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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.

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2 Answers 2

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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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. Looks like the comment exchange reached an end. If you need to revisit some of them, you will find them in that chatroom. $\endgroup$ May 18, 2018 at 12:32
  • $\begingroup$ With the help of José Carlos I've answered the question and begun to understand at least a little better the idea behind vector spaces. I've also edited my post to reflect this and so it can be seen for anyone who stumbles across the problem in the future. Thanks all $\endgroup$
    – Jake S
    May 18, 2018 at 20:59
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jake S
    May 17, 2018 at 20:37
  • $\begingroup$ @JakeS Yes, the fact that a typical element of $V$ cannot be represented by a couple of real numbers. $\endgroup$ May 17, 2018 at 20:40
  • $\begingroup$ 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. $\endgroup$
    – Jake S
    May 17, 2018 at 20:47
  • $\begingroup$ @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. $\endgroup$ May 17, 2018 at 20:59
  • $\begingroup$ 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? $\endgroup$
    – Jake S
    May 17, 2018 at 21:07

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