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Subspace problem

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine whether the following sets form subspaces of R2


    {(x1,x2)T|x1=3x2}

    So I rewrote the set in order for it to be homogenous ( i'm not sure why we would do that but I saw a problem saying if we can do it to do it.

    {(x1,x2)T|x1-3x2=0}

    So my logic when solving subspace problems is I want to make sure that what is on the right side is true, and will be closed under scalar multiplication and addition.

    using the matrix (1,1)T to check I get 1-3≠0 so it is not closed under addition.

    Not sure if I'm doing this correctly the concept is really confusing.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2013 #2

    Mark44

    Staff: Mentor

    Certainly x1 = 3x2 is equivalent to x1 - 3x2 = 0, so that's valid. I'm not sure how helpful it is, though.
    This makes very little sense.

    What you need to show is, that your set includes the zero vector, and, for any two vectors in your set, their sum is also in the set, and, for any scalar and any vector in your set, the scalar multiple is in the set.
    ??
     
  4. Feb 10, 2013 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    x1-3x2=0 isn't supposed to be true for ANY vector. It's only true for vectors in V={(x1,x2)^T: x1-3x2=0}. Suppose a=(a1,a2)^T is in V. What can you say about the relation between a1 and a2. Take b=(b1,b2)^T to be another point in V. a+b=(a1+b1,a2+b2)^T. You want to show that is also in V.
     
  5. Feb 10, 2013 #4
    So I kept staring at the problem a bit and realized that the left side will always be the same. So the right side must be manipulated in such a way to make it true under the two subspace rules.

    So let A=(3x2,x2)T and B=(3v2,v2)T is what the right side is saying now I am going to to check to make sure they follow the two rules.

    A is closed under scalar multiplication because (3(β)(x2),βx2)T
    Lies in the set, and is simply just a scalar multiple.

    if I add A+B I get (3(x2+v2),(x2+v2))T

    Which is closed under scalar addition.


    Is my reasoning correct?
     
  6. Feb 10, 2013 #5
    That's correct. Depending on how you learn it, I was taught you also have to show that the set is non-empty, ie, show that ##\vec 0## lives in the vector space, which is pretty trivial to show.
     
  7. Feb 10, 2013 #6

    Mark44

    Staff: Mentor

    We don't talk about the vectors being closed under these operations - we talk about the set they belong to being closed under scalar multiplication or vector addition.
    Start with a as you have defined it (capital letters are often used for matrices - you are dealing with vectors here) to show that βa is in the set.
    This is the idea, but you need to show what you're doing.

    a + b = (3x2, x2) + (3v2, v2) = ...

    Conclude that since a + b is also in the set, the set is closed under vector addition.
     
  8. Feb 10, 2013 #7
    To show that 0 lives in the vector space can I just show that it is homogeneous and thus has the trivial solution 0?
     
  9. Feb 10, 2013 #8

    Dick

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    Science Advisor
    Homework Helper

    The vector (0,0)^T is in the subspace because 3*0=0. That's all. You are thinking of this way too abstractly.
     
  10. Feb 10, 2013 #9
    Awesome thanks for the clarification guys.
     
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