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Triangle Inequality and Cauchy Inequality Proofs

  1. Oct 3, 2010 #1
    1. The problem statement, all variables and given/known data
    The question says to find a proof for Cauchy's Inequality and then the Triangle Inequality.

    This is an elementary linear algebra class I'm doing, so I can't use inner products or anything.
    2. Relevant equations

    3. The attempt at a solution
    I got the proofs using algebra, but I'm having trouble seeing them geometrically. I want to draw a picture for Cauchy's Inequality but can't quite visualize it.
    Last edited: Oct 3, 2010
  2. jcsd
  3. Oct 3, 2010 #2
    Edit: Misread the question. Seems like you are supposed to find both the algebraic and geometric proof for both the inequalities.
  4. Oct 3, 2010 #3
    Yeah, that's the way I read it too.

    But my proof for the first one is just that | A dot B| / (|A| dot |B|) = cos(x), and cos(x) ≤ 1, so |A dot B| ≤ |A| dot |B|.

    Given the definitions we got for this class, I don't see any way to prove this differently, unless I use inner products. I run in to the same problem with the other proof.
  5. Oct 3, 2010 #4
    | A • B | ≤ |A| |B|
    Algebraic: you’re right since cos is bound between 0 and 1 this shows this inequality must be true.
    Geometric: Think about what the dot product represents on a graph. It’s magnitudes of one vector projected onto another.

    | A + B| ≤ |A| + |B|
    Algebriac: Hint: Add |-a| <= a <= |a| and |-b| <= b <= |b|
    Geometric: This just means that no one leg of a triangle can be longer than the other two put together.
  6. Oct 3, 2010 #5
    For the geometric proof of | A • B | ≤ |A| |B|,

    I can use that Bcos(x) = (B dot A) / |A| = B dot n, where n is a unit vector in the direction of A. That's a projection.

    By definition n = A / |A|, so B dot (A / |A|) = Bcos(x) --> B dot A = ABcos(x), but that just brings me back to the proof I did first, so is it really any different?
  7. Oct 3, 2010 #6
    IMHO that's more of an algebraic proof. In this context I think geometric means proof by picture rather than from the axioms of geometry.
  8. Oct 3, 2010 #7
    I guess I'm having trouble seeing how you can prove it using a picture.

    I was trying to use a picture similar to the one on http://en.wikipedia.org/wiki/File:Dot_Product.svg

    But since |A dot B| and |A||B| are numbers, how can you show them on the diagram?
  9. Oct 3, 2010 #8
    What geometric shape would be represented |A||B|?
    What is geometricly represented by that picture.
  10. Oct 3, 2010 #9
    The picture geometrically represents the projection of A onto B, and that projection is A dot B. Is that right?

    But |A||B| is a number, how can it have a geometric interpretation?
  11. Oct 3, 2010 #10
    Don’t over think it, If I asked you to draw a picture of 3*5, or maybe build it with blocks, how would you do it?
  12. Oct 3, 2010 #11
    A vector with length 15?

    But how would you know which direction the vector would point?
  13. Oct 3, 2010 #12
    Don't think of |A|*|B| in terms of vectors, |A|*|B| is just some number, |A| is just some number, |B| is just some number. Who cares where we got them from.

    How do they teach little kids to understand multiplication? They always compare it to some shape.

    Now ask your self could the projection of A onto B ever be larger than that shape?
  14. Oct 3, 2010 #13
    A groups of B things?
  15. Oct 3, 2010 #14
    3*5 is often represented as a rectangle of height 3 length 5.
  16. Oct 3, 2010 #15
    Ah, I see it now. The rectangle describing |A dot B| will always have a smaller width than the other one, because the projection of B on A is smaller than B itself.

    Unless the angle is zero, in which case they're equal. Thanks for all your help!
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