Proof regarding self-adjoint linear operators.

Question

Let $H$ be a Hilbert space, and let $T : H \rightarrow H$ be a bounded self-adjoint linear operator, with $T \neq 0.$

I need to show that $T^{2^k} \neq 0$ $\forall k \in \mathbb{N}$. Here's what I've done so far:

$T^2x = T(Tx)$ and so $T^{2^k}x = T(T^{2k-1}x)$. Hence, as T is self-adjoint, $<Tx,y> = <x,T^*y>$ and so $<T^{2k}x,y> = <x,(T^{2k})^*y>$. However, I'm struggling to go from here, any help is appreciated.

Answer 1

Induction certainly does make this proof go by much smoother. Here's how one should proceed:

Since $T\neq0$, there is some nonzero $x\in H$ such that $Tx\neq0$. Hence $\|Tx\|>0$, $$\langle T^2x,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2>0,$$ and thus $T^2\neq0$. (Can you see how self-adjointness is used? )

For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.