# Thread: Nullity

1. ## Nullity

Please help me with this
Prove that if T is in L ( V ) and is normal, thennull ( Tk ) = null ( T ) and Im ( Tk ) = Im ( T ) for every positive integer k.
Thank you!

2. Originally Posted by maria_stoeva
Prove that if T is in L ( V ) and is normal, then null ( Tk ) = null ( T ) and Im ( Tk ) = Im ( T ) for every positive integer k.
I think you need the spectral theorem (the fact that a normal operator is diagonalisable) for this. If you take an orthonormal basis consisting of eigenvectors of T, then the null space of T is spanned by the eigenvectors corresponding to the eigenvalue 0, and the image of T is spanned by the eigenvectors corresponding to nonzero eigenvalues. It's fairly easy to see that the same description works for the null space and image of any positive power of T.