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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.

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    $\begingroup$ Have you tried using induction? $\endgroup$ – John Douma Nov 19 '18 at 15:43
  • $\begingroup$ @JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much. $\endgroup$ – Zombiegit123 Nov 19 '18 at 15:45
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    $\begingroup$ 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 $k+1$. $\endgroup$ – John Douma Nov 19 '18 at 15:46
  • $\begingroup$ @JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though. $\endgroup$ – Zombiegit123 Nov 19 '18 at 15:49
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    $\begingroup$ 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. $\endgroup$ – MaoWao Nov 19 '18 at 15:49
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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 $T$.

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

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