Let V be a finite-dimensional inner product space. Prove that the nullspace and range of a normal operator $f$ on V are unchanged by taking any positive power $f^m$ of the operator.
I've tried proving this using invariant subspaces but have had no look so far. Any help would be greatly appreciated.
If $A$ is normal, then there is a unitary (orthogonal) $Q$ such that $AQ = Q \Lambda$, where $\Lambda$ is diagonal.
Then ${\cal R} A = {\cal R} (AQ) = Q {\cal R} \Lambda = Q {\cal R} \Lambda^m = {\cal R} (A^mQ) = {\cal R} A^m$.
Similarly, $\ker A = \ker (QA) = \ker (\Lambda Q^*) = \ker (\Lambda^m Q^*) = \ker (Q A^m) = \ker A^m$.