I'm confused about proving vector spaces

In summary: A linear space is a vector space over {\mathbb R} (or {\mathbb C}, but let's stick to the reals for simplicity) that is not necessarily complete. A vector space over {\mathbb R} that is complete is, by definition, a Hilbert space. So a linear space is just like a Hilbert space, except that it is not necessarily complete. This is the way the term "linear space" is used in the field of functional analysis, for example. However, in the field of linear algebra, a linear space is usually defined to be exactly the same as a vector space. I agree that this is confusing, and for this reason I try to avoid the term "linear space
  • #1
astral
6
0
Let V denote the set of all differentiable real-valued functions defined on the real line. Prove that V is a vector space with the operations of addition and scalar multiplication as follows:

(f + g)(s) = f(s) + g(s) and (cf)(s) = c[f(s)]

---------------

I know I have to prove this by checking the 8 axioms of vector spaces, but I'm very confused as to how I actually go about it in a formal mathematical proof. If someone could just prove one of them (say commutativity of addition a+b=b+a), I'm sure I'll be able to figure out the rest.

I'm just confused about proving something so fundamental. I keep thinking there are no operations I can perform without first proving what I'm trying to prove. GAH
 
Last edited:
Physics news on Phys.org
  • #2
f(s) and g(s) are just numbers. You should invoke the comutivity and associativity of operations on real numbers, along with the definition you were given.
 
  • #3
I'm sorry if the answer is really obvious, but could you give me an example anyway?

Thanks.
 
  • #4
For instance, you want to show that (f+g)(s) = (g+f)(s). By the definition, (f+g)(s) = f(s) + g(s), and (g+f)(s) = g(s) + f(s). But f(s) and g(s) are just numbers, so (assuming the standard properties of real numbers) g(s) + f(s) = f(s) + g(s). From which the desired result obtains.
 
  • #5
I thought of doing it this way, which seems a bit better:

(f + g)(s) = f(s) + g(s)
(g + f)(s) = g(s) + f(s)

(f + g)(s) + f(s) = ( f(s) + g(s) ) + f(s) [by the definition]
(f + g)(s) + f(s) = f(s) + ( g(s) + f(s) ) [by the associativity axiom]
(f + g)(s) + f(s) = f(s) + (g + f)(s) (by the definition)

...

So simple. Thanks for your help.
 
Last edited:
  • #6
You make it more complicated than it is. Adding another f(s) as you have done is completely unneccessary. Commutativity in the function space follws directly from the commutativity of the real numbers under addition. Nothing else is needed.
 
  • #7
hmmmm.

It just seems to me like your method is simply proving commutativity by assuming commutativity, which intrinsically doesn't prove anything.

maybe I just don't understand.
 
  • #8
I'm telling you to assume cummutativity of the real numbers under addition, which is not what you are trying to prove. You are trying to prove commutativity of a functionspace under an "addition" rule defined by (f+g)(s)=f(s)+g(s). That is something completely different from real numbers being commutative under addition. Although, it is the case that this particular addition rule, together with the standard properties of real numbers, make it almost trivial to prove.

Assuming the standard properties of real numbers seems completely reasonable (to me) for a problem like this; of course you're always welcome to assume less. You could start from the Peano axioms if you really want to. . .
 
  • #9
Thanks for your patience, and I am definitely getting a better understanding of this now. I'm just new to vector spaces, and I just want to be sure I know what I can assume when dealing with these problems.

I think I'll completely understand if you show me how you would prove that an addition definition such as (f+g)(s)=f(s)+2g(s) is or is not commutative using the same method you used for proving the trivial definition (f+g)(s)=f(s)+g(s)
 
Last edited:
  • #10
Sure, no problem, first here's the original problem, in morbid detail, just to be clear:

[tex]
\begin{align*}
(f+g)(s) &= f(s) + g(s) \quad \text{definition of addition of \textbf{functions}} \\
& = g(s) + f(s) \quad \text{addition of \textbf{numbers} is commutative} \\
& = (g+f)(s) \quad \text{definition of addition of \textbf{functions}}
\end{align*}
[/tex]

As for proving something something is not commutative, it is usually just easiest to provide a counterexample. Let [tex]f(s) = s[/tex], and [tex]g(s) = s^2[/tex]. Then by your new rule, when, say, s = 2, we have

[tex](f+g)(2) = 2 + 2\cdot2^2 =10 \neq 8 = 2^2 + 2\cdot 2 = (g+f)(2)[/tex]

Therefore, since [tex](f+g)(s) \neq (g+f)(s)[/tex] in general, addition under this rule is not commutative.

