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Damm Proofs

  1. May 5, 2006 #1
    Suppose you had two vectors X and Y which are elements in in a n dimensional real space. Now why is |X+Y|<|X|+|Y|, i've been trying to understand spivak's proof but it boils down to why is |X||Y|>sum of(xy) where xy is the product of the individual components...
  2. jcsd
  3. May 6, 2006 #2
    If the norm is derived from an inner product (as, it would appear, in your case), there is the following standard geometric argument for the latter inequality.

    Consider the following self-evident fact: [tex]\Vert x + ty \Vert^2 \ge 0[/tex], where [tex]t[/tex] is a parameter. Now, the left-hand side is a quadratic function in [tex]t[/tex], i.e. a parabola, but the inequality says that this parabola has to lie above the x-axis (with at most one real zero).
    (If it were to have two real zeros, there would be a distinct interval where the parabola would go below zero, thus contradicting our initial inequality).

    The determinant of the quadratic function above is just [tex]\left<x, y\right>^2 - \left\Vert x \right\Vert^2 \left\Vert y \right\Vert^2[/tex], and expressing the fact that the parabola has no two distinct zeros is done by saying that this determinant should be smaller than or equal to zero, which is the inequality you need.
  4. Jul 17, 2006 #3


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    but the first ineqality is the rather obvious "triangle inequality", that says one side, namely X+Y, of a triangle cannot be longer than the sum of the lengths of the other two sides, namely X and Y.

    the second inequality is also easily seen from the law of cosines, that says

    X.Y = |X||Y|cos(t) where t is now the angle between the vectors X and Y. Thus obviously (since cos is never greater than 1), we have
    |X.Y|≤ |X||Y|.
  5. Jul 22, 2006 #4
    True, but (1) try giving a sound proof for the triangle inequality, and (2) the law of cosines is usually a consequence of the triangle inequality: i.e. one defines the angle between two vectors as the arccos of X.Y / |X||Y| (which is smaller than one and hence in the domain of arccos).
  6. Jul 22, 2006 #5

    matt grime

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    You can prove the law of cosines quite simply by using a little geometry: it is just the cosine rule for triangles.

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