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Verifying whether something is a vector space or not

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello, here is the question:

    "Rather than use the standard definitions of addition and scalar multiplication in R3, suppose these two operations are defined as follows. With these new definitions, is R3 a vector space? Justify your answers.

    a) (x1, y1, z1) + (x2, y2, z2) = (0,0,0)
    c(x, y, z) = (cx, cy, cz)

    b) (x1, y1, z1) + (x2, y2, z2) = (x1 + x2 + 9, y1 + y2 + 9, z1 + z2 + 9)
    c(x, y, z) = (cx, cy, cz)

    c) (x1, y1, z1) + (x2, y2, z2) = (x1 + x2 + 6, y1 + y2 + 6, z1 + z2 + 6)
    c(x, y, z) = (cx + 6c - 6, cy + 6c - 6, cz + 6c - 6)"


    2. Relevant equations
    Here are the axioms we are supposed to use:
    Assume u, v, w are in set V; let c and d be scalars.
    1. u + v is in V (Closure under addition)
    2. u + v = v + u (Commutative property)
    3. u + (v + w) = (u + v) + w (Associative property)
    4. V has a zero vector 0 such that for every u in V, u + 0 = u. (Additive identity)
    5. For every u in V, there is a vector in V denoted by -u such that u + (-u) = 0. (Additive inverse)
    6. cu is in V (Closure under scalar multiplication)
    7. c(u + v) = cu + cv (Distributive property)
    8. (c + d)u = cu + du (Distributive property)
    9. c(du) = (cd)u (Associative property)
    10. 1(u) = u (Scalar identity)

    3. The attempt at a solution
    Disclaimer: I don't know how to check the properties mechanically, but a forum moderator insisted that I show work.

    Our professor didn't get the opportunity to go over any examples with us, so I'm not even sure where to begin.

    b)
    Verification of property 1:
    (x_1 + x_2 + 9, y_1 + y_2 + 9, z_1 + z_2 + 9); all components are real, so property 1 fits.
    Verification of property 2:
    Switching the order of the terms makes components equivalent.
    Verification of property 3:
    Associating terms makes components equivalent.
    Verification of property 4:
    (x_1, y_1, z_1) + (0, 0, 0) = (x_1 + 9?, y_1 + 9?, z_1 + 9?)
    So it fails the additive identity? Not sure.
    Verification of property 5:
    (x_1, y_1, z_1) + (-x_1, -y_1, -z_1) = (x_1 - x_1 + 9?, y_1 - y_1 + 9?, z_1 - z_1 + 9)
    So it also fails the additive inverse? Not sure.
    Verification of property 6:
    Looks like this checks out. Not sure how to show work.
    Verification of property 7:
    Looks like this checks out. Not sure how to show work.
    Verification of property 8:
    Looks like this checks out. Not sure how to show work.
    Verification of property 9:
    Looks like this checks out. Not sure how to show work.
    Verification of property 10:
    Looks like this checks out. Not sure how to show work.
     
  2. jcsd
  3. Mar 26, 2015 #2

    Mark44

    Staff: Mentor

    To verify this property, you need to show that if u and v are arbitrary vectors in R3, then u + v is also in that vector space.
    You need to show this. And you need to show it for each of the following properties, or else show that a property fails to comply.
    Yes, this property fails. u + 0 should turn out to be equal to u, which isn't the case here.
    Yes, this one fails. If you add u and -u, what should you get?
    If c is an arbitrary scalar, and u is an arbitrary vector, is cu in the vector space? Yes or no?
    Since scalar multiplication has the usual definition for part b, there shouldn't be any surprises.
     
  4. Mar 26, 2015 #3

    Fredrik

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    You should write it all out:
    \begin{align}
    &(x_1,y_1,z_1)+(x_2,y_2,z_2)=(x_1+x_2+9,y_1+y_2+9,z_1+z_2+9)\\
    &=(x_2+x_1+9,y_2+y_1+9,z_2+z_1+9) =(x_2,y_2,z_2)+(x_1,y_1,z_1).
    \end{align} Edit: I just noted that I had typed a bunch of nonsense here. (I thought for a moment that we were writing an arbitrary vector as ##(x_1,x_2,x_3)## rather than ##(x_1,y_1,z_1)## and got the two notations mixed up). I have fixed the calculation so that it makes sense now.

    This way you're demonstrating that you are using the definition, and that this addition operation is commutative because the addition operation ##\mathbb R## is commutative.

    You're supposed to find out if there's a vector z such that x+z=x for all vectors x. You have only proved that (0,0,0) isn't such a z. So you have only checked one vector out of infinitely many. (Almost everyone who asks about this type of problem here makes this exact mistake).

