1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    (x1,y1)+(x2,y2)=(x1,0)
    c(x,y)=(cx,y)

    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

    radou

    User Avatar
    Homework Helper

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Vector space
  1. Vector spaces (Replies: 10)

  2. Vector space (Replies: 16)

  3. Vector space (Replies: 2)

  4. Vector spaces (Replies: 6)

  5. Vector space (Replies: 4)

Loading...