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Vector space

  1. Oct 8, 2007 #1
    rather than use the standard definitons of addition and scalar multiplication in R^2, suppose these two operations are defined as follows.

    what would the zero vector be? can it be (0,-y1)? why or why not? I think it should fail this axiom u+0=u.
  2. jcsd
  3. Oct 8, 2007 #2


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    First of all, your addition, as defined, is not commutative, i.e. (x1, y1) + (x2, y2) = (x1, 0) [itex]\neq[/itex] (x2, y2) + (x1, y1) = (x2, 0).
  4. Oct 8, 2007 #3


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    That is simply not a vector space. Not only is addition not commutative, as radou said, worse, there simply is NO additive identity. The equation (x1,y1)+ (x,y)= (x1,0)= (0,0) simply has no solution (x,y). This set is not a group with that addtion operation.
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