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Homework Help: Subspaces in R4

  1. Mar 7, 2006 #1

    Pengwuino

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    I'm so lost!

    1. W is the set of all vectors in R4 such that x1 + x3 = x2 + x4. Is W a subspace of R4 and Why?

    How do i get started here? i'm thoroughly confused on this whole idea of vector spaces and such.
     
  2. jcsd
  3. Mar 7, 2006 #2

    matt grime

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    What are the axioms you need to check, and what can you verify and how?
     
  4. Mar 7, 2006 #3

    HallsofIvy

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    I find looking at the definitions helpful in such a case.
     
  5. Mar 7, 2006 #4

    Pengwuino

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    So far we're looking at vector addition, scalar multiplication and the zero vector. Plugging in 0,0,0,0 produces a vector that is in the set. I'm not sure how to show vector addition however. Do i need to turn that equation into a function x1= x2 + x4 - x3?
     
  6. Mar 7, 2006 #5

    Hurkyl

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    What you need to do is to treat

    "(a, b, c, d) is in W"

    as being synonymous with

    "a + c = b + d"
     
  7. Mar 7, 2006 #6

    Pengwuino

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    Can i set a new vector equal to like (y1 + y3 = y2 + y4) and add the equations?
     
  8. Mar 7, 2006 #7

    Hurkyl

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    No. You need to treat

    "(a, b, c, d) is in W"

    as being synonymous with

    "a + c = b + d"


    So what, abstractly speaking, are you trying to show?
     
  9. Mar 7, 2006 #8

    matt grime

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    Take two elements that satisfy the condition, add them, does the result satisfy the condition?

    suppose i wanted to show that the subset V of R^2 defined by (x,y) in V if x=0, then let (x,y) and (s,t) be in V, the sum is (x+s,y+t), now is x+s zero when x and s are zero?
     
  10. Mar 7, 2006 #9

    Pengwuino

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    Ok so how can I show that they add up using two different vectors? I know for example, (1,3,4,2) are in W but how do i go about showing that all such vectors satisify vector addition and multiplication?



    I guess i'm trying to show that every vector that follows a + c = b + d is a vector in W. Is that right?
     
  11. Mar 7, 2006 #10

    Pengwuino

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    If I pick v = (1, 3, 4, 2) and u = ( 3, 6, 5, 2), they add up to be (4, 9, 9, 4) which is also in the set. I can also do cv, c being say, -3, to get cv = ( -3, -9, -12, -6) which is also in the set. How can i show that it works for all numbers?
     
  12. Mar 7, 2006 #11

    benorin

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    Formally, we have W is defined by

    [tex]W=\left\{ (x_1,x_2,x_3,x_4)\in\mathbb{R}^4 : x_1 + x_3 = x_2 + x_4 \right\}[/tex]

    The requirement [itex]x_1 + x_3 = x_2 + x_4 [/itex] may be written as [itex]x_4 = x_1 - x_2 + x_3 [/itex] and notice how once [itex]x_1, x_2, x_3 [/itex] are chosen, [itex]x_4[/itex] is determined, and so it would seem we only need three parameters to express points in W...

    If you need more, post on that wise.
     
  13. Mar 7, 2006 #12
    don't use numbers...use variables

    you are given a condition x1+x3=x2+x4, someone above
    said look at (a,b,c,d) and you yourself posted can i do x1=...
    NOW combine those 2 concepts to get a general vector satisfying your given condition. Note you should ask yourself how many unknowns should remain in your general vector
     
  14. Mar 7, 2006 #13

    matt grime

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    i am going to state categorically, and i'm sorry to do so, that you should ignore benorin.

    it's very easy, so stop making it difficult.

    let's do my example, since I refuse to answer your homework,

    Define W in R^2 by (x,y) is in W if and only if x=0.


    let's show it is a subspace.

    1. (0,0) is in W, clearly since the first coordinate is zero

    2. if (x,y) and (u,v) are in W the (x+u,y+v) is in W since x+u=0 because x=0 and u=0

    3. if (x,y) is in W then so is t(x,y)=(tx,ty) since x=0 implies tx=0

    and we are done.

    so how about it, now?

    You see how to do it?
     
  15. Mar 7, 2006 #14

    benorin

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    If [tex]\forall\alpha , \beta\in\mathbb{R},x,y\in W\Rightarrow (\alpha x+\beta y) \in W ,[/tex] then W is a subspace. That's it. Simple.

    Parts (1), (2), and (3) of your test correspond to the cases [tex]\alpha=\beta=0, \alpha=\beta=1, \mbox{ and }\alpha=t,\beta=0,[/tex] respectfully.
     
  16. Mar 8, 2006 #15

    matt grime

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    i meant it was this comment that should be ignored. i don't see what choosing and determing things has to do with proving its a subspace
     
  17. Mar 8, 2006 #16

    benorin

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    I was attempting to lead to the representation of vectors in W as vectors of the form (x1,x2,x3,x1-x2+x3) is all.
     
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