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Linear Algebra - Linearity of a transformation

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data

    Let be T : ℙ2 → ℙ2 a polynomial transformation (degree 2)
    Defined as

    T(a+bx+cx²) = (a+1) + (b+1)x + (b+1)x²

    It is a linear transformation?

    2. Relevant equations

    A transformation is linear if

    T(p1 + p2) = T(p1) + T(p2)


    T(cp1)= cT(p1) for any scalar c

    3. The attempt at a solution

    Let p1=(a+bx+cx²) and p2=(d+ex+fx²) degree 2 polynomials

    T(p1+p2)= (a+d+1) + (b+e+1)x + (b+e+1)x²


    T(p1) + T(p2)=[(a+1)+(d+1)] + [(b+1)+(e+1)]x + [(b+1)+(e+1)]x²
    T(p1) + T(p2)=(a+d+2) + (b+e+2)x + (b+e+1)x²


    T(p1+p2) ≠ T(p1) + T(p2)

    Making it non linear transformation.

    Yet, my answer key says it is linear, either the key is wrong or there is something here I am not understanding.

    Any advise would be appreciated.
    Last edited: Jun 10, 2017
  2. jcsd
  3. Jun 10, 2017 #2


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    Certainly, if ##P^2## is the second degree polynomials with the usual arithmetic, it is even easier to see your transformation is not linear because$$
    T(0 + 0x + 0x^2) \ne 0 + 0x + 0x^2$$Are you sure the destination ##P^2## has the usual operations?
  4. Jun 10, 2017 #3
    Indeed T(0) will not map it to P² zero, it will suffice to prove the answer key is wrong. Thank you.
  5. Jun 10, 2017 #4


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    I'm guessing that the answer key is not wrong and what is wrong is that you don't have the standard addition, additive identity, etc. in your destination. That is why I asked you about the operations in the destination of your mapping. Are you sure that your destination has the same operations as your domain? If I am right in my guess, you shouldn't call the destination simply ##P^2##. You might be mapping ##(P^2,+,\cdot)## to ##(P^2,\oplus,*)## where the operations are different.
  6. Jun 10, 2017 #5
    It is correct, both destination and domain have the usual operations.
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