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Let $H$ be a Hilbert space and let $T\in \mathcal{B}(H)$ such that $T$ is self-adjoint. I want to show that if $T$ is non-zero, then $T^n\neq 0$ for all $n\in \mathbb{N}$.
Suppose $n$ be the least positive integer such that $T^n=0$. Then for all $x,y\in H$, we have $\langle T^nx,y\rangle=0\implies \langle T^{n-2}x,T^2y\rangle=0$. Herefrom can I show that $T^{n-1}=0$? If it is possible, then I am done. Please suggest.
If $n$ is even then $0=\left<T^n x,x\right>=\left<T^{n/2} x,T^{n/2}x\right>$ so $T^{n/2}x=0$. What if $n$ is odd?
This may be a bit heavy-handed, but we can consider the spectral theorem for bounded self-adjoint operators on Hilbert spaces:
For self-adjoint $T\in \mathcal{L}(H, H)$, there is an $L^2(X, \mathcal{M}, \mu)$, a unitary $U : L^2(X, \mathcal{M}, \mu)\to H$, and an essentially bounded measurable function $f : X\to \mathbb{R}$ such that $U^*MU = T$, where $[M\varphi](x) = f(x)\varphi(x)$. As $T$ and $M$ are unitarily equivalent, we will have $$\|T\| = \|M\| = \|f\|_{\infty}$$ Assuming that $T$ is nonzero, we will therefore have $\|f\|_{\infty} > 0$, i.e. $\lvert f\rvert > 0$ on some set $S\subseteq X$ of positive measure.
As $U^*U = UU^* = I$, we have that $U^*M^nU = T^n$, i.e. $T^n$ and $M^n$ are unitarily equivalent. Therefore, $\|T^n\| = \|M^n\| = \|f^n\|_{\infty} > 0$, as $\lvert f^n\rvert > 0$ on $S$.
