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Math Help - Linear Math

  1. #1
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    Linear Math

    questions 1: the three vectors x, y, z have real-valued components and form the sides of a triangle. Prove the Law of Cosines for these vectors, i.e...that

    |z|^2 = |x|^2 + |y|^2 - 2*x*y = |x|^2 + |y|^2 - 2*x*y*cos(theta)

    where theta is the angle between vectors x and y. (can be using scalar product to solve)


    question 2: Let vector x = (1,1,0,0), y = (-1,1,0,0), z = ( 0,0, 1,1), t = (0,0,-1,1). Show that these vectors are linearly independent. Therefore, any subset of them is linearly independent. Discuss the space spanned by each pair of the vectors.


    anyone can help ?
    Last edited by linearmath; December 19th 2007 at 09:46 PM.
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  2. #2
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    1)
    |z|^2 = |x|^2 + |y|^2 - 2|x||y|cos(theta)

    Start from the following facts:
    |z|^2 = z \bullet z (definition)
    z = x-y (since the vectors make a triangle)

    2) Show that ax+by+cz+dt = (0,0,0,0) if and only if a=b=c=d=0. This is done by showing that the system of simultaneous equations has only this solution. You will discover this if you solve the system in the normal way.
    Last edited by badgerigar; December 18th 2007 at 06:45 PM. Reason: forgot modulus signs
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  3. #3
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    well, as you said I can read my notes. I just haven't seen any example of show vectors are linearly independent ( 4 vectors)
    Last edited by linearmath; December 19th 2007 at 09:48 PM.
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  4. #4
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    Quote Originally Posted by linearmath View Post
    question 2: Let vector x = (1,1,0,0), y = (-1,1,0,0), z = ( 0,0, 1,1), t = (0,0,-1,1). Show that these vectors are linearly independent. Therefore, any subset of them is linearly independent. Discuss the space spanned by each pair of the vectors.
    A set of vectors is linearly dependent if you can construct at least one member of the subset by a linear combination of the remaining vectors.

    So your question becomes four questions of the type:
    x = ay + bz + ct

    If no solution other than a = b = c = 0 exists then x is linearly independent from the rest. To prove that the whole set is linearly independent you have to do this for y, z, and t as well.

    -Dan
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  5. #5
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    If no solution other than a = b = c = 0 exists then x is linearly independent from the rest.
    If a = b = c = 0 is a solution then x = (0,0,0,0). My textbook says that any set containing this is considered linearly dependent. I think this is needed to keep all bases the same size? or maybe conventions differ?
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