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How is the following operator linear?

  1. Jun 9, 2012 #1
    How is the following operator linear??!

    1. The problem statement, all variables and given/known data

    Is the following a linear vector function?

    F(r) = r - ix


    2. Relevant equations

    A function is linear if:

    F(r1 + r2) = F(r1) + F(r2) AND F(ar) = aF(r)

    3. The attempt at a solution

    F(r1 + r2) = (r1 + r2) - ix

    F(r1) + F(r2) = (r1 - ix) + (r2 - ix) = (r1 + r2) - 2ix


    According to me, F(r1 + r2) ≠ F(r1) + F(r2), but Mary L Boas says otherwise. What gives?
     
    Last edited: Jun 9, 2012
  2. jcsd
  3. Jun 9, 2012 #2
    Re: How is the following operator linear??!

    Well you would agree that f(x) = mx + b is linear, right?
    But f(x1 + x2) != f(x1) + f(x2)
     
  4. Jun 9, 2012 #3
    Re: How is the following operator linear??!

    I'm not sure I understand your point? Yes, f(x) = mx + b is the function of a line, but it is NOT a linear function, since f(x1 + x2) ≠ f(x1) + f(x2). I'm not sure where you're getting at.

    According to the Mary L Boas book, the operator I posted is supposed to be a linear operator, which means f(r1 + r2) = f(r1) + f(r2) and f(kr) = kf(r). My results show otherwise, since I find f(r1 + r2) != f(r1) + f(r2).
     
  5. Jun 9, 2012 #4

    micromass

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    Re: How is the following operator linear??!

    f(x)=mx+b is not linear.

    OP: could you quote the entire problem. What is [itex]\mathbf{r}[/itex]? What is x?? What is [itex]\mathbf{i}[/itex] (I suspect it is just the first basis vector)?

    Could it be that [itex]\mathbf{r}=(x,y,z)[/itex]??
     
  6. Jun 9, 2012 #5
    Re: How is the following operator linear??!

    i is the first basis vector, and r has been defined earlier in the book as (x,y,z), or ix + jy + kz.

    I must be missing something, because it seems (according to your post) that r being (x,y,z) would change my solution, but I don't see how.
     
  7. Jun 9, 2012 #6
    Re: How is the following operator linear??!

    I still find f(r1+r2) = ix + 2jy + 2kz

    f(r1) +f(r2) = 2jy + 2kz
     
  8. Jun 9, 2012 #7

    Mark44

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    Re: How is the following operator linear??!

    You aren't distinguishing between the components of r1 and r2.

    Let r1 = (x1, y1, z1), and r2 = (x2, y2, z2).

    Now calculate f(r1 + r2) and compare that to f(r1) + f(r2).
     
  9. Jun 9, 2012 #8
    Re: How is the following operator linear??!

    Is this just a question about the multiple definitions of linear?

    y=Ax+b is linear in one sense, and not linear in another sense, right?
     
  10. Jun 9, 2012 #9

    vela

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    Re: How is the following operator linear??!

    Right. y=mx+b is the equation of a line in the xy-plane, so it's often called linear in that sense. It is not, however, linear in the linear algebra sense, which is what the OP's question is about.
     
  11. Jun 9, 2012 #10

    Ray Vickson

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    Re: How is the following operator linear??!

    What gives is that your x needs to be tied to your r, so if you have r1 and r2 you need x1 and x2.

    RGV
     
  12. Jun 9, 2012 #11

    micromass

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    Re: How is the following operator linear??!

    You may want to write your F as

    [tex]F(x,y,z)=(0,y,z)[/tex]

    Do you agree that this is your F?? Do you see that this is linear??
     
  13. Jun 9, 2012 #12
    Re: How is the following operator linear??!

    To make it clear, I think the proper expression of your linear operator is this:

    [itex]\underline F(a) = a - (a \cdot i) i[/itex]

    which clearly is linear.
     
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