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Help understanding what is/is not a linear transformation from R2->R3

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine whether the following are linear transformations from R2 into R3:

    2. Relevant equations
    a) L(x)=(x1, x2, 1)^t
    b) L(x)=(x1, x2, x1+2x2)^t
    c) L(x)=(x1, 0, 0)^t
    d) L(x)=(x1, x2, x1^2+x2^2)^t

    3. The attempt at a solution
    To show L is a linear transformation, I need to be able to show:
    1. L(a*x1+b*x2)=aL(x1)+bL(x2);
    2. L(x1+x2)=L(x1)+L(x2);
    3. L(a*x1)=aL(x1);

    By looking and playing around with this, I can see how d is not a transformation, since if I let a=-1 and b=-2, then rule 3 does not hold. If I'm wrong, please correct me.
    But as far as a, b, and c, it looks like all three rules hold. The appendix says a is not a transformation, but I'm not sure why. It seems to me that a also satisfies the 3 conditions, what am I missing? Is there an easier way to do this? Can someone explain this to me like I'm two years old?
     
    Last edited: Mar 22, 2009
  2. jcsd
  3. Mar 22, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Do you really think L(x1+x2)=L(x1)+L(x2) holds for a)? I mean try x1=(0,0) and x2=(0,0).
     
  4. Mar 22, 2009 #3
    I don't see it.
    I thought x1 and x2 would be single values such as x1=0 and x2=0, not coordinates in two space such as (0,0). Is it the case if I'm going from R2->R2 and not R2->R3? I have very few examples in my class notes (my professor spends most of the class time on proofs, not examples) but both examples I do have has a single value for x1, x2, so on.


    If I try L(x1+x2)=L(x1)+L(x2) for x1=(0,0) and x2=(0,0) do I get (0,0,2)=(0,0,1)+(0,0,1)? The only thing I'm sure about is that I'm probably way off.
     
  5. Mar 22, 2009 #4

    Mark44

    Staff: Mentor

    It's just a difference in notation. Dick used x1 and x2 to represent vectors in R^2, where you're looking at x1 and x2 as coordinates of a single vector.

    Here's what he said using different notation:
    Do you really think L(u+v)=L(u)+L(v) holds for a)? I mean try u=(0,0) and v=(0,0).

    Also, but more importantly, you are not understanding the definition of a linear transformation. You have this:
    In this definition x1 and x2 are vectors in R^2, not coordinates of a vector. It might be better for you to not to use indexes on vectors, but to use indexes on coordinates of vectors.

    This makes your definition look like this:
    1. L(a*u+b*v)=aL(u)+bL(v);
    2. L(u+v)=L(u)+L(v);
    3. L(a*u)=aL(u);

    In the above, u = (u1, u2) and v = (v1, v2)

    Got it?
     
  6. Mar 23, 2009 #5
    You sir, are correct.

    I should have guessed that he was referring to vectors and not coordinates since essentially the problem is asking which of the four are valid transformations of vectors from two space to three space. Thank you for the explanation. However, I may have one more question on this depending on how the next step turns out.

    So by rule 2, I have
    L(u+v)=L(u)+L(v) => L(u1+v1, u2+v2)=>L(u1,u2)+L(v1,v2);
    But if I go back to my original problem I'm not sure how the 1 in (u, v, 1)^t plays into this.


    I'm trying my best not to sound like a blithering idiot.
     
  7. Mar 23, 2009 #6

    Dick

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    That is exactly right. L takes a 2-vector and gives you a 3-vector. So L((0,0))=(0,0,1). But (0,0)+(0,0)=(0,0). So L((0,0))=(0,0,1) is NOT equal to L((0,0))+L((0,0))=(0,0,2). (Sorry I'm skipping putting the transpose symbol (0,0)^T on.)
     
  8. Mar 25, 2009 #7
    Mark44, Dick; thanks for the help. Clearly I need more practice, so I might be back in this thread this weekend.
     
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