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Homework Help: Help with norms

  1. Apr 11, 2010 #1
    If [tex]\lVert x \rVert=2[/tex] and [tex]\lVert y \rVert=3[/tex], what if anything, can we conclude about the possible values of [tex]\left\vert \mathbf{x}^T\mathbf{y} \right\vert[/tex]?

    I don't think anything can be concluded since the dot product can still end being positive or negative.
  2. jcsd
  3. Apr 11, 2010 #2
    Re: Norms

    Do you know the formula for the dot product involving cosine?
  4. Apr 11, 2010 #3


    Staff: Mentor

    Re: Norms

    x [itex]\cdot[/itex] y = ||x|| ||y|| cos([itex]\theta[/itex]).

    Can you conclude something about |x [itex]\cdot[/itex] y| now?
  5. Apr 11, 2010 #4
    Re: Norms

    [tex]u \cdot v=\lVert v \rVert\lVert u \rVert cos(\theta)[/tex]
  6. Apr 11, 2010 #5
    Re: Norms

    It is between 0 and 1, then?
    Last edited: Apr 11, 2010
  7. Apr 11, 2010 #6
    Re: Norms

    Theta is between -pi/2 and pi/2?
  8. Apr 11, 2010 #7
    Re: Norms

    Plug in all the values you know. Then consider the range of cosine. What values can it take? Knowing this, what values can ||x|| ||y|| cos theta take?
  9. Apr 11, 2010 #8
    Re: Norms

    For cosine to be positive, theta is between, and including, -pi/2 to pi/2. Therefore, the right side of equation will be between 0 to 1 times the magnitude of x times the magnitude of y?
  10. Apr 11, 2010 #9
    Re: Norms

    Yes, it is simpler to just write it out, however.

    xy = ||x|| ||y|| cos a = 6 cos a.
    cos a is between -1 and 1, so xy is in [-6, 6], and so |xy| is just the positive terms in that interval.
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