As an aside, so that you don't get so bogged down in the details of proving vector space properties that you lose track of the essence of what vector spaces are about, let me make one remark. Vector spaces are places where objects combine linearly to give new objects that are also vectors. That is, if [tex]A[/tex] and [tex]B[/tex] are vectors in some vector space [tex]V[/tex], and [tex]\alpha[/tex] and [tex]\beta[/tex] are (real or complex) scalars, then [tex]\alpha A + \beta B[/tex] is also a vector in [tex]V[/tex]. Vector spaces are "closed" under taking linear combinations (so they are also sometimes called "linear spaces"), and this is their most important property.

In this specific case, the important kernel of information is that if you add differentiable functions together pointwise, or multiply them by real numbers, you still get another differentiable function.
 
  • #11
Excellent. Thank you very much for your help.
 
  • #12
Originally posted by BigRedDot
Sure, no problem, first here's the original problem, in morbid detail, just to be clear:


As an aside, so that you don't get so bogged down in the details of proving vector space properties that you lose track of the essence of what vector spaces are about, let me make one remark. Vector spaces are places where objects combine linearly to give new objects that are also vectors. That is, if [tex]A[/tex] and [tex]B[/tex] are vectors in some vector space [tex]V[/tex], and [tex]\alpha[/tex] and [tex]\beta[/tex] are (real or complex) scalars, then [tex]\alpha A + \beta B[/tex] is also a vector in [tex]V[/tex]. Vector spaces are "closed" under taking linear combinations (so they are also sometimes called "linear spaces"), and this is their most important property.


A linear space is not the same as a vector space. I'd avoid using that particular label for them.There is at least one definition of a linear space that is commonly used that would cause confusion. ( I mean the goemeter's one)
 
  • #13
MathWorld disagrees, and so does http://wombat.doc.ic.ac.uk/foldoc/foldoc.cgi?linear+space [Broken], and every other reference I could turn up. So do I. I've never once encountered any context where "linear space" and "vector space" were not complete synonyms.

I'm certainly willing to have my opinion informed by hearing about a linear space that is not a vector space, though.
 
Last edited by a moderator:
  • #14
I'm wondering if Matt Grime wasn't referring to a "linear manifold" (In, say, R3, all subspaces are planes or lines that contain the origin. A linear manifold would be a plane or line that does not necessarily contain the origin.

Another possibility is the "affine space": R1, R2, Rn can be taken as examples of affine spaces in which we have points rather than vectors. We do not, in general, define addition or scalar multiplication of points. Given any affine space, we can turn it into a vector space by choosing a particular coordinate system.

I, however, am with BigRedDot. I would not call either a linear manifold or an affine space a "linear space". I would identify "linear space" with "vector space".

Matt, the ball's in your court!
 

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and operations, such as addition and scalar multiplication, that follow certain properties. These properties include closure, associativity, commutativity, and the existence of an identity element.

2. How do you prove that a set is a vector space?

To prove that a set is a vector space, you need to show that it satisfies all the properties of a vector space. This includes showing that the set is closed under vector addition and scalar multiplication, and that it contains an additive identity element and a multiplicative identity element. You also need to show that the set follows the properties of associativity, commutativity, and distributivity.

3. What is the importance of proving a set is a vector space?

Proving that a set is a vector space is important because it allows us to use the tools and properties of vector spaces to solve problems or analyze systems in a more efficient and organized way. It also helps us to understand the underlying structure and properties of the set, which can lead to further insights and applications.

4. Can a set be a vector space if it does not contain the zero vector?

No, for a set to be a vector space, it must contain the zero vector. This is because the zero vector is the additive identity element and is necessary for the set to follow the property of closure under vector addition.

5. What is the difference between a vector space and a subspace?

A vector space is a set that satisfies all the properties of a vector space, while a subspace is a subset of a vector space that also satisfies all the properties of a vector space. In other words, a subspace is a smaller vector space that is contained within a larger vector space.

Similar threads

  • Differential Geometry
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
628
  • Linear and Abstract Algebra
Replies
7
Views
882
  • Differential Geometry
Replies
21
Views
589
  • Linear and Abstract Algebra
Replies
3
Views
184
  • Precalculus Mathematics Homework Help
Replies
2
Views
833
  • Calculus and Beyond Homework Help
Replies
2
Views
154
Back
Top