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Prove the definitions of Linear Transformations

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that 2.1.1 is equivalent to the totality of 2.1.2 and 2.1.3.


    2. Relevant equations
    0Vlw4.jpg


    3. The attempt at a solution

    aTx + bTy = aT(x) + bT(y) = T(ax) + T(by) = T(ax + by) ?
     
  2. jcsd
  3. Oct 8, 2011 #2

    Dick

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    It's really pretty easy. Just think clearly about what you have prove. But there is a really nasty typo in (2.1.3). It should read T(ax)=a*T(x). Is that what's confusing you?
     
  4. Oct 8, 2011 #3
    I can 'get' that 2.1.1 follows from 2.1.2 and 2.1.3 and visa versa, but I'm not sure how to say it as a proof. I've never been good with proofs.
     
  5. Oct 8, 2011 #4

    Dick

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    You aren't bothered by (2.1.3) saying T(ax)=a*x?? You should be. But if you aren't good at proofs, maybe you aren't. Change (2.1.3) to T(ax)=a*T(x). The first stage of the proof is show that (2.1.1) implies (2.1.2) and (2.1.3). That's pretty easy. I'll give you a hint for the first part. Put a=1 and b=1 into (2.1.1). What do you conclude?
     
  6. Oct 8, 2011 #5
    I meant to say that I understood that T(ax)=aT(x), but only because I have other linear algebra books to help me get through this class.

    If a=b=1,

    T(ax +by) = T(1x + 1y) = T(1x) + T(1y) = 1T(x) + 1T(y) = T(x) + T(y)

    Edit:

    Now do I have to do a=a, b=0 to get:

    T(ax + by) = T(ax) =aT(x) ?
     
  7. Oct 8, 2011 #6

    Dick

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    Ok, that's good. So that shows (2.1.1) implies (2.1.2). Can you show it also implies (2.1.3)? The final step of the proof is showing (2.1.2) AND (2.1.3) imply (2.1.1).
     
  8. Oct 8, 2011 #7
    I'm not sure if you saw my edit or not, so I'll just give it its own post.

    To show 2.1.3, do I let a=a and b=0?

    T(ax + by) where a=a, b=0:

    T(ax + 0y) = T(ax) = aT(x).

    Conversely with a=0, b=b:

    T(0x + by) = T(by) = bT(y)
     
  9. Oct 8, 2011 #8

    Dick

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    Sure, that shows T(ax)=a*T(x). So (2.1.1) implies (2.1.3). Now you just have to show (2.1.2) AND (2.1.3) imply (2.1.1). Try it. You are doing fine so far.
     
  10. Oct 8, 2011 #9
    I'm stumped, I can see how you would prove that 2.1.1 implies 2.1.2 and that 2.1.1 implies 2.1.3, but I'm not sure how to combine them.
     
  11. Oct 8, 2011 #10

    Dick

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    Umm. You want to show T(ax+by)=aT(x)+bT(y), right? (2.1.2) says T of the sum of two vectors is T(first vector)+T(second vector), yes? Are ax and by vectors you could apply (2.1.2) to?
     
  12. Oct 8, 2011 #11
    If you let s = ax and t = by,

    then T(s + t) = T(s) + T(t) = T(ax) + T(by) = aTx + bTy,

    So, T(ax + by) = aTx + bTy ?

    Any closer?
     
  13. Oct 8, 2011 #12

    Dick

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    Way closer. In fact, you finished it. Just say at what point you used (2.1.3).
     
  14. Oct 8, 2011 #13
    So letting s=ax and t=by was the right thing to do?
    Are there other ways to say 2.1.1 stems from 2.1.2 and 2.1.3?
     
  15. Oct 8, 2011 #14

    Dick

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    You just did it. T(s+t)=T(s)+T(t)=T(ax)+T(by) by (2.1.2). Now (2.1.3) says T(ax)=aT(x) and T(by)=b*T(y). Doesn't that mean you are done??
     
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