My Math Forum Vector Spaces (Understanding the Basics)

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 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: 770 Thanks: 276 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 well-behaved 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 infinite-dimensional 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 non-trivial 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 $\mathbb{Q}$. Thanks from topsquark, studiot and BeneviereTheMathematician Last edited by skipjack; May 11th, 2018 at 03:24 AM.
 May 14th, 2018, 10:51 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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
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Quote:
 Originally Posted by BeneviereTheMathematician 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?.
With the usual definitions of addition and multiplication of real numbers, yes, the "Real Number Line" is a one dimensional vector space. We can take as basis the singleton set {1}. Every real number, r, is equal to the scalar, r, times 1.

Quote:
 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?
Any non-trivial vector space has positive integer dimension. The smallest possible dimension is 1. Any one dimensional vector space is isomorphic to the vector space of real numbers. (Any n-dimensional vector space is isomorphic to $\displaystyle R^n$.)

Quote:
 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.
We like to use scalars from a field so we can divide by scalars.

Last edited by skipjack; May 14th, 2018 at 10:15 PM.

May 14th, 2018, 11:40 AM   #5
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 Originally Posted by Country Boy The smallest possible dimension is 1.
The smallest possible dimension is 0 though.

 May 14th, 2018, 12:08 PM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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: 905 Thanks: 271 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
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 Originally Posted by Micrm@ss The smallest possible dimension is 0 though.
I said "Any non-trivial vector space" and the original post asked about "other than the trivial case of the singleton zero vector set {0}". A vector space with dimension 0 consists of the 0 vector only and is trivial.

May 15th, 2018, 07:41 AM   #9
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 Originally Posted by Micrm@ss The smallest possible dimension is 0 though.
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.

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 Originally Posted by BeneviereTheMathematician ........... 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? ............................. Or why do we demand that we use scalars from a field?
It's the definition of a vector space. Where does the definition come from? Generalization of vectors in mechanics.

May 15th, 2018, 08:46 AM   #10
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 Originally Posted by zylo 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.
It depends on what is meant by "smaller". If proper subsets of a set are considered smaller than that set, then there are many more examples than just the finite fields. For example, any proper subfield of the reals is naturally a vector space (over itself) and are smaller than $\mathbb{R}$ in this sense.

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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