# Nilpotent Linear Transformation

• Feb 3rd 2011, 07:03 PM
Shapeshift
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.
• Feb 3rd 2011, 08:36 PM
Jose27
Consider the operator \$\displaystyle I+T+T^2+...+T^{p-1}\$.
• Feb 4th 2011, 05:29 AM
Shapeshift
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?
• Feb 4th 2011, 05:58 AM
HallsofIvy
\$\displaystyle x^n- y^n= (x- y)\$ times what?
\$\displaystyle I- T^p= (I- T)\$ times what?
• Feb 4th 2011, 06:25 AM
Shapeshift
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?
• Feb 4th 2011, 07:42 AM
tonio
Quote:

Originally Posted by Shapeshift
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