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Intersect of U and U perpendicular; Orthogonality

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Let U be a subspace of ℝn. Show that if u[itex]\in[/itex]U[itex]\bigcap[/itex]U[itex]\bot[/itex], then u=0.


    2. Relevant equations



    3. The attempt at a solution
    I know that U[itex]\bot[/itex] will be orthogonal to U, so any vector u in U dotted with any vector in U[itex]\bot[/itex] will equal 0. But that does not necessarily mean that u = 0 so I must prove something else. I don't know where to include the intersect into all of this, we have never used that kind of thing before.
     
  2. jcsd
  3. Nov 5, 2011 #2
    Well, you've almost got it. Think about u^2 and remember that u is in Rn.
     
  4. Nov 5, 2011 #3
    u2? u is a vector, so I'm guessing you're insisting I think about u dot u... I'm a little confused how that helps me here
     
  5. Nov 5, 2011 #4

    Deveno

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    well one way to look at u.u is to think of "the left u" as being in U, and the "right u" as being in U (which we can do since u is in the intersection).
     
  6. Nov 5, 2011 #5
    Well if I do it that way then yes I will get the dot product is 0, because those two are orthogonal. I don't see how that makes u = 0, though. I'm missing something here
     
  7. Nov 5, 2011 #6

    Deveno

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    review the properties a dot product has to have....(positive definite is the property you're after)
     
  8. Nov 5, 2011 #7
    Yes, but also cancel out other cases using the orthogonal relation. Intuitively, if I rotate my u vector to make it orthogonal. What are the point(s) which are shared by both these vectors?
     
  9. Nov 5, 2011 #8
    Just looked through my textbook and lecture notes and could not find any property like that, do you think you could explain it?

    And now I know that u = 0 because it will be the only vector that is in U and U[itex]\bot[/itex] because, well obviously, they are at a 90° angle. I just don't understand how I'm supposed to show it.

    Actually come to think of it, can I use the formula [tex] cosθ = x[itex]\cdot[/itex]y/||x|| ||y|| [/tex] and plug in θ=pi/2, showing that u[itex]\cdot[/itex]w = 0 , so that u must be equal to the 0 vector? This answer seems a bit trivial...
     
  10. Nov 5, 2011 #9
    Deveno was probably alluding to the orthogonal inner product space. <x,y>=x.y. An inner product space must never be negative. Also, your method above is correct. Either, x OR y are the 0 vector. But in our example, that is u and u satisfy, u=0 and u=0.
    If we write out u=[u1,...,un] and use the fact that u.u=0 and solve for each ui
     
  11. Nov 5, 2011 #10

    Deveno

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    if we are talking about the standard dot product in Rn, then

    if u = (u1,u2,....,un)

    u.u = (u1)2 + (u2)2+...+(un)2

    now, if u.u = 0, we have:

    0 u.u = (u1)2 + (u2)2+...+(un)2

    one hopes that you know that the square of a real number is non-negative.

    the sum of two (or more) non-negative numbers is also ______?

    therefore.....

    *********

    you CANNOT conclude that just because u.w = 0 that one of u or w must be 0. for example, (0,-1).(1,0) = 0(1) + (-1)(0) = 0, but neither of these vectors is the 0-vector.

    and, geometrically, saying that the 0 vector is at "a right angle" to something, just doesn't make any sense. how do you choose θ?
     
  12. Nov 6, 2011 #11
    Didn't mean 0 vector is at a right angle. What I meant is that if two vectors are orthogonal (perpendicular), the only point they are going to have in common is 0 i.e. (0,0,0) in R3. And I chose θ to be pi/2 because that is a 90° angle and therefore the two vectors would be orthogonal... Is this logic not correct? I assumed this is what shaon0 was getting at. The thing is that I don't know how to "show" this, like the question asks. I can put it into words but not into equations.
     
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