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So I'm doing a review test and I have this problem:

Prove in full detail, with the standard operations in R2, that the set

{(x,2x): x is a real number}

is a vector space.

Attempt:

Given:

$(x_1, 2x_1) \in \mathbb{R}^2$

and

$(x_2, 2x_2) \in \mathbb{R}^2$

Addition:

$(x_1, 2x_1) + (x_2, 2x_2) = (x_1 + x_2, 2x_1 + 2x_2) \in \mathbb{R}^2$

Thus the set is closed under addition

Scalar multiplication:

$c(x_1, x_2) = (cx_1, cx_2) \in \mathbb{R}^2$.

Thus the set is closed under scalar multiplication


Are these operations enough to prove that the set is a vector space? Or do I have to go through each of the following (or in other words do I have to to the same thing for each property in the definition):

enter image description here

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6 Answers 6

up vote 4 down vote accepted

Write $$V=\{(x,2x)\mid x\in\mathbf{R}\}.$$

Here the main thing is to note that $(x_1,2x_1)+(x_2,2x_2)$ and $c(x_1,2x_1)$ [edit]a bad typo was here - it used to read c(x_1,2x_2)[/edit] can both be written in the form $(y,2y)$ for some $y\in\mathbf{R}$, in other words that they are elements of $V$. You have not shown that $V$ is closed under addition and scalar multiplication unless you do that.

As Vhailor pointed out, once you do this, you get the vector space axioms for free, because the set $V$ inherits them from $\mathbf{R}^2$, which is (hopefully) already known to you to be a vector space with respect to these very operations. So, to fix your proof, show that

1) $(x_1,2x_1)+(x_2,2x_2)\in V$ for all $x_1,x_2\in\mathbf{R}$.

2) $c(x,2x)\in V$ for all $x\in\mathbf{R}$.

Look up subspace criteria or something like that.

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So in other words I have to do what I said in Vhailor's first comment right? –  Virtuoso Jul 6 '11 at 4:13
    
Yes. Figuring out what $x_3$ should be will do part 1. For part 2) try to figure out, what $x'$ should be for the equation $c(x,2x)=(x',2x')$ to be true. Once you have done these, then you know that $V$ is closed under these operations. –  Jyrki Lahtonen Jul 6 '11 at 4:21
    
@Virtuoso the key phrase Prove in Full Detail suggests to me that the test wants you to show all the vector space axioms, you might be marked down on an exam if you prove by showing its a subspace. –  crasic Jul 6 '11 at 5:07
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@crasic: Hmm. As a teacher I would accept a proof referring to a theorem in the textbook as full detail. In this case a theorem giving a useful subspace criterion. I mean, when doing a geometry problem in `full detail' you are not expected to start from scratch and, say, prove Pythagoras' theorem along the way before you can use it. –  Jyrki Lahtonen Jul 6 '11 at 5:14
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@crasic: You're right. Cannot hurt to check with the professor. Occasionally they/we want the students to verify a set of axioms. Normally that would happen in a different setting, where other means are not available. That's why I, too, resort to those funny operations on $\mathbf{R}$, where you shift the origin somewhere else, or use an exponential mapping so that your vector space consists of positive numbers only. –  Jyrki Lahtonen Jul 6 '11 at 5:43
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Since you are working in a subspace of $\mathbb{R}^2$, which you already know is a vector space, you get quite a few of these axioms for free. Namely, commutativity, associativity and distributivity.

With the properties that you have shown to be true you can deduce the zero vector since $0 v=0$ and your subspace is closed under scalar multiplication, and same thing for the inverse, $-1 v=-v$.

You seemed to have skipped a few steps in your reasoning though, after doing the addition of two vectors from your subspace, in order to show that the resulting vector actually is in that same subspace you should show explicitly that it is of the form $(x,2x)$ (you're almost there).

For scalar multiplication, you seem to have taken a generic vector of $\mathbb{R}^2$ instead of a vector belonging to your subset, so it needs a bit of correction as well.

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So in other words these two are enough then, correct? (yes or no). For the addition, how am I supposed to show that the resulting vector is of the form (x,2x)? Can I say x1 + x2 = x3, and just substitute that back in so it looks like (x,2x)? For the scalar mult. please show me how I am supposed to do it, I don't know how to correct it. –  Virtuoso Jul 6 '11 at 2:51
    
@Virtuoso Typically you are not allowed to use theorems and concepts that have not been covered up to that point in class. If you are unfamiliar (i.e. it hasn't been covered yet) with the concept of a subspace then you should show all the axioms. Since a subspace is a vector space in its own right, you only need to prove that this set constitutes a subspace of $\mathbb{R}^2$ - it contains 0, closed under addition, and closed under scalar multiplication. –  crasic Jul 6 '11 at 3:09
    
Yes I've learned about subspaces. But you still haven't answered my questions. How am I supposed to fix what you said is wrong about my proof? Can you give me some equations please?? –  Virtuoso Jul 6 '11 at 3:18
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My interpretation is that the following level of detail should be given.

