Vector spaces

[vaidiarte]

How can I show the set of integers Z cannot be made into a vector space over R?

Z=(0, +-1, +-2,...)

 

For example I had to show how the set of integers Z cannot be made into a vector space over C by:

(i) exits in V and a vector (1) exits in Z

i*1=i therefore, i does not exists in Z

Hence it is not a vector space because it does not close under scalar mulitplication.

 

I have to do the same with rationals (Q) and with reals®

[Dave]

Well, surely you can modify your example slightly? For example, it's certainly true that [imath]\frac{1}{2} \in \R, \Q[/imath] and [imath]1 \in \mathbb{Z}[/imath]. But [imath]\frac{1}{2} \cdot 1 \notin \mathbb{Z}[/imath] so it's not closed.

 

Personally, I find the question a bit silly (without meaning to cause offense - I'm aware it's an assignment ). Every course I've seen has defined scalar multiplication as a map, [imath]F \times V \to V[/imath]. Since [imath]\mathbb{Z} \subset \R[/imath] and you're effectively using the same operation for scalar and vector multiplication it can't possibly be a well-defined operation. Perhaps this is what he's hoping to show.

[matt grime]

though there is nothing to suppose that 1/2.1=1/2, ie it might not be the multiplication inherited from Z being inside C/R/Q. equivalently there is nothing making us suppose that the vector sum of 1 and 1 is 2.

 

it is easy to show, even given that, that Z is not a v.s. over R or C. if s and t are distinct elements in R then s.1 and t.1 are distinct elements in Z, thus if it were a v.s. over R or C then there would have to be an uncountable number of elements in Z.

[ElijahJones]

I agree with Dave. Way to leverage cardinality in that proof Matt. I agree that the scalar multiplcation need not be the same as the operation used on elements of the vectors, but since it has to map into the same set its standard to have them be the same. In fact I cannot think of an example where those two operations are different.

 

You construct a vector space by taking a field as scalars, but that field also provides the elements of your vectors (typically). Its the only way it works (as far as I know). If your vectors are formed on a subset X then F.X must equal X or else you do not have closure. But look X must be a subset that is invariant under multiplication by F. The definition of vector space makes no assumptions about the multiplicative structure of X, it need only be an abelian group under vector addition and that may not necessarily be derived from addition on X(though I cannot think of an example where it is not). So the question seems to imply more restrictions on a vector space than the definition itself.

 

To get away from that apparent ambiguity lets say that the addition of vectors is derived from an addition defined on X an abelian group and that X is a subset of a field F that we will use as our set of scalars. The implication of this if we want a vector space is F.X=X. But if F.X=X then a.1 is in X for all a in F which contradicts the fact that X is a subset of F.

 

Suppose on the other hand X contains F, then again we require F.X=X. Now this might be possible. It might be possible to take your vectors over C and have the scalars be in R and have a vector space. I mean multiplying a complex number by a real number r simply increases its modulus by sqrt®. Hmm I never thought of that before. I do not know where it would be useful but it does seem to work. So it works if the field F is a subfield of the set X you take your vector elements from but not the other way around.

 

But trying to force Z to be closed under action of R is futile. Just one counter example will do the trick or a proof which I gave above.

 

Suppose you did the opposite. What if we take complex numbers as the vector elements and use Z as the scalar ring (remember Z is not a field but it is an integral domain which is a commutative ring among other things). Well then of course we get this: scalar multiplication can only take vectors to discrete locations meaning we have established a very interesting partition of C^n, n is the dimension. Now since Z is not a field we cannot call our construct a vector space it is properly called a Z-module (Dummit and Foote 2nd Ed p388)

 

The question turns out to be interesting though IMHO. BTW the vectors only need to be an abelian group under vector addition adding an inner product can yield a normed vector space. Adding a cross product can produce a field of vectors, whether you allow scalar multiplication or not.