If $T:H \rightarrow H$ is a bounded, self adjoint linear operator and $T\neq 0$ then $T^n\neq 0$

Question

If $T:H \rightarrow H$ is a bounded, self adjoint linear operator and $T\neq 0$ then $T^n\neq 0$.

Could someone please expand on the general case of the problem here for me? I see in the $n=2,4,8,\dots$ they used what looks like a contrapositive proof. I.e., they seemed to show that if $T^2 = 0$ then $T=0$.

My questions:

1) This doesn't seem very intuitive to me. I would think that if we have a T so that $\|T\|<1$ then, as $n$ gets large the norm would go to $0$. The proof seems to be saying that it won't.

2) I don't understand what is meant by

"For the general case, if $T^n=0$, then $T^m=0$ for all $m \geq n$, contradicting what you have shown."

Answer 1

For the second point, it really is only necessary to prove that $T\ne0$ implies $T^2\ne0$. As $T$ is self-adjoint, then so is $T^2$, and iterating the argument gives $T^4\ne0$, $T^8\ne 0$ etc.

For a general $n$, there is $2^k>n$ and $T^{2^k}\ne 0$. But $T^{2^k}=T^n T^{2^k-n}$ so $T^n$ must be nonzero.

Answer 2

For your question 1: there's a big difference between a finite $m$ and letting $m\to\infty.$ The proof to which you've linked doesn't say anything about letting $m\to\infty.$