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One more Linear Transformation

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    I've posted a few of these recently. I have one question about this problem -- hopefully my calculations are correct.

    f: R2 to P1, f(a,b)=b+a2x

    Is this a linear transformation?

    2. Relevant equations
    f(u+v) = f(u) + f(v)
    f(cu) = cf(u)

    where u and v are vectors in R2 and c is a scalar.

    3. The attempt at a solution
    let u=(u1,u2)
    let v=(v1,v2)

    f(u+v) = ((u2+v2) + (u12x+v12x))

    f(u) + f(v) = (u2 + u12x) + (v2 + v12x)

    So f(u+v) = f(u) + f(v) (First condition met)

    c(u1,u2) = (cu1,cu2)

    f(cu) = cu2 + cu12x

    cf(u) = c(u2 + u12x)
    = cu2 + cu12x

    Second condition met.

    However, my question is -- The question asks is this a linear transformation from R2 to p1. If my calculations are correct, both conditions are met, but the polynomial has a degree of 2, not 1, so it's not a linear transformation from R2 to p1.
     
  2. jcsd
  3. Dec 3, 2015 #2
    The squared u1 and v1 will have a c^2 coefficient, not c. So second condition is not met.
    if v1 -> c*v1
    then v1^2 -> c^2*v1^2
     
  4. Dec 3, 2015 #3

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    The way you compute ##f(u+v)## is not correct: in general, ##(u_1+v_1)² \neq {u_1}²+{v_1}²##.
    This is not correct, you should have a ##c²{u_1}²x## in the right hand side expression.
    The polynomial has a degree of 1 in ##x##.
     
  5. Dec 3, 2015 #4
    I see what you mean with the c2 on the right hand side of the equation, I forgot to put a parenthesis around it (cu12)x

    Just looking at the first part now. Thanks for your replies.
     
  6. Dec 3, 2015 #5
    f(u+v) = [(u2+v2)+(u1+v1)2x]

    I can see now (with @DuckAmuck's reply) that the second condition is not met.

    Oh yes, you are correct with the polynomial having a degree of 1 as well! I got my u,v and x variables mixed up there!

    I'm still making a few mistakes, but I'm a lot more confident with these now than I was a couple of days ago.
     
  7. Dec 3, 2015 #6

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Working that out will show that the first condition isn't met either.

    The never ending joy of learning. :smile:
     
  8. Dec 3, 2015 #7
    f(u+v) = [(u2+v2)+(u1+v1)2x]

    f(u) + f(v) = u2 + u12x + v2 + v12x

    That condition isn't met either! Totally skipped over that one.
     
  9. Dec 3, 2015 #8

    Mark44

    Staff: Mentor

    Pretend that you're trying to convince an idiot, and show why the first line above isn't equal to the second line.
     
  10. Dec 3, 2015 #9
    *Convincing myself*

    f(u+v) = [(u2+v2)+(u1+v1)2x]

    f(u) + f(v) = u2 + u12x + v2 + v12x

    (u1+v1)2 ≠ u12 + v12
     
  11. Dec 3, 2015 #10

    Mark44

    Staff: Mentor

    Better:
    f(u+v) = (u2+v2)+(u1+v1)2x = ##u_2 + v_2 + (u_1^2 + 2u_1v_1 + v_1^2)x##
    f(u) + f(v) = u2 + u12x + v2 + v12x = ##u_2 + v_2 + (u_1^2 + v_1^2)x##
    Since ## u_1^2 + v_1^2 \ne u_1^2 + 2u_1v_1 + v_1^2##, if ##u_1 \ne v_1##, then f(u + v) ##\ne## f(u) + f(v)
     
    Last edited: Dec 3, 2015
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