Proving Vector Space of Positive Quadruples of Real Numbers

asdf1
Messages
734
Reaction score
0
how do you prove the set of vectors "all ordered quadruples of positive real numbers" make a vector space?
 
Physics news on Phys.org
you show that the axioms for a vector space are satisfied. I'm sure they're in your book. it shouldn't be very hard, just direct verification. you put ""s around the phrase ordered quadruples... that just means 4-dimensional vectors whose entries are real numbers, that's all.
 
I had to memorize these in Linear Algebra:

http://www.math.niu.edu/~beachy/courses/240/vectorspace.html
 
The problem is that it isn't a vector space since there is no additive inverse vector, nor is scalar multiplication valid (unless you redefine vector addition, the zero vector and th field over which the vector space is defined).
 
how do you prove the set of vectors "all ordered quadruples of positive real numbers" make a vector space?
You don't because they don't.

To make a vector space, you need 3 things:
(1) A set of "vectors".
(2) A field of "scalars".
(3) An "addition" operation that takes two vectors and produces a vector.
(4) A "multiplication" operation that takes a scalar and a vector and produces a vector.

You've only specified (1), so that's certainly not enough to make a vector space. Once you specify the other three things, then we can start discussing the question of whether it's a vector space or not.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top