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Homework Help: Linear Algebra - one to one and onto question

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine whether each of the following mappings f is onto or one-to-one. Is f an isomorphism?

    1) f maps R2 to R3 defined by f(x, y) = (x, y, x+y)
    2) f maps R3 to R(1x3) defined by f(x,y,z) = [x^2, y^2, z^2]
    3) f maps R4 to P2(R) defined by f(a,b,c,d) = a+(b-c)x+dx^2

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to do these.. I understand somewhat about one-to-one and onto, but these notations kind of confuse me.. For onto, do I just need to look at the outcome and see if it spans the space? And for one-to-one, I look at rather every component from the source appears in every component in the result?

    1) Not one-to-one, onto.
    2) Not one to one, onto.
    3) Not one to one, onto.
     
  2. jcsd
  3. Feb 7, 2010 #2

    Mark44

    Staff: Mentor

    You need to understand the definitions of these terms more than just somewhat. Go back and look at the definitions of one-to-one and onto, and any examples there are in your book.
    Pretty much. More precisely, if f is a map from U to V, f is onto V if, for any v in V, there is a u in U such that f(u) = v. IOW, no matter what thing you pick in the output space, there is a thing in the input space that maps to it. For example, if f:R --> R is defined by f(x) = x2, f is not onto, since -1 is in R (the output space), but there is no real number x in R (the input space) such that f(x = -1.
    That's not how one-to-one-ness is defined. One definition is that if a != b, then f(a) != f(b). Using the same function as my previous example, f is not one-to-one, since f(2) = f(-2).
     
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