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Homework Help: Vector Subspace Question

  1. Oct 2, 2005 #1


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    I'm having a hard time understanding some things about vector subspaces.

    I have two problems that I am supposed to determine if the sets are subspaces of R^2.

    I know both sets are not subspaces but I dont really understand why or how to prove it for that matter. Can anyone shed some light for me?

    Problem 1

    {(x1,x2)^T | |x1| = |x2|}

    Problem 2

    {x1,x2)^T | x1^2 = x2^2}

    If anyone could help me with any part or both I would really appreciate it. I think I'm missing something relatively easily.
  2. jcsd
  3. Oct 3, 2005 #2


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    Do you know what the first set will look like? Perhaps you can see why the set of points such that |x1| = |x2| is the set of points such that x1 = x2 as well as the set of points such that x1 = -x2. So it's a union of two sets of points, one set being the set {(x,x) | x in R} and the other being {(x,-x) | x in R}. This is the set consisting of both the line y = x and the line y = -x. Now you should know that the only subspaces of R² are {0}, R², and any single line passing through the origin. What we have here is a union of two lines passing through the origin, and a union of two lines is not a single line, so it's not a vector subspace. How can you prove this? Well perhaps you can tell by looking at the graph what is wrong with this being a subspace. A subspace should be closed under addition. But this set contains both (x,x) and (x,-x), so if it is to be closed under addition, it should include (x,x) + (x,-x) = (2x, 0). Now for any x not equal to 0, you will see that this point is not in the set, so the set is not closed under addition, and hence isn't a subspace.

    The second problem is identical to the first (and I really mean identical), see if you can prove it.
  4. Oct 3, 2005 #3


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    As always with problems like these, look at the definitions.

    If V is a vector space and U is a subset of V then to be a subspace it must be true that:
    U is "closed" under addition: if x and y are in U then x+ y is in U.
    U is "closed" under scalar multiplication: if x is in U and a is a number then ax is in U.
    If x is in U then -x is in U.
  5. Oct 3, 2005 #4
    yeah, the trick is to think of an example that would demonstrate that those sets aren't closed under either addition or scalar multiplication.

    all you need to do is come up with one counterexample, and you're done!

    proving stuff false is easy! :biggrin:
  6. Oct 3, 2005 #5


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    Im thinking if the 2nd problem is just like the 1st problem, then the following will be true.

    (x^2,x^2) + (x^2, - x^2) = (2x^2, 0)

    So for all of X that doesnt equal 0, x1^2 wont equal x2^2.

    Am I correct?
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