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Linear Algebra - Is this set a vector space [Easy?]

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Does this set describe a vector space?

    Te set of all solutions (x,y) of the equation 2x + 3y = 0 with addition and multiplication by scalars defined as in R^2.

    2. Relevant equations

    Associativity of addition
    u + (v + w) = (u + v) + w.

    Commutativity of addition
    v + w = w + v.

    Identity element of addition
    There exists an element 0 ∈ V, called the zero vector, such that v + 0 = v for all v ∈ V.

    Inverse elements of addition
    For all v ∈ V, there exists an element w ∈ V, called the additive inverse of v, such that v + w = 0. The additive inverse is denoted −v.

    Distributivity of scalar multiplication with respect to vector addition  
    a(v + w) = av + aw.

    Distributivity of scalar multiplication with respect to field addition
    (a + b)v = av + bv.

    Compatibility of scalar multiplication with field multiplication
    a(bv) = (ab)v [nb 3]

    Identity element of scalar multiplication
    1v = v, where 1 denotes the multiplicative identity in F.

    3. The attempt at a solution

    Could not find one counterexample. Since I obviously can not go through every solution to prove this is a valid vector space, there must be a counterexample or some equation (that I'm not aware of) that proves that every solution of this equation is a valid vector space.
    Last edited: Oct 19, 2009
  2. jcsd
  3. Oct 19, 2009 #2


    Staff: Mentor

    You don't need to check every solution of the equation 2x + 3y. All you need to do is to show that all 10 of your axioms are satisfied. Each solution of the equation is a vector (x, y). For example, (3, -2) is one solution.

    Every solution (x0, y0) is such that 2x0 + 3y0 = 0.

    Is there a solution that acts as the additive identity? Yes, (0, 0).
    Is addition commutative for your solutions? I.e., if u and v are solutions, is u + v = v + u?

    And so on, for all of the axioms.
  4. Oct 19, 2009 #3
    But aren't there infinitely many solutions for that equation?

    If I come up with 1 or 2 random solutions to the equations, and prove all 10 axioms with them, that doesn't necessarily mean that the next 1 or 2 random solutions will be not counterexamples. Am I wrong?

    I say 1 or 2 because some axioms require two solutions.
    Last edited: Oct 19, 2009
  5. Oct 19, 2009 #4
    Have you seen proof methods before?

    It's technically called universal generalization in logic, but if you show that the axioms are satisfied for a GENERAL solution, (Mark44 picked [tex]\left(x_0,y_0\right)[/tex]) then they are satisfied for ALL solutions.

    Also, some of the axioms are existential proofs that require you to show that something with a certain property exists...which means you must be specific in these cases.

    For example, showing closure requires you pick a general solution. On the other hand, the existence of an additive identity requires you to show that a specific object exists.
    Last edited: Oct 19, 2009
  6. Oct 19, 2009 #5
    Ok, I will look that up.

    I was just a little suspicious since there are about 40 problems that require you to do the same for different rules and all solutions to odd exercises had counterexamples.

  7. Oct 19, 2009 #6
    Here's a hint: check closure first. Usually that's what fails in these types of problems.

    For instance, if


    was your equation instead, it's fairly simple to show that closure fails.
  8. Oct 19, 2009 #7
    I forgot about this part (quote from my book):

    "...to disprove a general rule, all we need is one counterexample. On the other hand, to prove a general rule is more difficult, we must prove it for all cases, which we usually do by algebraically providing an arbitrary typical case with letter symbols..."

    You implied that this is true for any case, I just have to figure out a proof without a specific case.

    Edit: I'll try out your hint.
  9. Oct 19, 2009 #8
    Yup. That's what Mark44 started with. For closure under scalar multiplication, assume that [tex] \left(x_0,y_0\right) [/tex] is a solution, then take a look at [tex] c \left(x_0,y_0\right) [/tex].

    Closure under addition is very similar, it involves you assuming another solution...and then...

    Just a side note: this is from an intro linear algebra class, right? When you get to the chapter on nullspaces, come back and look at this problem...because you're actually proving the nullspace of the associated matrix is a vector space. But that's for later!
  10. Oct 19, 2009 #9
    Oh, haha, Mark you were right. It is valid for every solution (I didn't read your reponse correctly). Thanks.

    You too union.
  11. Oct 20, 2009 #10


    User Avatar
    Science Advisor

    Since it is well known that [itex]R^2[/itex], the set of ordered pairs with coordinate-wise addition and scalar multiplication is as vector space, and this is just a subset of that, you are really only proving that it is a subspace of [itex]R^2[/itex].

    That is, you already know that addition and scalar multiplication are associative, comutative, etc. because that is true of [itex]R^2[/itex] in general.

    All you really need to do to show this is a subspace is show that it is closed under addition and multiplication. If (x, y) is such that 2x+ 3y= 0, is that also true of (ax, ay) for every scalar a? If (x, y) and (x', y') are such that 2x+ 3y= 0 and 2x'+ 3y'= 0, is that also true of (x+ y, x'+ y')?
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