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Math Help - Rank and Nullity

  1. #1
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    Rank and Nullity

    Let T : P3 -> R

    T(a0 + a1x + a2x^2 + a3x^3) = a0 - a1 + a2 - a3

    Find the Rank and Nullity of T and find a basis for the kernel of T.



    For my work :

    I set a3 = a0 - a1 + a2
    Plugged this into a3 for the left hand side and after some simplifications, I get

    a0(1 + x^3) +a1(x - x^3) + a2(x^2 + x^3)

    Basis = [ (1 + x^3), (x - x^3), (x^2 + x^3) ]

    Now my professor says the nullity is 3 which makes the rank = 1 but i'm not understanding WHY it's 3. Also, my professor chose initially to set a3 = a0 - a1 + a2. If I had chosen to set say a1 = a0 + a2 - a3 would it still be correct even though the Basis would look different?
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  2. #2
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    Quote Originally Posted by JonathanEyoon View Post
    Let T : P3 -> R

    T(a0 + a1x + a2x^2 + a3x^3) = a0 - a1 + a2 - a3

    Find the Rank and Nullity of T and find a basis for the kernel of T.



    For my work :

    I set a3 = a0 - a1 + a2
    Plugged this into a3 for the left hand side and after some simplifications, I get

    a0(1 + x^3) +a1(x - x^3) + a2(x^2 + x^3)

    Basis = [ (1 + x^3), (x - x^3), (x^2 + x^3) ]

    Now my professor says the nullity is 3 which makes the rank = 1 but i'm not understanding WHY it's 3. Also, my professor chose initially to set a3 = a0 - a1 + a2. If I had chosen to set say a1 = a0 + a2 - a3 would it still be correct even though the Basis would look different?
    you found a basis for \ker T with 3 elements 1+x^3, x - x^3, x^2 + x^3, this exactly means that the nullity of T is 3. (remember: \text{nullity of} \ T = \dim \ker T)
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  3. #3
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