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.
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.
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:
$\displaystyle 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