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SImple Linear Algebra, Subspace Problem

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Let 'S' be the collection of vectors [x;y;z]
    in R3 that satisfy the given property. Either prove that 'S' forms a subspace of R3 or give a counterexample to show that it does not.

    |x-y|=|y-z|

    2. Relevant equations



    3. The attempt at a solution
    First I tested the 0 vector, and it passed that test. Now I have to test if it is closed under scalar multiplication and addition. This is where I am getting confused. I don't know if it is the absolute value signs that are confusing me, or if I am just generally unsure how to proceed. Thanks.
     
  2. jcsd
  3. Mar 1, 2010 #2
    Let (x1, y1, z1) and (x2, y2, z2) be in S. Then, S is closed under scalar multiplication if c(x1, y1, z1) is in S (for all scalars c). And S is closed under addition if (x1, y1, z1) + (x2, y2, z2) is in S.

    Of course, if (x, y, z) is in S, then |x-y| = |y-z|, so you must check if the resulting vectors are of this form to decide if they are in S.

    You might want to think about a few examples of vectors of this form.
     
  4. Mar 1, 2010 #3
    I understand that much, but I am confused how to apply that. My other examples were of a different form, and the way you described is how I did those others. For this one, I am confused how to apply that method.
     
  5. Mar 1, 2010 #4
    Think of a few examples of vectors of this form. Take a bunch of them and add them together, scale them, etc. Are they still in this form? If they are not, you have a counterexample. And even if they are still in this form, it gives you an idea of why that might be, which is useful for these types of problems.

    You may want to consider vectors like (5, 7, 9), etc, but also, if you already input x and y, then you have narrowed done the choices for z, so you can think of it as in some general form depending on x and y. And similar for the other variables.
     
  6. Mar 1, 2010 #5
    Yes. If you test 2(5,7,9) it still checks. If you do (5,7,9) + (6,8,10) it still checks.
     
  7. Mar 1, 2010 #6
    So consider the general forms. Can you factor scalar c out of |cx-cy| = |cy-cz|?

    For addition of vectors, take a general form such as (x1 + x2, y1 + y2, z1 + z2 and consider something like |(x1 + x2) - (y1 + y2)|. Use the individual properties of x and y being in S.
     
  8. Mar 1, 2010 #7
    Would it just be....c|x-y|=c|y-z|
    and..... |(x+x1)-(y+y1)|=|(y+y1)-(z+z1)|
     
  9. Mar 1, 2010 #8
    For your first line, you will need to take the absolute value of c.
    For the last part, that is what you want to prove, you cannot simply state that. You must use the properties that (x1, y1, z1) and (x2, y2, z2) are in S and thus satisfy that equation.
     
  10. Mar 1, 2010 #9
    Ok so the first line becomes... |c||x-y|=|c||y-z|?

    For the second part, I see what you are saying, but I am still confused on how to demonstrate that. Do I prove it with real numbers?
     
  11. Mar 1, 2010 #10

    Dick

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    You folks aren't being very creative here. (0,1,0), (1,2,3).
     
  12. Mar 1, 2010 #11
    Ok so (0,1,0), (1,2,3) shows that 'S' does not form a subspace since when those two vectors are added, the result is not in 'S'. I still need help in actually proving this though, without real numbers. Thanks.
     
  13. Mar 1, 2010 #12

    Dick

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    You don't need to prove it without real numbers. A single counterexample proves it not a subspace.
     
  14. Mar 1, 2010 #13
    Oh ok. So just by showing why (0,1,0), (1,2,3) don't work should suffice? Since it is proving that a counterexample exists?
     
  15. Mar 1, 2010 #14

    Dick

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    Well, yeah. Being a subspace means addition has to work for ALL pairs. A single pair that doesn't work proves the 'ALL' part is wrong. It's not a subspace.
     
  16. Mar 1, 2010 #15
    Ok got it. I will use that example and show that it does not work. Thanks!
     
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