# Prove $\operatorname{Range}(T^k) = \operatorname{Range}(T)$ and $\operatorname{Ker}(T^k) = \operatorname{Ker}(T)$ for all natural numbers k

Here, we assume that $T$ is a normal operator on a complex finite dimensional inner product space $(V, <,>)\DeclareMathOperator{\Range}{Range}\DeclareMathOperator{\Ker}{Ker}$. Now, I know that for any operator, $\Ker(T) \subset \Ker(T^2)$ and $\Range(T^2) \subset \Range(T)$ and by the rank-nullity theorem, $\Ker(T^2) = \Ker(T)$ if and only if $\Range(T^2) = \Range(T)$. I used the spectral theorem to obtain an orthonormal basis $(v_1, \dots ,v_n)$ for $T$ where $(v_{k+1}, \dots, v_n)$ is a basis for $\Ker(T)$. Now, how do I prove $\Range(T) = \Range(T^2) = \operatorname{Span}(v_1, \dots, v_k)$?

This orthonormal basis has another property, guaranteed by the spectral theorem: the $v_i$ are all eigenvectors. That is, for all $i$, each $Tv_i = \lambda_i v_i$ for some $\lambda_i$. The kernel vectors will have $\lambda_i = 0$. All the other vectors will map to a non-zero multiple of themselves, hence \begin{align*} \operatorname{Range}(T) &= \operatorname{span}(Tv_1, \ldots, Tv_n) \\ &= \operatorname{span}(\lambda_1 v_1, \ldots, \lambda_k v_k, 0 \ldots, 0) \\ &= \operatorname{span}(v_1, \ldots, v_k). \end{align*} Similarly, \begin{align*} \operatorname{Range}(T^2) &= T(\operatorname{span}(v_1, \ldots, v_k)) \\ &= \operatorname{span}(\lambda_1 v_1, \ldots, \lambda_k v_k) \\ &= \operatorname{span}(v_1, \ldots, v_k). \end{align*}