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Checking for linearity 2

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-07-19at24040AM.png

    here is theorem 4.5

    Screenshot2012-07-19at23441AM.png


    3. The attempt at a solution

    How can theorem 4.5 even relate to the question? The question deals with p's and q's and the theorem deals with u v w.
     
  2. jcsd
  3. Jul 19, 2012 #2
    The short version of what I said in the other thread is this:

    The theorem deals with three arbitrary vectors that are elements of a given vector space. They do not need to be called u, v, and w.
     
  4. Jul 19, 2012 #3
    well how do i even find out if
    p(x) = a0 + a1x + a2x2
    q(x) = b0 + b1x + b2x2 and P2
    meets the condition (u+v) *w = u*w + v*w?
    What's the first question I ask myself? I don't even know what step 1 is, much less 2 and 3
     
  5. Jul 19, 2012 #4
    [itex]P_2[/itex] denotes the space of second-order polynomials. All you need to test this condition is to come up with a third "vector" (another second-order polynomial) that you can plug in as w.

    This condition you're testing is the distributive property. It's the same thing as saying that (5+10)x6 = 5x6 + 10x6. I can plug in any three numbers there instead of 5, 10, and 6 and it works because the real line is a (very simple) vector space of sorts.

    What you do to test this is simple. If I want to test the distributive property on real numbers, I go to what I just wrote and say, hm, is (5+10)x6 -> 15x6 really equal to 5x6 + 10x6?

    At that point, you just evaluate both sides until you get something that's obviously (un)equal.
     
  6. Jul 19, 2012 #5
    ok, i got it.
     
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