"Properties $2$, $3$, $7$, $8$, $9$, $10$ follow easily from the fact that they hold for $\mathbb{R}^2$." That is all one needs to say about these. But it must be said.

We now deal with the remaining axioms.

What you wrote for $1$ is not bad, but to be totally explicit it should have ended with $$(x_1,2x_1)+(x_2,2x_2)=(x_1+x_2, 2x_1+2x_2)=(x_1+x_2,2(x_1+x_2)).$$ So the sum of our two vectors is of the shape $(u,2u)$, and therefore is in our set.

It is handy to give our set a name, such as $U$, because we will be referring to it often.

What you ended that part of the argument with, namely $\in\mathbb{R}^2$ could be a problem. Of course the thing is in $\mathbb{R}^2$, but that's not what needs to be shown. The fact you wrote it down could be interpreted as indicating confusion about what it means to be a subspace.

Next, for $4$, we need to show that the $0$ vector is in $U$. So all we need to do is to show it has the right shape. This is easy, $(0,0)=(0,2(0))$. The properties of the zero vector do not need proof, they are inherited from $\mathbb{R}^2$.

For $5$, we need to show that the ordinary additive inverse of an element of $U$ is in $U$. So look at the additive inverse of $(x,2x)$. It is $(-x,-2x)$, which is equal to $(-x,2(-x))$, so it is in $U$. Everything else in $5$ is inherited.

Finally, $6$. We need to show that if $(x,2x)\in U$ then $c(x,2x) \in U$. This is just as easy as all the others: $$c(x,2x)=(cx,c(2x))=(cx,2(cx))$$ and obviously (love that word) $(cx,2(cx))\in U$.

Your proposed handling of scalar multiplication was not good. What needs to be shown is that $U$ is closed under scalar multiplication, and there was no apparent attempt to do that.

As soon as what actually needs to be verified is clear, the verifications themselves are unchallenging. What is being tested in the problem is whether you fully know the meaning of subspace.

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Alternatively show the isomorphism $\rm\ (1,2)\:\mathbb R\: \cong \mathbb R\ $ via the linear map $\rm\: r\mapsto r\:(1,2)$

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Sorry I don't understand the notation, could you explain a little? Thanks. –  user38268 Jul 6 '11 at 3:46
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Dear Bill, the question asks to prove that $(1,2)\mathbb{R}=\{(x,2x):x\in\mathbb{R}\}$ is a vector space. A priori you cannot speak of linear maps and isomorphisms (of vector spaces) if you do not know/have not proven that $(1,2)\mathbb{R}$ is a vector space; this is especially true if you are using linear maps to "prove" that it is a vector space. On the other hand, the assertion is correct as it stands but if the assertion itself is the answer (e.g., an additional remark), then this should be made clear. –  Amitesh Datta Jul 6 '11 at 6:43
    
@Amitesh: I agree with your comment. It would however be possible to argue by transport of structure: i.e., that the given operations of addition and scalar multiplication on $V$ are exactly those obtained by "transporting" the standard operations on $\mathbb{R}$ via the given bijection $(x,2x) \mapsto x$. That would work and still save a lot of time verifying the axioms, although if I ever had an actual linear algebra student give an answer like this, I would send her immediately into the next course. (And for such an argument the word "linear" seems out of place, as you say.) –  Pete L. Clark Jul 6 '11 at 8:05
    
@Pete: Dear Pete, I completely agree and am familiar with transport of structure but Bill's usage of the words "isomorphism" and "linear map" suggest he was thinking about something else. I would approve of the answer if he used "bijection" and "map" instead of "isomorphism" and "linear map", respectively, and then mentioned that the algebraic structure on $(1,2)\mathbb{R}$ is induced by that of $\mathbb{R}$ via the map in question (as you have already pointed out, of course). However, the high quality of most of Bill's answers on this website suggests that this answer was just a minor lapse. –  Amitesh Datta Jul 6 '11 at 8:38
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@Amitesh and downvoter: The idea I had in mind was is essentially as Pete has elaborated above. This is a more conceptual and less brute-force way of deducing the vector space structure. As you may have inferred by now, often my hints and answers do require students to think more deeply about the matter. But the reward in insight is usually well worth the effort. As always, I recommend asking questions before casting a downvote. –  Bill Dubuque Jul 6 '11 at 12:40
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If you have reached that point, your book should have a theorem which says that basically, "subspaces of vector spaces are vector spaces" at which point you only have to prove your set is closed under scalar multiplication and vector addition, which you have done.

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Lots of necro-posting going on, but - strangely - not a lot of new information is added with each new answer. –  The Chaz 2.0 Sep 16 '11 at 4:53
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If I got the question right, the idea is that the set of real numbers is closed under the operation of addition and multiplication, i.e. if $z_1 \in \mathbb{R}, z_{2} \in \mathbb{R}$, then $z_1 +z_2=y \in \mathbb{R}$. You do not have to prove other axiom, because they follow from these three (closed under addition, scalar multiplication and includes the origin).

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