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Why y=2x+1 is not a vector space?

  1. Nov 3, 2007 #1
    I'm trying to solve a problem ice109 recommended. I'm trying to show how y=2x+1 is not a vector space. Here I go.

    Let u=(x,2x+1) v=(x',2x'+1) w=(x",2x"+1)

    1. If u and v are objects in V, then u + v is in V.

    u+v=(x+x',2x+2x'+2) fails because 2 is not in V?

    2. u + v = v + u

    passes by inspection

    3. u + (v + w) = (u + v) + w

    also passes by inspection

    4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u+ 0 = u for all u in V

    If x=0, the zero vector is (0,1)?

    (0,1)+(x,2x+1)=(x,2x+2) which is not u, so it fails this test.

    5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.

    It passes obviously

    6. If k is any scalar and u is any object in V, then ku is in V.

    I don't know about this one... I don't know the meaning of this axiom...

    7. k(u + v) =ku + kv

    k[(x,2x+1)+(x',2x'+1)]=(kx,k2x+k)+(kx',k2x'+k) passes

    8. (k + m)u = ku + mu

    k(x,2x+1) + m(x,2x+1)=(k+m)(x,2x+1)=[(k+m)x,(k+m)2x+(k+m)]


    9. k(mu) = (km)(u)


    10. 1u=u

    I can say for sure it fails axiom 1 and 4, but how do I check to see if it fails 6?
  2. jcsd
  3. Nov 3, 2007 #2


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    It'll help if you're a bit more strict with your notation. So, we have the set V={(x,y):y=2x+1} and want to see if it's a vector space or not.

    Axiom 4: The zero vector is not (0,1) but is the standard zero vector (0,0). Is this in V?
    Axiom 5: The vector -u is the vector (-x,-y). Is this in V?
    Axiom 6: ku=(kx,ky)=(kx,k(2x+1)). Is this in V for all k?
    The other axioms after this hinge on 6.

    Note that, in order to say that V is not a vector space, you only need to find one axiom that it does not satisfy.
  4. Nov 3, 2007 #3
    I agree, there cannot be a vect of (0,0), so it fail axiom 4. I believe (-x,-y) is in V because (-x,-y)+(x,y) for any real numbers, should =0. So if this condition passes 7,8,9,10 then it cannot fail axiom 6? Can you eplain further? So if it fails axiom 6, it fails the rest? Axiom 6 is very abstact to me, I cannot get its meaning...
  5. Nov 3, 2007 #4


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    Axiom 4: There are two things that a zero vector must satisfy here. Firstly, there must exist a 0 s.t. 0+u=u+0=u for all u in V. Secondly, the zero vector must be in V.

    Axiom 5: Again, there are two points here. Firstly, there exists a -u s.t. u+(-u)=0, and secondly that -u is in V. -u=-(x,y)=(-x,-y)=(-x,-2x-1). Is this in V?

    Axiom 6: ku is in V. Well, this says that, for all k, ku=k(x,y)=k(x,2x+1)=(kx,2kx+k) is in V. Is this true?

    Finally, I meant that if axiom 6 does not hold, then 7 through 9 cannot, since they depend on the definition of ku.
  6. Nov 3, 2007 #5
    axiom 6 cannot work if k is any real number, is k any real number? if k=0 then ku is not in V. I'm still having trouble with axiom 5. Why is -u not in V?
    Last edited: Nov 3, 2007
  7. Nov 3, 2007 #6


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    I wanted to comment on this problem when ice109 posted it, but decided not to.

    The question whether or not the set S = {(x,y) in R^2 : y=2x+1} = {(x,2x+1) in R^2 : x in R} doesn't make sense. To have a vector space, we need to specify a scalar field and two operations. What are they in this case?

    It is true that S is not a subspace of R^2 under the usual definitions of addition and scaling (because it doesn't contain (0,0) -- that's all you need to say). But who is to say that S cannot be made a vector space under some other operations?

    Exercise: Give suitable definitions of addition and scalar multiplication on S that make it a vector space (over R).
  8. Nov 3, 2007 #7


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    Well, yes, k is a real number, so in general, the point (kx,2kx+k) does not satisfy the given condition.
    -u= (-x,-2x-1)=(x',2x'-1), on letting -x=x'. This is clearly not in V.

    And yes, to reply to morphism's point, one does need to define addition and multiplication in V. However, when there is no definition of these operations, I at least assume the standard operations.
  9. Nov 4, 2007 #8


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    When showing that an axiom fails, give a specific counter-example.

    For example axiom 1... let u = (0,1) and v = (0,1)

    both u and v belong to the set described because (0,1) satisfies the equation y = 2x + 1.

    But u + v = (0,1) + (0,1) = (0,2)

    Does (0,2) satisfy the equation y = 2x + 1... left side comes out to 2... right side comes out to 1. left side does not equal right side... so (0,2) does not satisfy the equation... so (0,2) does not belong to the given set... so u + v does not belong to the given set.

    so axiom 1 is violated...

    for axioms 2, 3 pass.

    axiom 4 is violated. For axiom 4... I'd recommend the extra step of showing that the zero vector if it exists must be (0,0)... Then since (0,0) does not belong to the described set, axiom 4 is violated... ie:

    let a = (x1,y1) be a vector such that u + a = u where u is an object in the described set...

    I can write u = (x2,y2)

    u + a = u

    (x1+x2, y1+y2) = (x2,y2)

    x1+ x2 = x2 => x1 = 0.
    y1+y2 = y2 =>y1 =0.

    so if the described object has a zero vector, then it must be (0,0) (this is the only possibility for a zero vector).

    But (0,0) does not satisfy y = 2x + 1. so it does not belong to the described set. So the described set has no zero vector...

    So axiom 4 is violated.

    For axiom 5... actually axiom 5 is ill-defined now because there's no zero vector for the set (since axiom 4 is violated)... If we ignore that and if we define -u by the standard operation ie: if u = (a,b) and if we define -u as (-a,-b)... then take a specific example u = (0,1). so -u = (0,-1).... -u = (0,-1) does not satisfy the equation y = 2x + 1.

    we can't say that axiom 5 passes or fails... axiom 5 simply doesn't make sense now.

    Axiom 6 is violated... u = (0,1) belongs to the set. But 2u = (0,2) does not belong to the set (check and see that (0,2) does not satisfy the given equation).

    Axioms 7, 8, 9 and 10 pass. I think cristo's point is that in 7, 8 and 9 we may no longer be dealing with objects in the described set... ie ku may not be in the described set since axiom 6 is violated... but the axioms only state that u and v need to belong to the set... not that ku or kv must belong to the set.
    Last edited: Nov 4, 2007
  10. Nov 4, 2007 #9
    recommend me a good book guys, if you guys know of a good book a person like me can gain great insights from. A good book that can help me understan mathematical concepts like these. Anything that can complement my Linear Algebra book, "Elementary Linear Algebra" by Howard Anton, or replace it when it comes to explainations. Thanks.
  11. Nov 4, 2007 #10

    matt grime

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    You get 'great insights' by working through the theory and becoming very adept at it. There are no great mysteries in mathematics, just hard work.
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