# Thread: [SOLVED] Vector spaces over division rings.

1. ## [SOLVED] Vector spaces over division rings.

Hi:
I have the following definition: let V be a vector space over
a division ring D. A mapping a of V into V is called a linear trans-
formation of V if it has the followiwng two properties:
(x+y)a = xa+ya for x,y $\displaystyle \in$ V,
(x$\displaystyle \alpha$)a = (xa)$\displaystyle \alpha$ for x $\displaystyle \in$ V, $\displaystyle \alpha$ D

And here I find an odd thing. If a is the mapping multiplication by
a scalar (that is, by an element of D), then a is not in general a linear trans-
formation of V according to the definition, because D needs not be
commutative. Any hint will be welcome.

2. First, in English, at least, a set with operations of addition and "scalar multiplication" with the scalars from a division ring rather than a field is usually called a "module", not a "vector space".

Second, If you define scalar multiplication by "$\displaystyle x\alpha$" for $\displaystyle \alpha$ in D, then you must either define $\displaystyle \alpha x= x\alpha$ or leave [tex]\alpha x[tex] undefined. The fact that multiplication of scalars is not commutative has nothing to do with the relationship between $\displaystyle x\alpha$ and $\displaystyle \alpha x$.

3. Originally Posted by HallsofIvy
First, in English, at least, a set with operations of addition and "scalar multiplication" with the scalars from a division ring rather than a field is usually called a "module", not a "vector space".
Actually, Thomas Hungerford defines a vector space using division rings.

4. Originally Posted by ENRIQUESTEFANINI
Hi:
I have the following definition: let V be a vector space over
a division ring D. A mapping a of V into V is called a linear trans-
formation of V if it has the followiwng two properties:
(x+y)a = xa+ya for x,y $\displaystyle \in$ V,
(x$\displaystyle \alpha$)a = (xa)$\displaystyle \alpha$ for x $\displaystyle \in$ V, $\displaystyle \alpha$ D

And here I find an odd thing. If a is the mapping multiplication by
a scalar (that is, by an element of D), then a is not in general a linear trans-
formation of V according to the definition, because D needs not be
commutative. Any hint will be welcome.
multiplication by an element of D is a linear transformation iff that element is in the center of D.

5. Originally Posted by roninpro
Actually, Thomas Hungerford defines a vector space using division rings.
And Neal H. McCoy too.

6. Originally Posted by NonCommAlg
multiplication by an element of D is a linear transformation iff that element is in the center of D.
Quite understandable. And if D is a field the center of D is D. So in this case that mapping is always a linear transformation. Thanks a lot, NonCommAlg.