If T:V-->V is normal.
Prove a.) null ($\displaystyle T^k$) = null (T)
b.) range($\displaystyle T^k$)=range(T)
ok so i settled on this as my way to solve the first part.
Since, T is normal, there exists an ONB of V so that the matrix representation of T is diagonal (by the complex spectral theorem)
Since A is diagonal, the null(A)=0 (the zero vector).
$\displaystyle T^k$ can be represented as A but with its entries to the power of k.
Since $\displaystyle T^k$ can be represented as A' which is also diagonal, the null($\displaystyle T^k$)=0 the zero vector.
Therefore the null(T)=null($\displaystyle T^k$)=0 the zero vector.
****AM I ON THE RIGHT TRACK?!?!?!!*******
also, how do i prove the range side? thanks all