# Proving vector spaces

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)]

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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

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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.

I'm sorry if the answer is really obvious, but could you give me an example anyway?

Thanks.

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.

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.

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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.

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.

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. . .

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)

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Sure, no problem, first here's the original problem, in morbid detail, just to be clear:

\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*}

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

$$(f+g)(2) = 2 + 2\cdot2^2 =10 \neq 8 = 2^2 + 2\cdot 2 = (g+f)(2)$$

Therefore, since $$(f+g)(s) \neq (g+f)(s)$$ 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 $$A$$ and $$B$$ are vectors in some vector space $$V$$, and $$\alpha$$ and $$\beta$$ are (real or complex) scalars, then $$\alpha A + \beta B$$ is also a vector in $$V$$. 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.

Excellent. Thank you very much for your help.

matt grime
Homework Helper
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 $$A$$ and $$B$$ are vectors in some vector space $$V$$, and $$\alpha$$ and $$\beta$$ are (real or complex) scalars, then $$\alpha A + \beta B$$ is also a vector in $$V$$. 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)

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.

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