Self adjoint operators on a Hilbert space

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.

• @MichaelLee FYI: I deleted my previous comment. This comment will self-destruct within the next 12 hours Aug 6, 2017 at 14:33

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?
• Very nice. This doesn't seem to require $T$ to be bounded, either, just defined everywhere (or on some invariant subspace). Jul 29, 2017 at 8:40
• @daw That's an "exercise for the reader" $\smile$ Jul 29, 2017 at 14:49
• If $n$ is odd, then $n+1$ is even and $\langle T^{n+1},x\rangle=0\implies T^{\frac{n+1}{2}}x=0$. But $\frac{n+1}{2}<n$. Jul 29, 2017 at 15:25
• No need to hit all possible exponents. $T$ is nonzero, so $T^2$ is nonzero by the above logic. Thus, $T^4$ is nonzero, $T^8$ is nonzero, etc. Of course, if $T^{2^k}$ is nonzero, then $T^n$ can't be identically zero for any $n < 2^k$. Jul 29, 2017 at 16:26
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 $S$.