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So I have a final tomorrow and I have no clue how to determine whether a set is vector space or not. I've looked online on how to do these proofs but I still don't understand how to do them. Can any one help me with a question like this?

Let V be the set of all positive real numbers. Determine whether V is a vector space with the operations below.

$x + y = xy$

$cx=x^c$

If it is, verify each vector space axiom; if not, state all vector space axioms that fail.

Edit: Turns out I'm going to fail the exam based on what you guys are saying.

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  • $\begingroup$ Hint: it is a vector space. Which one of the axioms are you having difficulty with? $\endgroup$ Feb 14, 2014 at 21:13
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    $\begingroup$ It helps if you write the (alleged) vector space operations as something else than $+$ and juxtaposition, so you won't get confused about what is really addition/multiplication and what is the vector space operation. So, say, define $\oplus \colon V^2 \to V$ by $x \oplus y := xy$ and scalar multiplication $\otimes \colon {\mathbb R} \times V \to V$ by $c \otimes x := x^c$. $\endgroup$ Feb 14, 2014 at 21:14
  • $\begingroup$ @JustinCampbell Perhaps if you could show me how to do it with Closure under addition [(u+v) is in V] I may be able to do all of them. $\endgroup$
    – Mark
    Feb 14, 2014 at 21:15
  • $\begingroup$ @Mark, that appears to be a wholly different problem. The OP is closed under $+$ because the product of two positive real numbers is a positive real number. $\endgroup$
    – vadim123
    Feb 14, 2014 at 21:16
  • $\begingroup$ @vadim123 I don't understand. $\endgroup$
    – Mark
    Feb 14, 2014 at 21:17

5 Answers 5

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Choose another notation $x \oplus y := xy$ and $c \otimes x := x^c$. Then the exponential map gives an isomorphism of structures $(\mathbb{R},+,*) \cong (\mathbb{R}^+,\oplus,\otimes)$. Since the first is a vector space, the same is true for the latter. And this way the creator of this "exercise" came up with this artificial vector space (he wanted that you waste your time with computations ...).

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  • $\begingroup$ It's not the fact, that I have to do all the axioms. It's that I should know how to do them. I really don't know how to do any of the axioms really. $\endgroup$
    – Mark
    Feb 14, 2014 at 21:22
  • $\begingroup$ You haven't read my answer carefully. Anyway, you should have an idea how to prove $x \oplus (y \oplus z) = (x \oplus y) \oplus z$. $\endgroup$ Feb 14, 2014 at 21:24
  • $\begingroup$ If you really want to do it the hard way: just write the axioms down (using $\oplus$ and $\otimes$) and then expand the definition (of $\oplus$ and $\otimes$). $\endgroup$ Feb 14, 2014 at 21:24
  • $\begingroup$ @MartinBrandenburg I don't get what you're saying. $\endgroup$
    – Mark
    Feb 14, 2014 at 21:28
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    $\begingroup$ I think the OP will understand nothing because he doesn't make any effort to think. No offense, but this is my observation. $\endgroup$ Feb 14, 2014 at 22:38
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Here is a start.

  • Additive axioms: For every $x,y,z \in \mathbb{R^+}$,

i) $x+y=xy = yx = y+x\, $ ( since real numbers commute)

ii) $(x+y)+z=xy+z = xyz = x+yz= x + (y+z).$

Can you continue?

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  • $\begingroup$ I'm starting to get it now. But what about $x+0 = x$? That doesn't quite workout because wouldn't it be $(0)x= x^0 = 1$? $\endgroup$
    – Mark
    Feb 14, 2014 at 21:49
  • $\begingroup$ Am I missing something here? $\endgroup$
    – Mark
    Feb 14, 2014 at 21:50
  • $\begingroup$ $0 \not\in V$; the neutral element with respect to the 'addition' on $V$ is $1$. $\endgroup$ Feb 14, 2014 at 22:12
  • $\begingroup$ @Magdiragdag Can you show me how to do the additive identity? $\endgroup$
    – Mark
    Feb 14, 2014 at 22:16
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    $\begingroup$ I like Mhenni Benghorbal's response because it's simple. I'm trying to understand what you're saying but I only learn well by examples. $\endgroup$
    – Mark
    Feb 14, 2014 at 22:37
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You need to translate the usual vector space axioms, expressed using '+' and '$\cdot$', into your problem's notation. For example, the vector space axiom $$c\cdot(x+y)=c\cdot x+c\cdot y$$becomes$$(xy)^c=(x^c)(y^c)$$ As other comments have indicated, the closure axioms need to be dealt with. If $x,y\in V$, then you need to show that $xy$ is in $V$, since $xy$ is the 'vector sum' of $x$ and $y$.

One last point: do you know about vector space isomorphisms? There is an isomorphism between your vector space and a much more familiar vector space (call it $W$). If $\psi:V\rightarrow W$ stands for this isomorphism, then we have $$\psi(xy)=\psi(x)+\psi(y)$$for example.

By the way, you didn't not explicitly state what possible values $c$ could have.

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  • $\begingroup$ I was never taught anything about vector space isomorphisms in my linear algebra course. The reason I didn't state the possible values of $c$ is because that question is literally in my textbook so its not explicitly stated. $\endgroup$
    – Mark
    Feb 14, 2014 at 21:32
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Lucky I got the same question in one of my Assignments. And here's what I got:

http://www.math.tamu.edu/~dallen/linear_algebra/chpt3.pdf

The same question in here!

Hope it helped :)

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    $\begingroup$ While that link may answer the question, it's better to include the main ideas here, as the content could become inaccessible later. $\endgroup$ Oct 5, 2014 at 8:15
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What about closure under scalar multiplication? For example, if c = 1/2 and x = -1, we have: 1/2*(-1) = (-1)^(1/2) which is not real. So, it's not closed under scalar multiplication, correct? Therefore, not a vector space.

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  • $\begingroup$ Note than in the OP $V$ is the set of all positive real numbers. $\endgroup$
    – Ludolila
    May 25, 2014 at 16:40

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