    Your attempt to check property 5 has a similar problem. Note that property 5 doesn't make sense if there's no zero vector, so you can only check property 5 if property 4 turns out to hold.

    Just use the definitions. I'll do 10:
    $$1(x,y,z)=(1x,1y,1z)=(x,y,z).$$
     
    Last edited: Mar 27, 2015
  5. Mar 26, 2015 #4

    Dick

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    You should stay open to the possibility that (0,0,0) isn't the additive identity under your modified addition rule. But there may still be one, in fact there is. I think Fredrik has already hinted at that.
     
    Last edited: Mar 26, 2015
  6. Mar 26, 2015 #5
    Okay, let me try this again:
    Part b)
    Verification of 4:
    v_1 + v_2 = v_1
    (x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1, y_1, z_1)

    So, v_2 could be (-9, -9, -9)?
    Therefore, property 4 holds?

    Verification of 5:
    v_1 + (-v_1) = 0
    The additive inverse could just be any vector in the set based on the properties. Is this correct?
    So could -v_1 be (-x_1 - 9, -y_1 - 9, -z_1 - 9)?
    Does that mean property 5 also holds?

    I'm still missing why this isn't a vector space though. Which property fails if the scalar multiplication hasn't changed?
     
    Last edited: Mar 27, 2015
  7. Mar 27, 2015 #6

    Mark44

    Staff: Mentor

    According to the definition of vector addtion in part b, the left side above is ##(x_1 + x_2 + 9, y_1 + y_2 + 9, z_1 + z_2 + 9)##. So what does v2 have to be so that the above is equal to ##(x_1, y_1, z_1)##?
    You tell us. Have you found a vector that passes for the zero vector?
    You are not using the definition of vector addition as given in part b.
    Property 5 has to hold for all vectors in the space.
     
  8. Mar 27, 2015 #7
    I still don't understand. It still looks like (-9, -9, -9) satisfies it.
    ##(x_1, y_1, z_1) + (-9, -9, -9) = (x_1 - 9 + 9, y_1 - 9 + 9, z_1 - 9 + 9)##
    Does this use the definition we were given?

    I'm getting frustrated, because I do not understand. It seems like two people here are telling me that the additive identity doesn't have to be the zero vector. Yet, you seem to be telling me that it has to be a zero vector? Which is it? Why is it the way it is?

    Please. Please, enlighten me. I don't understand how I'm not using the definition, because to me, it seems like I am.

    So, was I wrong? I can't tell by what you said.

    Thank you.
     
  9. Mar 27, 2015 #8

    Mark44

    Staff: Mentor

    Yes. So (-9, -9, -9) acts as the "zero" vector.

    What I said was "passes for the zero vector." That doesn't mean that it has to be literally (0, 0, 0).

    You have to show that for any vector v1, v1 + (-v1) = 0. Can this happen? Why or why not. Expand the left side according to the formula you are given for vector addition.
    See above.
     
  10. Mar 27, 2015 #9

    Fredrik

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    An identity element of a binary operation ##\star## on a set ##X## is an ##e\in X## such that ##x\star e=e\star x=x## for all ##x\in X##. A binary operation can never have more than one identity element, because if ##e## and ##f## are both identity elements, we have ##f=f\star e=e##. So once we have proved that a binary operation has an identity element, it's always safe to talk about "the" identity element.

    You showed that (-9,-9,-9) is an identity element of the addition operation in problem b). So (-9,-9,-9) is the identity element of this addition operation.

    The "zero vector" in a vector space is the identity element of the associated addition operation. It's not clear at this point whether (-9,-9,-9) can be called "the zero vector", because we still don't know if we're dealing with a vector space. But it's clear that (0,0,0) shouldn't be called "the zero vector" here. It's the zero vector of a vector space that's irrelevant to this problem. If the set we're dealing with turns out to be a vector space, its zero vector is going to be (-9,-9,-9), not (0,0,0).

    Edit: I realized that I had written some nonsense in the calculation in post #3. I have fixed it now, so check it out again.
     
    Last edited: Mar 27, 2015
  11. Mar 27, 2015 #10

    Fredrik

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    Yes.

    No. This would make ##v_1+(-v_1)\neq (-9,-9,-9)##.

    The axioms that involve only addition seem to hold. The ones that involve only scalar multiplication must hold, since it's the standard scalar multiplication operation on ##\mathbb R^3##. So perhaps you should be looking at the two axioms that involve both addition and scalar multiplication.
     
    Last edited: Mar 27, 2015
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