Prove null $T^k$ = null $T$ and range $T^k$ = range $T$ I'm trying to prove that if $T$ is a normal operator, then null $T^k$ = null $T$ and range $T^k$ = range $T$. Showing null $T$ $\subset$ null $T^k$ is simple, so I'm working on the other inclusion. So far I've been able to deduce that for a vector $v \in$ null $T^k$ we have $TT^\star v = T^\star Tv$ $\implies$ $T^k T^\star v = T^\star T^k v$ $\implies$ $T^k T^\star v = 0$ $\implies$ $T^\star v \in$ null $T^k$. I'm not sure if this is useful though, and I'm stuck on where I should go from here.
 A: First for $S$ self-adjoint: suppose $S^kx = 0$. Then 
$$0 = \langle S^{k}x, S^{k-2}x \rangle = \langle S^{k-1}x, S^{k-1}x\rangle$$
so by positive definiteness of inner product, $S^{k-1}x = 0$, and we can continue down to $Sx=0$.
If $T$ is normal, suppose $T^kx=0$. Then 
$$ (T^*T)^kx = (T^*)^k (T^kx) = 0$$ 
(The key is $(T^*T)^k = (T^*)^k T^k$ since $T$ is normal). So by the first part (since $T^*T$ is self adjoint)
$$0 = \langle T^* Tx, x \rangle = \langle Tx, Tx \rangle$$
so $Tx = 0$.
To show $\operatorname{Rg}(T^k)=\operatorname{Rg}(T)$, note using the above result,
$$\operatorname{Rg}(T^k) = \operatorname{Ker}((T^k)^*)^\perp = \operatorname{Ker}((T^*)^k)^\perp\\ = \operatorname{Ker}(T^*)^\perp = \operatorname{Rg}(T).$$
Comment: this says that any normal operator has the same kernel as any of its powers. If $T$ is normal, then $T-\lambda I$ is normal, which shows that $(T-\lambda I)^kx = 0 \Rightarrow (T-\lambda I)x=0$. This shows that any normal operator in finite dimensions is diagonalizable over $\mathbb{C}$. Some short additional work is needed to show it is orthogonally diagonalizable.
A: Let $T^{k}v=0$. Then
$(T^*T)^kv=0$ and
$[(T^*T)^\dagger T^*T]^kx=0$.
Then $T^*Tx=0$ and so $Tx=0$, premultiplying both sides with Moore-Penrose inverse of $T^*$.
(Note that Moore-Penrose inverse of a self adjoint element commutes with that element.)
