If $T^m(\alpha)=0,$ with $\alpha \in V$ and $m \in \mathbb{N}$, then $T(\alpha)=0$

Question

I am not so sure how to prove this exercise, please help me:

Let $V$ be a finite dimensional vector space with inner product $\langle,\rangle$. Let $T$ be a linear normal operator on $V$, then:

If $T^m(\alpha)=0,$ with $\alpha \in V$ and $m \in \mathbb{N}$, then $T(\alpha)=0 $

Answer 1

If $T^*T^m\alpha = 0$ then $T^*T^{m-1}\alpha = 0$ since $||T^*T^{m-1}\alpha||^2 = \langle T^*T^{m-1}\alpha,T^*T^{m-1}\alpha \rangle = \langle TT^*T^{m-1}\alpha,T^{m-1}\alpha \rangle = \langle T^*T^m\alpha, T^{m-1}\alpha \rangle$ (using $T^*T = TT^*$ in the last equality). So inductively $T^*T^m\alpha = 0$ implies $T^*\alpha = 0$; hence $||T\alpha|| = ||T^*\alpha|| = 0$ which shows $T\alpha = 0$.