May 10th, 2018, 05:31 PM  #1 
Newbie Joined: May 2018 From: California, United States Posts: 1 Thanks: 0 
Hello everyone, I'm studying for the Math GRE subject test and I'm currently going over Linear Algebra. I'm carefully rereading my course book "Linear Algebra Done Right" By Sheldon Axler (Third Edition). I have a few questions in regards to the fundamental concepts: Going over the definition of a Vector Space, it doesn't seem obvious to me why we need the scalars, adjoined to the vector set, to be members of a Field. If we loosen this definition of a Vector Space a little bit, I think we can get algebraic structures that are analogous to that of a Vector Space and thus meaningful. Hence my firstquestion: Is the Real Number Line a Vector Space?. If so, then this would imply that numbers themselves can be interpreted as vectors. Then, in trying to consider "small" Vector Spaces (other than the trivial case of the singleton zero vector set {0}) I wonder, in this sense, my second question: can there be Vector Spaces "smaller" than the Real Number Line? It is here where loosening the definition of a vector space can be of merit. If we allow our scalars to be Integers and our vector set to be all the multiples (positive and negative) of a Natural Number then that vector set adjoined with the operation of vector addition and scalar multiplication would manifest all the characteristics that define a Vector Space. Try using the set that contains all the multiples of three  {x  x = α3 where α∈Z}. Thus why would such a structure not be a vector space. Is it missing something? Or why do we demand that we use scalars from a field? I'm interested in knowing what you think. Last edited by skipjack; May 11th, 2018 at 01:05 AM. 
May 11th, 2018, 01:20 AM  #2 
Senior Member Joined: Oct 2009 Posts: 456 Thanks: 148 
If the scalars are not from a field, but merely from a ring, then the stucture is called a module. If the scalars are from $\mathbb{Z}$, then you have a $\mathbb{Z}$module. Modules are very interesting objects, but are much less wellbehaved than vector spaces. For example, every vector space over the field $k$ is isomorphic to $k^n$. But this is not at all true for a module. 1) Is the real number line a vector space? Depends on your field of scalars. If your field is $\mathbb{R}$, then yes it is a vector space of dimension $1$. If your field is $\mathbb{Q}$, then it is an infinitedimensional vector space. But other fields such as $\mathbb{F}_2$ would not make the reals into a vector space. 2) Can there be vector spaces smaller than the real number line? Again depends on your field of scalars. If your field is $\mathbb{R}$, then no, there is no nontrivial vector space smaller than $\mathbb{R}$. If your field is $\mathbb{Q}$, then there are plenty of vector spaces smaller than $\mathbb{R}$, for example $\mathbb{Q}$. Last edited by skipjack; May 11th, 2018 at 03:24 AM. 
May 14th, 2018, 10:51 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,437 Thanks: 106 
If you want a shot at the GRE, the place to start is with a definition: A vector space is a set closed under asddition and scalar multiplication over F. Is S a vector space? 1) Define addition and ck: Closed, x+y in V Associativity, commutativity, zero, and subtraction, 2) Define scalar multiplication and ck: Closed, ax in V (a+b)x = ax+bx a(x+y)= ax+ay For ex, is R a vector space over C? No. cr is not in R Is C a vector space over R? rc is in C and check other conditions . With reference to the definition, you can define any variation you like and give it a name and examine the implications. 
May 14th, 2018, 11:25 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
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Quote:
Last edited by skipjack; May 14th, 2018 at 10:15 PM.  
May 14th, 2018, 11:40 AM  #5 
Senior Member Joined: Oct 2009 Posts: 456 Thanks: 148  
May 14th, 2018, 12:08 PM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,437 Thanks: 106  https://en.wikipedia.org/wiki/Examples_of_vector_spaces discusses finite vector spaces over the unique finite field Fq . For n=1 this would be smaller than R, if by smaller you mean fewer members. EDIT: Unfortunately, these days finite may mean countable, in which case the above is correct if you go where I don't want to go, namely the real numbers are uncountable. Last edited by skipjack; May 14th, 2018 at 10:13 PM. 
May 14th, 2018, 03:26 PM  #7 
Senior Member Joined: Jun 2015 From: England Posts: 853 Thanks: 258 
Why do we need scalars huh? Well the short answer is we don't. But then the mathematics of the resulting vector space would be so much poorer and less general. We would have to exclude most of the useful reasons for having vector spaces in the first place, which may be laid at the door of applied mathematicians. We could not have the dot product so could not say work = force x distance. We could not generate dual spaces because we could not create functionals. I think we would be in trouble with vector based geometry because we couldn't generate tangents using the dot product. 
May 15th, 2018, 02:56 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  
May 15th, 2018, 07:41 AM  #9  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,437 Thanks: 106  The OP didn't ask for a vector space of smaller dimension than the real line, he asked for one smaller than the real line, which is only possible for a finite field. Quote:
 
May 15th, 2018, 08:46 AM  #10  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 219 Thanks: 70 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Furthermore, many of these subfields (in fact, uncountably many of them*) have smaller cardinality than the reals. e.g. for any real number $\alpha$, the subfield $\mathbb{Q}(\alpha) \subsetneq \mathbb{R}$ is countable. *There's a small subtlety with this: while there are uncountably many $\alpha \in \mathbb{R}$, some of them give rise to the same field $\mathbb{Q}(\alpha)$. For example, $\mathbb{Q}(\alpha) = \mathbb{Q}$ for every rational number $\alpha$. Despite this, it's still the case that there are uncountably many distinct fields of the form $\mathbb{Q}(\alpha)$ with $\alpha \in \mathbb{R}$. (Indeed, if there were only countably many of them, then the union of all of them would be countable, because countable unions of countable sets are countable. But their union is $\mathbb{R}$, which is not countable.) Yet another interpretation of "smaller": every subfield of $\mathbb{R}$ (including $\mathbb{R}$ itself) is a vector space over $\mathbb{Q}$. The dimension of $\mathbb{R}$ as a vector space over the rationals, $\dim_\mathbb{Q}(\mathbb{R})$, is infinite, while $\dim_\mathbb{Q}(F)$ is finite for countably many subfields $F$ of $\mathbb{R}$. (In fact, these $F$ are precisely the fields $\mathbb{Q}(\alpha)$ where $\alpha$ is an algebraic number, as follows from the primitive element theorem.)  

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