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Show that U intersection W does not equal ((0,0,0))

  1. Oct 13, 2008 #1
    show that U intersection W does not equal ((0,0,0)) and hence that U+W is not a direct sum

    U being (a,0,a) and W being (c,d,c+2d) and we know that c d and a are elements of R, why are there not a direct sum? is there a rule that prevents any of a c or d from being the zero vector? if this is the case obviously its easy to solve???
     
  2. jcsd
  3. Oct 13, 2008 #2

    CompuChip

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    It is only a direct sum if they have no elements in common, that is: you can only write every element from U+W in the form u+w (with u in U and w in W) in one and only one way.
    This follows from the first part, namely: that the intersection is not empty. Because, if x is an element in the intersection it can be written both as x + 0 and 0 + x, so the decomposition is not unique.

    To show that it is non-empty, you only need to give a single non-zero element that is in both, which is not hard.
     
  4. Oct 13, 2008 #3
    sorry that went right over my head, can someone dumb it down a bit please?
     
  5. Oct 13, 2008 #4
    To show that [tex]U\cap W\neq\{(0,0,0)\}[/tex], you need to find some other vector (x,y,z) which belongs in both U and W where at least one of x, y, or z is not zero.

    In other words, if [tex](a, 0, a) = (c,d,c+2d)[/tex], what equations can you set up to solve for a, c, and d? Do all three variables have to equal 0?
     
  6. Oct 13, 2008 #5
    Just a note: unless the above is a typo, it seems that you have a pretty severe misconception about vectors. a, c, and d are elements of [tex]\mathbb{R}[/tex], i.e. scalars. Thus none of them can "be the zero vector." The zero vector in this case (since we are working in [tex]\mathbb{R}^3[/tex]) is the vector (0, 0, 0).
     
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