1. Limited time only! Sign up for a free 30min personal 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!

How can I prove this function is a linear function?

  1. Jul 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove f(r) = A [itex]\bullet[/itex] (r - kz) is a linear operator, but I don't understand why it is linear.

    2. Relevant equations
    A function of a vector is called linear if:
    f(r1+r2) = f(r1) + f(r2) and f(ar) = af(r)


    3. The attempt at a solution

    f(r1+r2) = A [itex]\bullet[/itex] (r1+r2 - kz) = A[itex]\bullet[/itex]r1 + A[itex]\bullet[/itex]r2 - A[itex]\bullet[/itex]kz

    However, f(r1) + f(r2) = A[itex]\bullet[/itex]r1 - A[itex]\bullet[/itex]kz
    + A[itex]\bullet[/itex]r2 - A[itex]\bullet[/itex]kz

    So, I just proved that the function is NOT a linear function, how great... :(

    By the way, r = (x,y,z) in this book, if that changes anything.
     
  2. jcsd
  3. Jul 20, 2012 #2
    Is it safe to assume this is an error in my book's manual? I even tried developing the vectors, here's my attempt. (Disregard the 3.7.1 part)
     

    Attached Files:

  4. Jul 20, 2012 #3
    Ok, upon further investigation, the only way this can be linear is if I have to apply the -kz part BEFORE writing down my r vectors, which means

    f(r1+r2) = (a1,a2,a3) [itex]\bullet[/itex] (x1+x2,y1+y2,0)

    If that was the case, how in the hell was I supposed to know I had to do that (as opposed to what I did in my scanned solution)??!
     
  5. Jul 20, 2012 #4

    Zondrina

    User Avatar
    Homework Helper

    I'm assuming by you saying r is a vector, you're saying r[itex]\in[/itex]ℝn. Notice that what you're trying to prove is f is closed under addition and scalar multiplication which implies f is linear.

    Suppose that : r1,r2[itex]\in[/itex]ℝn

    Then the way you would go about proving something like f(r1+r2)=f(r1) + f(r2) is to literally ADD the two vectors together since adding two objects together from the same field should result in something from the same field ( If it doesn't then you simply don't have anything to prove here ) and THEN you perform the operator f on the NEW object. Simply separating the newly formed object into its two founding objects is a simple task at this point which should result in your answer.
     
  6. Jul 20, 2012 #5
    k is the unit vector in the z direction, and r is given by:

    r = x i + y j + z k

    Therefore, f (r) = A [itex]\bullet[/itex] (x i + y j)

    Alternately,

    f(r1) = A[itex]\bullet[/itex] (r1 - k z1)

    f(r2) = A[itex]\bullet[/itex] (r2 - k z2)

    Try it now.
     
  7. Jul 21, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Note: [itex]z[/itex] is the third component of [itex]\textbf{r}[/itex], so if you have [itex]\textbf{r}_1[/itex] and [itex]\textbf{r}_2[/itex] you need [itex]z_1[/itex] and [itex]z_2[/itex].

    RGV
     
  8. Jul 21, 2012 #7
    Yeah thanks everyone. I was assuming the kz wasn't the "same z" as the z component from the r vectors, which means I considered r1-z as (x1,y1,z1)-(0,0,z) = (x1,y1,z1-z).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook