# Proving a structure is a vector space

I understand that in order to confirm whether or not a given structure is a vector space, I must check to see if it holds to a number of properties that define vector spaces, among the most important of which are that the structure must be closed for vector addition and scalar multiplication.

I will try not to overload this question by proving each individual property, but I would like to know if I'm heading in the right direction.

EDIT I have edited to include proofs for each property.

$$V$$ = {$$f: \mathbb R \to \mathbb R$$}, $$(f+g)(x) = f(x) + g(x), \forall x \in \mathbb R, \forall f,g \in V,$$

$$(λ \cdot f) (x) = f(λx), \forall x, λ \in \mathbb R, \forall f \in V.$$

I first test:

(1) the set $$V$$ is closed under vector addition, that is, $$v+w \in V$$

By definition, $$f(x) \in V+ g(x) \in V = (f+g)(x) \in V$$ thus the set is closed under addition.

(2) vector addition is commutative, $$v+w=w+v$$

$$f(x) + g(x) = (f+g)(x) = (g+f)(x) = g(x) + f(x)$$

(3) vector addition is associative, $$(v+w)+u=v+(w+u)$$

$$(f(x)+g(x))+h(x) = (f+g)(x)+h(x)$$

$$(f+g)(x)+h(x)=f(x)+(g(x)+h(x))=f(x)+(g+h)(x)=f(x)+g(x)+h(x)$$

(4) there is a zero vector $$0_v \in V$$ such that $$v + 0_v = v$$ $$\forall v \in V$$

$$f(x) + (0 \cdot g)(x) = f(x) + g(0x) = f(x) + 0 = f(x)$$

(5) each $$v \in V$$ has an additive inverse $$w \in V$$ such that $$w + v = 0_v$$

$$f(x) + (-1 \cdot f)(x) = f(x) + f(-1x) = f(x) + (-1)f(x) = f(x) - f(x) = 0$$

(6) the set V is closed under scalar multiplication, that is, $$r \cdot v \in V$$

The definition provides this.

(7) addition of scalars distributes over scalar multiplication, $$(r+s) \cdot v = r \cdot v + s \cdot v$$

$$(a+b) \cdot f(x) = ((a+b)\cdot f)(x) = a \cdot f(x) + b \cdot f(x) \neq f(a \cdot x) + f(b \cdot x)$$

Property (7) does not hold.

(8) scalar multiplication distributes over vector addition, $$r \cdot (v+w) = r \cdot v + r \cdot w$$

$$(a \cdot f)(x) = f(ax) = a \cdot f(x)$$

$$f(x) + g(x) = (f+g)(x)$$

$$a \cdot (f+g)(x) = (a \cdot (f+g))(x) = (f+g)(ax) = f(ax) + g(ax) = a \cdot f(x) + a \cdot g(x)$$

(9) ordinary multiplication of scalars associates with scalar multiplication, $$(rs) \cdot v = r \cdot (s \cdot v)$$

$$a(b \cdot f)(x) = a \cdot f(bx) = (a \cdot f)(bx) = f((ab)x) = f(x) \cdot (ab)$$

(10) multiplication by the scalar 1 is the identity operation, $$1 \cdot v = v$$.

$$(1 \cdot f)(x) = f(1x) = f(x)$$

CONCLUSION

The addition of scalars does not distribute over multiplication (Property (7) does not hold). $$V$$ is not a vector space.

No, you are not heading in the right direction. The elements of $V$ are the functions from $\mathbb R$ to $\mathbb R$. Where are they? When you try to prove that $V$ is a vector space, you starting talking about matrices with two rows and one column, instead of functions.
Your attempt disregards completely the kind of objects the elements of $V$ are. You can't just write down the proof that $\Bbb R^2$ is an $\Bbb R$-vector space whenever you want to prove that some $V$ is a vector space.
• I apologize, I forgot to include that my question defines all of the structures as $\mathbb R$-vector spaces. Nonetheless I am clearly not understanding something. May 17, 2018 at 20:37
• @JakeS Yes, the fact that a typical element of $V$ cannot be represented by a couple of real numbers. May 17, 2018 at 20:40
• I understand (that I've related things which are not related). By "a typical element of $V$ cannot be represented by a couple of real numbers," are you referring to the fact that I represented $x$ and $f(x)$ as a matrix? I was looking at an example in my textbook of how the author proves a given subset is a vector space, but it is a line in $\mathbb R^2$. Could you provide as an example the proofs of the first property I've used for my problems? I think once I've seen how to start working out the exercise I should be able to proceed. May 17, 2018 at 20:47
• @JakeS Oh was that supposed to be $x$ and $f(x)$? It kind of makes more sense but it is still wrong. An element of $V$ is a function from $\Bbb R$ to $\Bbb R$, period. There is no particular way to represent such a function when it's arbitrary. Just give it a name. So say, let $f$ and $g$ be two functions, we have to check that $f+g$ is a new function. May 17, 2018 at 20:59
• OK, I understand. So if $f$ and $g$ are functions, then by the definition of my problem $f(x) + g(x)$ = $(f+g)(x)$ and since $f, g \in V$ this new function also belongs to $V$ and thus the set is closed under vector addition? May 17, 2018 at 21:07