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Math Help - Nilpotent Linear Transformation

  1. #1
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    Nilpotent Linear Transformation

    A linear transformation T: V -> V is called nilpotent of order p if T^p = 0 and T^(p-1) is not equal to 0. If T is nilpotent, show that 1 - T is an isomorphism, where 1 denotes the identity map on V.

    Any help or hints would be greatly appreciated.
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  2. #2
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    Consider the operator I+T+T^2+...+T^{p-1}.
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  3. #3
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    I'm not quite sure I still understand. If I add all of the T's, will I be getting a summation of T from i=1 to i=p-1?
    Last edited by Shapeshift; February 4th 2011 at 05:42 AM.
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  4. #4
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    x^n- y^n= (x- y) times what?
    I- T^p= (I- T) times what?
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  5. #5
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    so x^n -y^n = (x-y)(x^n-1 + x^n-2y + x^n-3y^2 +...+ xy^n-2 + y^n-1)

    so I - T^p = (I-T)(I^0 + I^-1T + I^-2T^2 +...+ IT^p-2 + T^p-1)

    so then I get I - T^p = (I-T)(I + T + T^2 +...+ T^p-1)

    (I-T) = (I - T^p)/(I+T+T^2+...+T^p-1)

    and T^p=0, so (I-T) = I/(I+T+T^2+...+T^p-1)= I + T^-1+T-2+T^-3+...+T^1-p

    which means that T has an inverse, T^-1 for T, T^-2 for T^2, etc, so T is invertible, and it is invertible iff T is one-to-one and onto, so it is isomorphic. Is this correct?
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  6. #6
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    Quote Originally Posted by Shapeshift View Post
    so x^n -y^n = (x-y)(x^n-1 + x^n-2y + x^n-3y^2 +...+ xy^n-2 + y^n-1)

    so I - T^p = (I-T)(I^0 + I^-1T + I^-2T^2 +...+ IT^p-2 + T^p-1)

    so then I get I - T^p = (I-T)(I + T + T^2 +...+ T^p-1)

    (I-T) = (I - T^p)/(I+T+T^2+...+T^p-1)

    and T^p=0, so (I-T) = I/(I+T+T^2+...+T^p-1)= I + T^-1+T-2+T^-3+...+T^1-p

    which means that T has an inverse, T^-1 for T, T^-2 for T^2, etc, so T is invertible, and it is invertible iff T is one-to-one and onto, so it is isomorphic. Is this correct?


    I think you succeeded to make pretty cumbersome and messy something that is supposed to be pretty simple:

    T^p=0\Longrightarrow I=I-T^p=(I-T)(I+T+T^2+\ldots+T^{p-1})\Longrightarrow I-T is invertible since

    it multiplied by all that stuff withing the long parentheses is the unit transformation. Period.

    The only thing still left is to explain why the second equality above (from the left) holds, since product

    of transformations isn't usually commutative, but in this case...

    Tonio
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