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, $$ = $$ and so $$ = $$. However, I'm struggling to go from here, any help is appreciated.

• Have you tried using induction? – John Douma Nov 19 '18 at 15:43
• @JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much. – Zombiegit123 Nov 19 '18 at 15:45
• Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$. – John Douma Nov 19 '18 at 15:46
• @JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though. – Zombiegit123 Nov 19 '18 at 15:49
• Additionally, you should really use self-adjointness. For the equality $\langle Tx,y\rangle=\langle x,T^\ast y\rangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators. – MaoWao Nov 19 '18 at 15:49

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 $$T$$.
• I follow this quite well, but how do you get the first $T^2$ from that norm? – Zombiegit123 Nov 19 '18 at 17:44
• Read it backwards. Since $\|Tx\|>0$, $\|Tx\|^2>0$, and thus $\langle T^2x,x\rangle>0$. – Aweygan Nov 19 '18 at 17:46