       z_i d = x_i     (i = 1, 2, ...)
.

Here is what I do not understand. Def.2 says that
given any ordered set x_1, x_2,..., x_k of k lin. ind.
vectors in V and a set y_1, y_2,..., y_k of vectors in
V, there exists a\in T such that x_i a = y_i for i = 1,
2, ..., k, and that this is true for any positive k.
An entirely different thing is to say that given a set
x_1, x_2,... of lin. ind. vectors and a set y_1,
y_2,... there is a\in T such that x_i a = y_i for i =1,
2,... That is to say, now he is saying T is w-fold
transitive, where w is omega, the set of natural
numbers. Any suggestion will be welcome.

2. Originally Posted by ENRIQUESTEFANINI
Hi:

First, I must make some definitions, following Neal H.
McCoy, The Theory of Rings.

Def.1: A vector space V over a division ring D is a
D-module with the additional requirement that if 1
is the unity of D and x \in V, then x1 = x.

Def.2: Let k be a positive integer. Then the ring R of
linear transformations is said to be k-fold
transitive if given any ordered set x_1, x_2,...,x_k
of k linearly independent vectors and any ordered
set y_1, y_2,...,y_k of k arbitrary vectors, there
exists a \in R such that x_i a = y_i (i=1,2,...,k).
If R is k-fold transitive for every positive
interger k, R is said to be a dense ring of linear
transformations.

Now, I have this theorem:
Theor.1: Let T be the ring of all linear
transformations of a vector space V with a denumerable
basis. Then...

At a certain point in the proof, he says: Since Ve is
now assumed to have infinite dimension, it has a
denumerable basis X = {x_1, x_2,... } [because V has a
denumerable basis; e \in T; that is, Ve is a subspace
of V].

[T being the ring of ALL linear transformations of V, T
is dense.]

Further on he says: Let Z = {z_1, z_2, ...} be a basis
of V. Since there exists an element of T which maps the
basis elements z_i (i = 1, 2,...) onto any specified
elements of V, there exists an elemnt d of T such that
Code:
       z_i d = x_i     (i = 1, 2, ...)
.

Here is what I do not understand. Def.2 says that
given any ordered set x_1, x_2,..., x_k of k lin. ind.
vectors in V and a set y_1, y_2,..., y_k of vectors in
V, there exists a\in T such that x_i a = y_i for i = 1,
2, ..., k, and that this is true for any positive k.
An entirely different thing is to say that given a set
x_1, x_2,... of lin. ind. vectors and a set y_1,
y_2,... there is a\in T such that x_i a = y_i for i =1,
2,... That is to say, now he is saying T is w-fold
transitive, where w is omega, the set of natural
numbers. Any suggestion will be welcome.

I think you're right: the leap from "for any k" to "infinite many", whether denumerable or not, seems too high to give it just like this and, perhaps, it doesn't work according to the given definition.

Nevertheless the general claim is true: there's ALWAYS a UNIQUE linear transformation (operator, in fact) that carries ANY basis to ANY set with the same cardinal as a basisin your case, simple define $\displaystyle v_id:=x_i$ and extend this definition by linearity.

Tonio

3. Thank you very much, Tonio.

4. the important point you're missing is that we're dealing with $\displaystyle T$, the ring of all linear transformations of $\displaystyle V,$ here. $\displaystyle T$ is $\displaystyle k$-fold transitive for any $\displaystyle k,$ finite or infinite. this is easy to see:
for any set $\displaystyle I$, any linearly independent vectors $\displaystyle z_i, \ i \in I,$ and any vectors $\displaystyle x_i, \ i \in I,$ you can easily define $\displaystyle \left(\sum_{i \in I} \alpha_i z_i \right)d=\sum_{i \in I} \alpha_i x_i,$ where all but finitely many of the scalars $\displaystyle \alpha_i$ are zero. the map $\displaystyle d$ is well-defined because $\displaystyle z_i$ are linearly independent and it is obviously linear. thus $\displaystyle d \in T.$ but if $\displaystyle R$ is some subring of $\displaystyle T,$ then $\displaystyle d$ might not be in $\displaystyle R.$

5. Yes, in particular, if I = the natural numbers, then I have the result I was looking for. And "where all but finitely many of the scalars are zero", because any element of V is a lin. comb. of a finite subset of the basis {z_1, z_2, ...}.

This is a generalization of Example 1 on page 96 (section 27) of the book. Or better: there is a condition that is stronger than density although, in the case of T, both are equivalent.
Thank you so much.